Important Questions for CBSE Class 11 Physics Chapter 15 - Waves

Very Short Answer Type Questions (1 Mark)

1. Explosions on other planets are not heard on earth. Why?

Ans: Explosions on other planets are not heard on earth because there is no material medium between the earth and the planets over a long distance, and without a material medium for propagation, sound waves cannot travel.

2. Why longitudinal waves are called pressure waves?

Ans: Longitudinal waves are called pressure waves because the propagation of longitudinal waves through a medium consists of the variations in the volume and the pressure of the air, these variations in volume and air pressure result in the formation of compressions and rarefactions.

3. Why do tuning forks have two prongs?

Ans: The tuning fork has two prongs because the two prongs of a tuning fork produce resonant vibrations that help to keep the vibrations going for longer.

4. Velocity of sound increases on a cloudy day. Why?

Ans: Velocity of sound increases on a cloudy day, because the air is wet on a cloudy day, it contains a lot of moisture, the density of air is lower, and because velocity is inversely proportional to density, velocity increases.

5. Sound of maximum intensity is heard successively at an interval of 0.2 second on sounding two tuning fork to gather. What is the difference of frequencies of two tuning forks?

Ans: The beat period is 0.2 second so that the beat frequency is ${\text{fb}} = \dfrac{1}{{0.2}} = 5\;{\text{Hz}}$ Therefore, the difference of frequencies of the two tuning forks is $5{\text{HZ}}$.

6. If two sound waves have a phase difference of ${60^0}$, then find out the path difference between the two waves?

Ans, Phase difference, $\phi  = {{60}^{0}} = \dfrac{\pi }{3}{\text{rad}}$

Now, in general for any phase difference, $(\Phi )$, the path difference (x) :-

$\phi  = \dfrac{{2\pi }}{\pi }x$

Given $\phi  = \dfrac{\pi }{3},x = ?$

$\dfrac{\pi }{3} = \dfrac{{2\pi }}{\pi } \times x$

$x = \dfrac{\pi }{3} + \dfrac{{2\pi }}{\pi }$

$x = \dfrac{\pi }{3} \times \dfrac{\pi }{{2\pi }}$

$x = \dfrac{{\pi m}}{6}$

7. If the displacement of two waves at a point is given by: -

${{\text{Y}}_1} = asin{\text{wt}}$

${{\text{Y}}_2} = {\text{an}}\sin \left( {wt + \dfrac{\pi }{2}} \right)$

Calculate the resultant amplitude?

Ans: Given data :

If ${{\text{a}}_1} = $ amplitude of first wave

${a_2} = $ amplitude of second wave

${a_r} = $ resultant amplitude

$\phi  = $ phase difference between 2 waves

Then,

${{\mathbf{a}}_t} = \sqrt {{a_1}^2 + {a_1}^2 + 2{a_1}{a_2}\cos \phi } $

In our case,

${{\text{a}}_1} = a;{{\text{a}}_2} = a;\phi  = \dfrac{\pi }{2}50$

${{\text{a}}_t} = \sqrt {{a^2} + {a^2} + 2a \times a\operatorname{Cos} \left( {\dfrac{\pi }{2}} \right)} $

$ = \sqrt {2{a^2}} \quad \left( {\because \operatorname{Cos} \left( {\dfrac{\pi }{2}} \right) = 0} \right)$

${a_t} = \sqrt {2a} $

8.A hospital uses an ultrasonic scanner to locate tumors in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is $1.7\;{\text{km}}_5^{ - 1}$ ? The operating frequency of the scanner is $4.2{\text{MHz}}$.

Ans: Given data :

Speed of sound in the tissue, $v = 1.7\;{\text{km}}/{\text{s}} = 1.7 \times {10^3}\;{\text{m}}/{\text{s}}$

Operating frequency of the scanner, ${\text{v}} = 4.2{\text{MHz}} = 4.2 \times {10^5}\;{\text{Hz}}$

Now,

The wavelength of sound in the tissue is given as:

$\lambda  = \dfrac{v}{V}$

$ = \dfrac{{1.7 \times {{10}^2}}}{{4.2 \times {{10}^8}}} = 4.1 \times {10^{ - 4}}\;{\text{m}}$

9. Given below are some functions of $x$ and $t$ to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a traveling wave, (ii) a stationary wave or (iii) none at all:

(a) $y = 2\cos (3x)\sin (10t)$

(b) $y = 2\sqrt {x - 13} $

(c) $y = 3\sin (5x - 0.5n) + 4\cos (5x - 0.5t)$

(d) $y = \cos x\sin t + \cos 2x\sin 2t$

Ans: (a) It is a stationary wave because the harmonic terms are present individually in the equation, the preceding equation indicates a stationary wave $kx$ and $\omega t$ appear separately in the equation.

(b) There is no harmonic term in the provided equation. As a result, it doesn't represent either a moving or stationary wave. 

(c) The harmonic terms in the preceding equation describe a moving wave and are in the combination of$kx$and $\omega t$are in the combination of $kx - \omega t$.

(d) The given equation represents a stationary wave because the harmonic terms $kx$ and $\omega t$ appear separately in the equation. This equation actually represents the superposition of two stationary waves.

10. A narrow sound pulse (for example, a short pap by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) if the pulse rate is 1 after every $20\;{\text{s}}$, (that is the whistle is blown for a split of second after every $20\;{\text{s}}$ ), is the frequency of the note produced by the whistle equal to $\dfrac{1}{{20}}$ or $0.05{\text{Hx}}$ ?

Ans: (a) No, the pulse doesn’t have a definite . 

(ii) No, It doesn’t have frequency.

(iii) Yes, It have a wavelength.

(b) No, the frequency of the note produced by the whistle is not equal .

Explanation:

(a) There is no defined wavelength or frequency for the narrow sound pulse. However, the sound pulse's speed remains constant, i.e., it is equal to the speed of sound in that medium. 

(b) The short pip produced after every 20 s does not mean that the frequency of the whistle is $\dfrac{1}{{20}}$ or $0.05{\text{H}}$. It means that $0.06{\text{H}}$ s is the frequency of the repetition of the pip of the whistle.

2 Marks Questions

1. A pipe $20\;{\text{cm}}$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430\;{\text{Hz}}$ source? will this source be in resonance with the pipe if both the ends are open?

Ans: Given data :

 Length of pipe $ = {\text{L}} = 20\;{\text{cm}} = 0.2\;{\text{m}}$

Frequency of ${{\text{n}}^{{\text{th }}}}$ node $ = {v_{\text{n}}} = 430\;{\text{Hz}}$

Velocity of sound $ = {\text{v}} = 340\;{\text{m}}\mid {\text{s}}$

Now, ${v_n}$ of closed pipe is :-

${v_n} = \dfrac{{(2n - 1)v}}{{4L}}$

$430 = \dfrac{{(2n - 1) \times 340}}{{4 \times 0.2}}$

$2n - 1 = \dfrac{{430 \times 4 \times 0.2}}{{340}}$

$2n - 1 = 1.02$

$2n = 1.02 + 1$

$2n = 2.02$

$n = 1.01$

Hence, it will be the first normal mode of vibration, ln a pipe, open at both ends We have ,

${v_n} = \dfrac{{H \times {\text{V}}}}{{2L}} = \dfrac{{{\text{A}} \times 340}}{{2 \times 0.2}} = 430\quad {\text{So}},430 = \dfrac{{{\text{A}} \times 340}}{{2 \times 0.2}}$

$n = \dfrac{{430 \times 2 \times 0.2}}{{340}}$

$w = 0.5$

As ${\text{n}}$ has to be an integer, open organ pipe cannot be in resonance with the source.

2. Can beats be produced in two light sources of nearly equal frequencies?

Ans: No, because light emission is a random and fast occurrence, and we obtain uniform intensity instead of beats.

3. A person deep inside water cannot hear sound waves produces in air. Why?

Ans: Person deep inside water cannot hear sound produces in air because the speed of sound in water is roughly four times that of sound in air, refractive index is four times that of sound in air.

index \[u = \dfrac{{{\text{ Sin }}i}}{{\sin r}} = \dfrac{{Va}}{{Vw}} = \dfrac{1}{4} = 0.25\]

For, refraction ${r_{\max }} = {90^0}$, ${i_{\max }} = {90^0}$ , ${i_{\max }} = {14^0}$ . 

Since ${i_{\max }} \ne {r_{\max }}$

Thus, Sounds are only reflected in the air, and people deep in the water are unable to hear them.

4. If the splash is heard 4.23 seconds after a stone is dropped into a well. 78.4 meters deep, find the velocity of sound in air?

Ans: Given data,

depth of$well = s = 78.4m$

Total time after which splash is heard $ = 4.23\;{\text{s}}$

Assume that,

If ${t_1}$ = time taken by stare to hit the water surface is the well

${t_2}$ = time taken be splash of sound to reach the top of well.

then ${t_1} + {t_2} = 4.23{\text{sec}}$.

Now, for downward journey of stone;

$u = 0,a = 9.8m\mid {s^2},s = 78.4m$

${\text{t}} = {\text{t}}1 = 2$

$As,s = ut + \dfrac{1}{2}a{t^2}$

$\therefore 78.4 = 0 + \dfrac{1}{2} \times 9.8 \times t_1^2$

$78.4 = 4.9t_1^2$

$t_1^2 = \dfrac{{78.4}}{{4.9}}$

$t_1^2 = 16$

${t_1} = \sqrt {16}  = 4{\text{sec}}$

Now, $t1 + t2 = 4.23$

$4 + {t_2} = 4.23$

${t_2} = 4.23 - 4.00$

${t_2} = 0.23\;{\text{s}}$

If ${\text{V}} = $ velocity of sound in ${\text{air}}$.

${\text{v}} = \dfrac{{{\text{ dis }}\tan ce(t)}}{{\operatorname{tim} (t)}} = \dfrac{{78.4}}{{0.23}} = 340.87\;{\text{m}}\mid {\text{s}}$

5. How roar of a lion can be differentiated from bucking of a mosquito?

Ans: The roaring of a lion produces a low-pitched, high-intensity sound, but the buzzing of mosquitoes produces a high-pitched, low-intensity sound, and therefore the two noises may be distinguished.

6. The length of a sonometer wire between two fixed ends is $110\;{\text{cm}}$. Where the two bridges should be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio of 1: 2: 3?

Ans: Assume that,

$Let\,\,{{\text{l}}_1},{{\text{l}}_2}$ and ${{\text{L}}_{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{3} }}$ be the length of the three parts of the wire and

 ${{\text{f}}_1},{{\text{f}}_2}$ and ${t_3}$ be their respective frequencies.

Since,

 T and m are fixed quantities, and 2 are constant

$f = \dfrac{1}{{2t}}\sqrt {\dfrac{T}{m}} $

$f = \alpha \dfrac{1}{l}$

$or\,{{\text{f}}_1} = $ constant

$50,{{\text{f}}_1}{{\text{l}}_1} = $ Constant $ \to (1)$

${{\text{f}}_2}{{\text{l}}_2} = $ Constant $ \to (2)$

${{\text{f}}_3}{{\text{l}}_3} = $ Constant $ \to (3)$

Equating equation 1), 2) & 3)

${{\text{f}}_1}{{\text{l}}_1} = {{\text{f}}_2}{{\text{l}}_2} = {{\text{f}}_3}{{\text{l}}_3}$

${\text{No}}{{\text{w}}_n}{{\text{l}}_2} = \dfrac{{{f_1}}}{{{f_2}}}{l_{\text{I}}}$

${l_2} = \dfrac{1}{2}{l_1} \Rightarrow (4)\left( {\dfrac{{{f_1}}}{{{f_2}}} = \dfrac{1}{2}} \right)G$

Also, ${l_3} = \dfrac{{{f_1}}}{{{f_i}}}{l_1}$

${l_1} = \dfrac{1}{3}{l_1}\left( {\dfrac{{{f_1}}}{{{f_1}}} = \dfrac{1}{3}(} \right.$ given $\left. ) \right)$

Now, Total length $ = 110\;{\text{cm}}$

1.e ${{\text{l}}_1} + {{\text{l}}_2} + {{\text{l}}_3} = 110\;{\text{cm}}$

${l_1} + \dfrac{1}{2}{l_1} + \dfrac{1}{3}{l_1} = 110$

$\dfrac{{11\,{l_1}}}{6} = 110$

i. e. ${{\text{l}}_1} = 60\;{\text{cm}}$

${l_2} = \dfrac{{{l_1}}}{2}$

${l_2} = 30\;{\text{cm}}$

${l_2} = \dfrac{{60}}{2}$

Now_

${l_1} = \dfrac{{{l_1}}}{3}\quad {l_3} = \dfrac{{60}}{3}$

${l_3} = 20\;{\text{cm}}$

7. If string wires of same material of length $l$ and $2i$ vibrate with frequencies $100{\text{H}}$ and$150\;{\text{Hz}}$. Find the ratio of their tensions.

Ans: Given data ,

Since frequency $ = {\text{f}}$ at a vibrating string of mass and

 Tension $ = {\text{T}}$ is given by:

$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $

Let for first case, 

${{\text{f}}_1} = 100\;{\text{Hz}};{{\text{l}}_1} = 1;{{\text{T}}_1} = $ Initial Tension.

For second case, ${t_2} = 150Hz;{1_2} = 21;{{\text{T}}_2} = $ Final Tension

so, 

${f_1} = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_1}}}{m}} $

${f_1} = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_1}}}{m}} $

$100 = \dfrac{1}{{2l}}\sqrt {\dfrac{{{T_1}}}{m}}  \to ({\text{l}})$

and ${f_2} = \dfrac{1}{{2{l_2}}}\sqrt {\dfrac{{{T_1}}}{m}} $

$150 = \dfrac{1}{{{2_1}l}}\sqrt {\dfrac{{{T_2}}}{m}}  \to (2)$

Divide equation 1) by equation 2)

$\dfrac{{100}}{{150}} = \dfrac{{\dfrac{1}{{2l}}\sqrt {\dfrac{{{T_1}}}{m}} }}{{\dfrac{1}{{4l}}}}\sqrt {\dfrac{{T2}}{m}} $

$\dfrac{{1\phi  \times 2}}{{153}} = 2\sqrt {\dfrac{{{T_1}}}{{{T_2}}}} $

$\dfrac{2}{3} = 2\sqrt {\dfrac{{{T_1}}}{{{T_2}}}} $

$\dfrac{{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{Z} }}{{3 \times {2^\prime }}} = \sqrt {\dfrac{{{{\bar T}_1}}}{{{T_2}}}} $

$\dfrac{1}{3} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}}  \to $ Squaring both sides

or $\dfrac{1}{9} = \dfrac{{{T_1}}}{{{T_2}}}$

Hence, the ratio of tensions is 1: 9 .

8. Two similar sonometer wires of the same material produces 2 beats per second. The length of one is $50\;{\text{cm}}$ and that of the other is $50.1{\text{em}}$. Calculate the frequencies of two wives?

Ans: Given data, 

The frequency (f) of a Sonometer wire of length (L), mass (m) and 

Tension (T) is given by

$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $

Let, 

$k = \dfrac{1}{2}\sqrt {\dfrac{T}{m}} $

so,

$f = \dfrac{k}{l}$

Now, 

In first case; 

$\left. {{{\text{f}}_1} = \dfrac{k}{{{l_1}}} \to 1} \right$ and 

$f2 = \dfrac{k}{{{l_2}}} \to 2$ 

Subtract equation $1$ & $2$

${{\text{f}}_1} - {{\text{f}}_2} = \dfrac{k}{{{l_1}}} = \dfrac{k}{{{l_3}}} = k\left( {\dfrac{1}{{{l_1}}} - \dfrac{1}{{{l_1}}}} \right)$

Now, given ${l_1} = 50\;{\text{cm}};{{\text{l}}_2} = 50.1\;{\text{cm}}$

${{\text{f}}_1} \cdot {f_2} = 2$

$50,2 = k\left[ {\dfrac{1}{{50}} = \dfrac{1}{{50.1}}} \right]$

\[2=k\dfrac{50.1{}^\circ -50{}^\circ }{50\times 50.1}\]

$2 = K\left[ {\dfrac{{0.1}}{{2505.0}}} \right]$

$\dfrac{{2 \times 2505}}{{0.1}} = k$

$\dfrac{{5010 \times 10}}{{01.1}} = k$

$50100 = k$

so, ${f_1} = \dfrac{k}{{{l_1}}} = \dfrac{{501009}}{{5\phi }} = 1002HZ$

9.  Why are all stringed instruments provided with hollow boxes?

Ans: The sound box is a hollow box that comes with stringed instruments. Forced vibrations are produced in the sound box when the strings are set into vibration. The enormous surface area of the sound box causes a big amount of air to vibrate. The result is a loud sound with the same frequency as the string.

10. Two waves have equations:

${X_1} = a\operatorname{Sin} (wt + \phi )$                ${X_2} = a\operatorname{Sin} \left( {wt + {\phi _2}} \right)$

If in the resultant wave, the amplitude remains equal to the amplitude of the super posing waves. Calculate the phase difference between ${{\text{x}}_1}$ and ${{\text{x}}_2}$ ?

Ans: Given data,

The first wave $ \to {{\text{x}}_1} = a\sin (wt + \phi )$

The second wave $ \to {X_2} = a\operatorname{Sin} \left( {wt + {\phi _2}} \right)$ 

Where ,

${{\text{x}}_1}$ = wave function first wave

${{\text{X}}_2} = $ wave function in second wave

a = amplitude

$w = $ Angular frequency

${\text{t}} = $ time

${\phi _1} = $ phase difference of first wave

${\phi _2} = $ phase difference of second wave.

Assume that,

The resultant amplitude $={{\text{a}}^{\prime }}$ and phase difference$= \phi s0$,

$a = \sqrt {a_1^2 + a_2^2 + 2a,a2\operatorname{Cos} \phi } $

${a_1} = $ amplitude of first wave

${a_2} = $ amplitude of second wave

$\phi  = $ phase difference between two waves.

Now, in oar case,

$a1 = a$

$a2 = a$

so, $a = \sqrt {{a^2} + {a^2} + 2a \times a\operatorname{Cos} \phi } $

$a = \sqrt {2{a^2} + 2{a^2}\operatorname{Cos} \phi } $

$a = \sqrt {2{a^2}(1 + \operatorname{Cos} \phi )} $

$a = \sqrt {2{a^2} \times 2\operatorname{Cos} \dfrac{\phi }{2}} \quad \left( {\because 1 + \operatorname{Cos} \theta  = 2\operatorname{Cos} \dfrac{\theta }{2}} \right)$

$a = \sqrt {4{a^2}\operatorname{Cos} \dfrac{\phi }{2}} $

$a = 2a\operatorname{Cos} \dfrac{\theta }{2}$

$\not x = 2,a\operatorname{Cos} \dfrac{\phi }{2}$

$\dfrac{1}{2} = \operatorname{Cos} \dfrac{\phi }{2}$

$\frac{\theta }{2}= \cos ^{-1}(\frac{1}{2})\\z = 6 \theta ^4 \frac{\theta}{2} \\ =60^{\circ}$

$\phi \text{ }\!\!~\!\!\text{ }=2\times 60{}^\circ $

$\Phi \text{ }\!\!~\!\!\text{ }=120{}^\circ $

So, the phase difference between ${{\text{X}}_1}$ and ${{\text{X}}_2}$ is $120{}^\circ $.

11 A Tuning fork of frequency $300\;{\text{Hz}}$ resonates with an air column closed at one end at ${27^9}{\text{C}}$. How many beats will be heard in the vibration of the fork and the air column at ${0^ \circ }{\text{c}}$ ?

Ans: According to the question,

Frequency of air column at ${27^ \circ }{\text{C}}$ is $300\;{\text{Hz}}$

Let 1 = length of air column and speed of sound $ = {{\text{V}}_{27}}$

For a pipe, closed of one end, the frequency of ${{\text{n}}^{{\text{th }}}}$ harmonic is:

${f_n} = n\left( {\dfrac{v}{{4l}}} \right)$

${\text{n}} = 1,3,5,7$

1 = length of air column

${\text{v}} = $ velocity

So, Let at ${0^9}{\text{C}}$, the speed of sound $ = {{\text{v}}_{\text{a}}}$ then,

${f_1} = {r_1}\left( {\dfrac{{{v_1}}}{{4t}}} \right)$ and ${f_1} = {n_2}\left( {\dfrac{{{v_2}}}{{4i}}} \right)$

${f_1} = $ frequency at ${{27}^{0} }{\text{C}} = 300{\text{HZ}}$

${{\text{f}}_2} = $ frequency at ${{0}^{0} }{\text{C}} = ?$

${\text{So}},{{\text{f}}_1} = \dfrac{{{V_2}}}{4}$

$\left. {300 = \dfrac{{{V_{27}}}}{{4!}} \Rightarrow 1} \right)$

$\left. {{f_2} = \dfrac{{Vo}}{{4!}} \Rightarrow 2} \right)$

Now, $\dfrac{{{V_1}}}{{{V_2}}} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}} $

$\dfrac{{{V_{27}}}}{{{V_0}}}\sqrt {\dfrac{{(27 + 273)k}}{{(273)k}}} $

$\dfrac{{{V_{27}}}}{{{V_0}}} = \sqrt {\dfrac{{300}}{{273}}} $

$\dfrac{{{V_{21}}}}{{{V_0}}} = \not \rho  \cdot g$

or $\dfrac{{{V_0}}}{{{V_{27}}}} = \sqrt {\dfrac{{273}}{{300}}} $

$\dfrac{{{V_O}}}{{{V_{27}}}} = 0.954$

${{\text{v}}_0} = (0.954){{\text{V}}_{27}}$

$\therefore $frequency of air column at ${0^ \circ }{\text{C is}}: - $

${f_2} = \dfrac{{{V_0}}}{{4l}}($ equation 2)

Using ${\text{Vo}} = (0.954){{\text{v}}_2}$, We get

${{\text{f}}_2} = \dfrac{{(0.954){V_{27}}}}{{4t}}$ Now $\dfrac{{{V_{11}}}}{{4l}} = 330($ equation $A)$

${f_2} = 0.954 \times 330 = 286{\text{HZ}}$

The frequency of tuning fork remains $300 + {\text{H}}$. Number of beats $ = {{\text{f}}_1} - {t_2}$

$ = 300 - 286 = 14$ Per second

12. A vehicle with horn of frequency ' $^\prime {{\mathbf{n}}^\prime }$ is moving with a velocity of $30\;{\text{m}}\mid {\text{s}}$ in a direction perpendicular to the straight Line joining the observer and the vehicle. If the observer perceives the sound to have a frequency of ${\text{n}} + {{\text{n}}_1}$. Calculate ${{\text{n}}_1}$ ?

Ans:  The apparent change in wave frequency caused by relative motion between the source of waves and the observer is known as the Doppler effect.

If,

 ${v_2} = $ velocity of the listener

${{\text{v}}_{\text{x}}} = $ velocity of the source

$v = $ velocity of sound

$\gamma $ = frequency of sound reaching from the source to the listener.

${\gamma ^1} = $ Apparent frequency (Le. Changed frequency due to movement of source and listener)

${\gamma ^1} = \left( {\dfrac{{V - {V_2}}}{{V - {V_s}}}} \right)x\gamma $

But in our case, the source and observer more at right angles to each other. The Doppler Effect is not observed when the source of the sound and the observer are moving at right argyles to each other.

S0, if ${\text{n}} = $ original frequency of sound the observer will perceive the sound with a frequency of ${\text{n}}$ because of no Doppler effect) Hence the ${{\text{n}}_1} = $ charge frequency $ = 0$.

13. We cannot hear echo in a room. Explain?

Ans: We all know that in order for an echo to be heard, the obstruction must be hard and vast in size. In addition, the obstruction must be at least a distance from the source. The parameters for the generation of Echo are not met since the length of the room is usually less than. As a result, there is no echo in the room.

14. Why do the stages of large auditoriums give curved backs?

Ans: The backs of big auditorium stages are curved because a speaker's voice is rendered parallel following reflection from a concave or parabolic seer face when he stands at or near the focal of a curved surface. As a result, the voice can be heard from afar.

15. Show that Doppler effect in sound is asymmetric?

Ans: It is seen that apparent frequency of sound when source is approaching the stationary listener (with velocity ${{\text{v}}^1}$) is not the same as the apparent frequency of sound when the listener is approaching the stationary source with a velocity ${{\text{v}}^1}$. This shows that Doppler Effect in sound is asymmetric.

Apparent frequency $ = {{\text{f}}^1} = \dfrac{{\text{V}}}{{{\text{V}} - {\text{us}}}} \times {\text{f}}$

This is when source approaches station any listener

${\text{V}} = $ Velocity of sound in air

${{\text{u}}_{\text{n}}} = $ Velocity of sound source $ = {{\text{v}}_1}$

${{\text{f}}^1} = $ Apparent frequency

${\text{f}} = $ Original frequency of sound

Apparent frequency ${f^{\prime \prime }} = \dfrac{{V + w0}}{V} \times f$

since ${f^{\prime \prime }} = {f^{\prime \prime }}$ (Hence Doppler effect is asymmetric)

16. An organ pipe ${{\text{P}}_1}$ closed at one end vibrating in its first overtone and another pipe ${{\text{P}}_2}$open at both the ends vibrating in its third overtone are in resonance with a given tuning fork. Find the ratio of length of ${{\text{P}}_1}$ and ${{\text{P}}_2}$ ?

Ans: Given data, 

Length of pipe closed at one end for first overtone, ${l_1} = \dfrac{{3\pi }}{4}$

Length of pipe closed at both ends for third overtone; ${1_2} = \dfrac{{3\pi }}{4}$

$\pi  = $ wavelength

We know that,

${l_2} = \dfrac{{4\pi }}{2} = 2\pi $

$\therefore \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{\dfrac{{3\pi }}{4}}}{{2\pi }} = \dfrac{{3\pi }}{{4 \times 2\pi }} = \dfrac{3}{8}$

${l_1};{l_2} = 3:8$

17. A simple Romanic wave has the equation

${\text{Y}} = 0.30{\text{Sin}}(314{\text{t}} - 1.57{\text{x}})$

${\text{t}} = \sec ,{\text{x}} = $ meters, ${\text{y}} = {\text{cm}}$. Find the frequency and wavelength of this wave.

Another wave has the equation.

${{\text{Y}}_1} = 0.1\sin (314{\text{t}} - 1.57{\text{x}} + 1.57)$

Deduce the phase difference and ratio of intensities of the above two waves?

Ans: Given data,

  If y is in meters, then equation becomes: $ \to $

$y = \dfrac{{0.30}}{{100}}\sin (314{\text{t}} - 1.57{\text{x}}) \to (1)$

The standard equation of plane progressive wave is

$y = a\operatorname{Sin} (wt - kx) \Rightarrow (2)$

Now, 

Comparing equation 1) & 2)

$w = 314,k1.57;{\text{a}} = \dfrac{{0.30}}{{100}} = 3 \times {10^{ - 3}}\;{\text{m}}$

frequency $ = {\text{f}} = \dfrac{w}{{2\pi }} = \dfrac{{314}}{{2\pi }} = 50HZ$

wave velocity, ${\text{v}} = \dfrac{w}{k} = \dfrac{{314}}{{1.57}} = 200\;{\text{m}}/{\text{s}}$

wavelength $ = \pi  - {\text{r}}\dfrac{{\text{v}}}{f} = \dfrac{{200}}{{50}} = 4\;{\text{m}}$

On inspection of the equations of the given two waves,

Phase difference,

$\Delta \phi  = 1.57{\text{rad}}$

$ = 1.57 \times \left( {\dfrac{{180}}{\pi }} \right) = {90^2}$

Ratio of amplitudes of two waves $ = \dfrac{{{a_1}}}{{{a_1}}} = \dfrac{{0.3}}{{0.1}} = 3$

Ratio of intensities $ = \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{a_i^2}}{{a_2^2}} = {\left( {\dfrac{3}{1}} \right)^2} = \dfrac{9}{1}$

18. The component waves producing a stationary wave have amplitude, Frequency and velocity of $8\;{\text{cm}},30{\text{HZ}}$ and $180\;{\text{cm}}/{\text{s}}$ respected. Write the equation of the stationary wave?

Ans: Since the ware equation of a travelling wave

 $y = a\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\pi }} \right)$

${\text{a}} = $ Amplitude

$t = $ time

${\text{T}} = $ Time Period

${\mathbf{x}} = $ Path difference

$\pi  = $ wavelength

Let ${y_1} = a\sin 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\pi }} \right)$

${{\text{Y}}_2} = {\text{a}}\sin 2\pi \left( {\dfrac{t}{T} + \dfrac{x}{\pi }} \right)$

$(\because ){\text{t}}$ is travelling in opposite direction)

By principle of superposition, wave equation for the resultant wave $ = y = {y_1} + {y_2}$

$y = a\operatorname{Sin} 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\pi }} \right) + a\sin 2\pi \left( {\dfrac{t}{T} + \dfrac{x}{\pi }} \right)$

$ = a\left( {\operatorname{Sin} 2\pi \left( {\dfrac{t}{T} - \dfrac{x}{\pi }} \right) + a\operatorname{Sin} 2\pi \left( {\dfrac{t}{T} + \dfrac{x}{\pi }} \right)} \right)$

Using $\operatorname{Sin} C + \sin D = a\cos \dfrac{{C - D}}{z} \cdot \dfrac{{\operatorname{Sin} C + D}}{z}$

$y = 2a\operatorname{Cos} \dfrac{{2\pi x}}{\pi } \cdot \dfrac{{\sin 2\pi t}}{T}$

Here ${\text{a}} = 8\;{\text{cm}};{\text{f}} = 30\;{\text{Hz}},\;{\text{V}} = 180\;{\text{cm}}\;{\text{s}}$

$T = \dfrac{1}{{30}}$ sec, $\pi  = $ wavelength $ = \dfrac{v}{f} = VT$

$\pi  = 180 \times \dfrac{1}{{30}} = 6\;{\text{cm}}$

$y = 2a\operatorname{Cos} \dfrac{{2\pi x}}{\pi }\operatorname{Sin} \dfrac{{2\pi t}}{T}$

$y = 16\dfrac{{{\text{ Cos }}\pi x}}{3} \cdot \dfrac{{Sin60\pi t}}{{15}}$

19. A wine of density a $g$/cm $^3$ is stretched between two clamps $1\;{\text{m}}$ apart while subjected to an extension of $0.06\;{\text{cm}}$. What is the lowest frequency of transverse vibration in the wire? Let young's Modulus $ = {\text{y}} = 9 \times {10^{10}}\;{\text{N}}\;{{\text{m}}^2}$ ?

Ans: The lowest frequency of transverse vibrations is given by: -

Area $ = A$

Density $ = {\text{P}}$

$f = \dfrac{1}{{21}}\sqrt {\dfrac{T}{m}} $

Here ${\text{m}} = $ mass Per unit length = area $ \times $ Density

because Density $ = \dfrac{{{\text{ Mass }}}}{{{\text{ Volwme }}}}$

${\text{m}} = A \times P$

F=35.3 v/sec

20. Given two cases in which there is no Doppler effect in sound?

Ans: The two circumstances in which there is no Doppler effect in sound (i.e. no change in frequency) are as follows: -

1) When both the sound source and the listener are moving in the same direction and at the same speed.

2) When one of the source  listeners is in the circle's center and the other is travelling around it at a constant speed.

21. A string of mass $2.50\;{\text{kg}}$ is under a tension of $200\;{\text{N}}$. The length of the stretched string is $20.0\;{\text{m}}$. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Ans: Given data,

 Mass of the string, $M = 2.50\;{\text{kg}}$

Tension in the string, ${\text{T}} = 200\;{\text{N}}$

Length of the string, $l = 20.0\;{\text{m}}$

Mass per unit length, $\mu  = \dfrac{M}{l} = \dfrac{{2.50}}{{20}} = 0.125\;{\text{kg}}{{\text{w}}^{ - 1}}$

The velocity (v) of the transverse wave in the string is given by the relation:

$v = \sqrt {\dfrac{T}{\mu }} $

$ = \sqrt {\dfrac{{200}}{{0.125}}}  = \sqrt {1600}  = 40\;{\text{m}}/{\text{s}}$

Time taken by the disturbance to reach the other end, $t = \dfrac{l}{v} = \dfrac{{20}}{{40}} = 0.50\;{\text{s}}$

22.$A$ steel wire has a length of $12.0\;{\text{m}}$ and a mass of $2.10\;{\text{kg}}$. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at ${20^{ - 1}}{\text{C}} = 343\;{\text{m}}\;{{\text{s}}^{ - 1}}$ ?

Ans: Given data ,

 Length of the steel wire, $I = 12\;{\text{m}}$

Mass of the steel wire, $m = 2.10\;{\text{kg}}$

Velocity of the transverse wave, $v = 343\;{\text{m}}/{\text{s}}$

Mass per unit length, $\mu  = \dfrac{m}{l} = \dfrac{{2.10}}{{12}} = 0.175\;{\text{kg}}\;{{\text{m}}^{ - 1}}$

For tension $T$, velocity of the transverse wave can be obtained using the relation:

$v = \sqrt {\dfrac{{\bar I}}{H}} $

$\therefore T = {v^2}\mu $

$ = {(343)^2} \times 0.175 = 20588.575 \approx 206 \times {10^4}\;{\text{N}}$

23. A bat emits ultrasonic sound of frequency $1000{\text{kHz}}$ in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is $340\;{\text{m}}\;{\text{s}} - 1$ and in water $1486\;{\text{m}}_5^{ - 1}$.

Ans:  (a) Frequency of the ultrasonic sound, ${\text{v}} = 1000{\text{kHz}} = 106{\text{H}}$ ?

Speed of sound in air, ${v_a} = 340\;{\text{m}}/{\text{s}}$

The wavelength $({h_r})$, of the reflected sound is given by the relation:

${\lambda _r} = \dfrac{V}{V}$

$ = \dfrac{{340}}{{{{10}^5}}} = 3.4 \times {10^{ - 6}}n$

(b) Frequency of the ultrasonic sound ${\text{v}} = 1000{\text{kHz}} = 106\;{\text{Hz}}$

Speed of sound in water, $vw = 1486\;{\text{m}}/{\text{s}}$

The wavelength of the transmitted sound is given as:

${\lambda _{\text{r}}} = \dfrac{{1486}}{{{{10}^8}}} = 1.49 \times {10^{ - 3}}\;{\text{m}}$

24. (1) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (i) What is the amplitude of a point 0.375 m away from one end?

Ans: (1)

(a) Yes, except at the nodes

(b) Yes, except at the nodes

(c) No

(ii) 0.042 m

Explanation:

(i)

(a) All the points on the string oscillate with the same frequency, except at the nodes which have zero frequency.

(b) All the points in any vibrating loop have the same phase, except at the nodes.

(c) All the points in any vibrating loop have different amplitudes at vibration.

(ii) The given equation is:

$y(x,t) = 0.06\sin \left( {\dfrac{{2\pi }}{3}x} \right)\cos (120\pi t)$

For $x = 0.375\;{\text{m}}$ and $t = 0$

Amplitude = Displacement $ = 0.06\sin \left( {\dfrac{{2\pi }}{3}x} \right)\cos 0$

$ = 0.06\sin \left( {\dfrac{{2\pi }}{3}0.375} \right) \times 1$

$ = 0.06\sin (0.25\pi ) = 0.06\sin \left( {\dfrac{\pi }{4}} \right)$

$ = 0.06 \times \dfrac{1}{{\sqrt 2 }} = 0.042\;{\text{m}}$

25.  A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45\;{\text{Hz}}$ The mass of the wire is $3.5 \times {10^{ - 2}}\;{\text{kg}}$ and its linear mass density is $4.0 \times {10^{ - 2}}\;{\text{kg}}\;{\text{m}} - 1.$ What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Ans: Given data,

 (a) Mass of the wire, $m = 3.5 \times {10^{ - 2}}\;{\text{kg}}$

Linear mass density, $\mu  = \dfrac{m}{l} = 4.0 \times {10^{ - 1}}\;{\text{kg}}\;{{\text{m}}^{ - 1}}$

: Length of the wire, $l = \dfrac{m}{\mu } = \dfrac{{3.5 \times {{10}^{ - 1}}}}{{4.0 \times {{10}^{ - 2}}}} = 0.875\;{\text{m}}$

The wavelength of the stationary wave $(2)$ is related to the length of the wire by the relation:

$\lambda  = \dfrac{2}{n}$

Where, ${\text{n}} = $ Number of nodes in the wire

For fundamental node, $n = 1$ :

$\lambda  = 2l$

$\lambda  = 2 \times 0.875 = 1.75\;{\text{m}}$

The speed of the transverse wave in the string is given as:

$v = v\lambda 45 \times 1.75 = 78.75\;{\text{m}}/{\text{s}}$

(b) The tension produced In the string is given by the relation:

$T = {V_1}\mu $

$ = {(78.75)^2} \times 4.0 \times {10^{ - 1}} = 248.06\;{\text{N}}$

26. Two sitar strings $A$ and B playing the note "Ga' are slightly out of tune and produce beats of frequency $6{\text{H}}$. The tension in the string ${\text{A}}$ is slightly reduced and the beat frequency is found to reduce to $3\;{\text{Hz}}$ if the original frequency of ${\text{A}}$ is $324\;{\text{Hz}}$, what is the frequency of $\mathbb{B}$ ?

Ans: Given data ,

Frequency of string A, ${f_A} = 324\;{\text{Hz}}$

Frequency of string ${\text{B}} = {f_{\text{B}}}$

Beat's frequency, $n = 6\;{\text{Hz}}$

Beat's frequency is given as:

$n = {f_A} \pm {f_B}$

$6 = 324 \pm {f_B}$

\[{f_B} = 330{\text{Hz or }}318\;{\text{Hz}}\]

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:

$v \propto \sqrt T $

Hence, the beat frequency cannot be $330\;{\text{Hz}}$

$\therefore {f_S} = 318\;{\text{Hz}}$

3 Marks Questions

1. Explain briefly the analytical method of formation of heats?

Ans: Assume that,

Let us consider two wares’ trains at equal amplitude 'a' and with different frequencies ${v_1}$4 and ${v_2}$ in same direction.

Let displacements are ${{\text{y}}_1}$ and ${{\text{y}}_2}$ in time ${\text{t}}$.

${w_1} = $ Angular vel. of first ware

${w_1} = 2\pi {v_1}$

${y_1} = a\sin {w_1}t$

${y_1} = a\operatorname{Sin} 2\pi {v_1}t$

${y_2} = a\operatorname{Sin} 2{v_2}t$

According  to superposition principle, the resultant displacement ${y^\prime }$ at the same time t' is-

$y = y1 + yz$

$y = a\left[ {\operatorname{Sin} 2\pi {u_1}t + \sin 2\pi {v_1}t} \right]$

Using $\sin C + \sin D = 2\cos \dfrac{{C - D}}{2}\sin \dfrac{{C + D}}{2}$

we get,

$y = 2a\cos \dfrac{{2\pi \left( {{u_1} - {v_i}} \right)t}}{2}\sin \dfrac{{2\pi \left( {{v_1} + {v_2}} \right)t}}{2}$

$y = 2a\operatorname{Cos} \pi \left( {{v_1} - {v_2}} \right)t\sin \pi \left( {{v_2} + {v_2}} \right)t$

$y = A\operatorname{Sin} \pi \left( {{v_1} + {v_2}} \right)t$

Where, ${\text{A}} = 2{\text{a}}\cos \pi \left( {{v_1} - {v_2}} \right)t$.

Now, amplitude A is maximum when,

$\operatorname{Cos} \pi \left( {{u_1} - {v_2}} \right) +  = \max  +  \pm 1 = \operatorname{Cos} {\text{k}}\pi $

$\pi \left( {{v_1} - {v_2}} \right)t = k\pi ,k = 0,1,2 - $

${\text{t}} = \dfrac{k}{{\left( {{v_1} - {v_1}} \right)}}$

i.e. resultant intensity of sound will be maximum at times, $t = 0,\dfrac{1}{{\left( {{u_1} - {v_2}} \right)}} \cdot \dfrac{3}{{\left( {{u_1} - {v_2}} \right)}} -  -  -  - $

Time interval between 2 successive Maximax’s $ = \dfrac{1}{{\left( {{v_1} - {v_2}} \right)}} - 0 = \dfrac{1}{{\left( {{v_1} - {v_2}} \right)}} \to $ (1)

similarly, a will be minimum,

$\operatorname{Cos} \pi \left( {{v_1} - {v_2}} \right)t = 0$

$\operatorname{Cos} \pi \left( {{u_1} - {v_1}} \right)t = \operatorname{Cos} (2k + 1)\dfrac{\pi }{2},k = 0,1,2,3 - $

$\pi \left( {{v_1} - {v_2}} \right)t = (2k + 1)\dfrac{\pi }{2}$

$t = \dfrac{{(2k + 1)}}{{2\left( {{v_1} - {v_2}} \right)}}$

i.e. resultant intensity of sound will be minimum at times

$t = \dfrac{1}{{2\left( {{v_1} - {v_2}} \right)}},\dfrac{3}{{2\left( {{v_1} - {v_2}} \right)}},\dfrac{5}{{2\left( {{u_1} - {v_2}} \right)}}$

Hence time interval between 2 successive minim as are

$ = \dfrac{3}{{2\left( {{v_1} - {v_2}} \right)}} - \dfrac{1}{{2\left( {{v_1} - {u_2}} \right)}} = \dfrac{1}{{\left( {{q_1} - {v_2}} \right)}} \to (2)$

Combining 1) &: 2) frequency of beats $ = \left( {{v_1} - {v_1}} \right)$

$\therefore \dfrac{{{\text{ No}}{\text{. of beats }}}}{{{\text{ seconds }}}} = $ Difference in frequencies of two sources of sound.

2. Show that the frequency of nth harmonic mode in a vibrating string which is closed at both the end is 'n" times the frequency of the first harmonic mode?

Ans: When a sting under tension is placed into vibration, transverse harmonic waves propagate along its length, and reflected waves exist when the length of the sting is fixed. In sting, the incident and reflected waves will superimpose on one other, resulting in transverse stationary waves.

 Let a harmonic wave be set up in a sting of length $ = \lambda $ which is fixed at 2 ends: $ \to {\text{x}} = 0$ and $x = L$

Let the incident wave travels from left to right direction, the wave equation is:

$\left. {{y_1} = r\operatorname{Sin} \dfrac{{2\pi }}{\pi }(Vt + x) \to 1} \right)$

The reflected wave equation will have the same amplitude, wavelength, velocity, and time as the incident wave equation, but the only difference between the incident and reflected waves will be in their propagation direction.

${\text{So}},: \to $

${y_2} = r\operatorname{Sin} \dfrac{{2\pi }}{\pi }(v - x)$

$\because $ Reflected wave travels from right to left

Reflection, the wave will suffer a phase change of $\pi .S0$,

$y2 = r\operatorname{Sin} \dfrac{{2\pi }}{\pi }[vt - x + \pi ]$

$y2 =  - r\operatorname{Sin} \dfrac{{2\pi }}{\pi }[{\text{vt}} - x] \to (2)$

According to the principle of superposition, the wave equation of resultant stationary wave will be-

$y = {y_1} + {y_2}$

Using equation 1) & 2)

$y = r\operatorname{Sin} \dfrac{{2\pi }}{\lambda }[vt + x] - r\sin \dfrac{{2\pi }}{\lambda }(v - x)$

$ = r\left[ {\sin \dfrac{{2\pi }}{\lambda }(vt + x) - \sin \dfrac{{2\pi }}{\lambda }(vt - x)} \right]$

Using $\sin C \cdot \sin D = 2\operatorname{Cos} \dfrac{{C + D}}{Z}\sin \dfrac{{C - D}}{Z}$

$y = r \times 2\operatorname{Cos} \left[ {\dfrac{{\dfrac{{2\pi }}{2}(v + x) + \dfrac{{2\pi }}{\lambda }(v - x)}}{2}} \right]$

$\operatorname{Sin} \left[ {\dfrac{{2\pi }}{\lambda }(v + x) - \dfrac{{2\pi }}{\lambda }(v - x)} \right]$

$y = 2r\cos \left( {\dfrac{{\dfrac{{2\pi }}{2}vt + \dfrac{{2\pi }}{7}x + \dfrac{{2\pi }}{2}vt - \dfrac{{2\pi }}{7}x}}{2}} \right)$

$y = 2r\operatorname{Cos} \dfrac{{2\pi }}{\lambda }\left( {\dfrac{{{2^\prime }vt}}{{{7^\prime }}}} \right) \cdot \sin \dfrac{{2\pi }}{\lambda }\left( {\dfrac{{2x}}{2}} \right)$

$y = 2r\operatorname{Cos} \dfrac{{2\pi }}{2}vt.\sin \dfrac{{2\pi x}}{\lambda }$

Now, at $x = 0{\text{ & }}\,x = L;y = 0$

1) At $x = 0,y = 0$ is satisfied

2) $Atx = {L_n}y = 0$

$y = 2{\text{r}}\operatorname{Cos} \left( {\dfrac{{2\pi }}{\lambda }vt} \right) \cdot \sin \left( {\dfrac{{2\pi }}{\lambda } \times L} \right)$

$0 = 2{\text{t}}\operatorname{Cos} \left( {\dfrac{{2\pi }}{\lambda }vt} \right)\operatorname{Sin} \left( {\dfrac{{2\pi }}{\lambda } \times L} \right)$

Now, $r \ne 0,\dfrac{{2\pi }}{\lambda }vt \ne 0$ so

$o = \operatorname{Sin} \left( {\dfrac{{2\pi L}}{\lambda }} \right)$

Sin ${\text{n}}{z^\prime } = \operatorname{stn} \dfrac{{2\pi 2}}{\lambda }$

i.e ${\text{L}} = {\text{n}}\left( {\dfrac{x}{2}} \right)$

1) Let ${\text{n}} = 1$ (Le first harmonic mode or fundamental frequency

$\lambda  = {\lambda _1}$

$L = \dfrac{{{\lambda _1}}}{2}$

${\lambda _1} = 2L$

Let ${\text{v}} = $ velocity, ${\gamma _1} = $ frequency of first harmonic Mode ${\text{v}} = {\gamma _1}{\lambda _1}$

$v = {\gamma _1} \times 2L$

$\dfrac{v}{{2L}} = {\gamma _1}$

If ${\text{n}} = 2$ (second harmonic Mode or first overtone)

$L = \lambda 2$

$v = $ velocity, ${\gamma _2} = $ ferq4 of second harmonic Mode

${\text{v}} = {\gamma _1}{\hat h_2}$

${\gamma _2} = \dfrac{v}{{{\lambda _2}}} = \dfrac{v}{2}$

$\gamma 2 = \dfrac{{v \times 2}}{{L \times 2}}\quad \left( {\because \dfrac{v}{{2L}} = {\gamma _1}} \right)$

${\gamma _2} = 2{\gamma _1}$

i.e. frequency of second harmonic Mode is twice the frequency of first harmonic Mode Similarly, $m = n{\% _1}$ and frequency of ${{\text{n}}^{{\text{th }}}}$ harmonic Mode is n times the frequency of first harmonic Mode.

3. Differentiate between the types of vibration in closed and open organ pipes?

Ans: 1) In closed pipe, the wavelength of nth mode $ = \pi n = \dfrac{{4l}}{n}$ 

where ${\text{n}} = $ odd integer

where as in open pipe, $\pi N = \dfrac{{2l}}{n}$ and $n = $ all integer

2) The fundamental frequency of open pipe is twice that of closed pipe of same length.

3) A closed pipe of length $\dfrac{2}{2}$ produces the same fundamental frequency as an open pipe of length L .

4) For an open pipe, harmonics are present for all integers and for a closed pipe, harmonics are present for only odd integers hence, open pipe gives richer note.

Put the value of ${\text{n}}$ in equation for frequency

$f = \dfrac{1}{{21}}\sqrt {\dfrac{T}{{AP}}} $

$f = \dfrac{1}{{2!}}\sqrt {\dfrac{{T/A}}{P}} $

Now, Force = Tension and $\dfrac{{{\text{ Farce }}}}{{{\text{ Area }}}} = $ Stress

$\left. {{\text{f}} = \dfrac{1}{{2\;{\text{J}}}}\sqrt {\dfrac{{s\operatorname{tress} }}{{\text{P}}}}  \to 2} \right)$

Now, young's Modulus $ = {\text{y}} = \dfrac{{{\text{ Stress }}}}{{{\text{ Strairt }}}}$

Or Stress $ = yx$ strain

Stress $ = y \times \dfrac{{\Delta L}}{L}$

Put the value of Stress in equation

 2)

$f = \dfrac{1}{{2l}}\sqrt {\dfrac{{y \times \Delta L}}{{Z \times P}}} $

Put $y = 9 \times {10^{10}}N/m,\vartriangle L = 0.05 \times {10^{ - 2}}ms$

$L= 1.0 m , P = 9 \times 10^3 \frac{kg}{m^3} f =\frac{1}{2 \times 1}\sqrt{\frac{9 \times 10^9 \times 0.05 \times 10^{-2}}{1 \times 9 \times 10^3}}$

Since fundamental frequency of a stretched spring $\dfrac{{\alpha 1}}{l}$

$\frac{f_2}{f_1} = \frac{l_2}{l_1} = \frac{50.1}{49.9}\\f_2 = \frac{50.1}{49.9}f_1 \\\text{now} ,f_2-f_1 = 1 \text{or} \frac{50.1}{49.9}f_1 -f_1\\l =\frac{50.1 f_1 - 49.9 f_2}{49.9}\\l = \frac{0.2 f_1}{49.9} \\l \times 49.9 = 0.2 f_1\\ f_1 = \frac{49.9}{0.2}\\f_1 = 249.5 \text{HZ} \\f_2 - f_1 =1 \\f_2 = 1+f_1 \\=1+249.5$

${{\text{f}}_2} = 250.5{\text{HZ}}$

4. A bat is flitting about in a eave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is $40{\text{kHz}}$. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

Ans: Ultrasonic beep frequency emitted by the bat, ${\text{V}} = 40{\text{k}}{{\text{H}}_z}$

Velocity of the bat, ${v_2} = 0.03v$

Where, $v$ = velocity of sound in air

The apparent frequency of the sound striking the wall is given as:

${v^\prime } = \left( {\dfrac{v}{{v - {v_b}}}} \right)v$

$ = \left( {\dfrac{v}{{v - 0.03v}}} \right)40$

$ = \dfrac{{40}}{{0.97}}{\text{kHz}}$

This frequency is reflected by the stationary wall $\left( {{v_4} = 0} \right)$ toward the bat.

The frequency ( $V''$) of the received sound is given by the relation:

${v^{\prime \prime }} = \left( {\dfrac{v}{{v + {v_z}}}} \right){v^\prime }$

$ = \left( {\dfrac{{v + 0.3v}}{v}} \right) \times \dfrac{{40}}{{0.97}}$

$ = \dfrac{{1.03 \times 40}}{{0.97}} = 42.47{\text{kHz}}$

5. A stone dropped from the top of a tower of height $300\;{\text{m}}$ high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is $340\;{\text{m}}\;{{\text{s}}^{ - 1}}?\left( {\;{\text{g}} = 9.8\;{\text{m}},{5^{ - 1}}} \right)$

Ans: Given data ,

Height of the tower, $s = 300\;{\text{m}}$

Initial velocity of the stone, $u = 0$

Acceleration, $a = g = 9.8\;{\text{m}}{5^{ - 1}}$

Speed of sound in air $ = 340\;{\text{m}}/{\text{s}}$

The time $\left( {{t_1}} \right)$ taken by the stone to strike the water in the pond can be calculated using the

second equation of motion, as:

$s = s{t_1} + \dfrac{1}{2}gt_1^2$

$300 = 0 + \dfrac{1}{2} \times 9.8xt_1^2$

$\therefore {t_1} = \sqrt {\dfrac{{300 \times 2}}{{9.8}}}  = 7,82s$

Time taken by the sound to reach the top of the tower, ${t_1} = \dfrac{{300}}{{340}} = 0.88s$

Therefore, the time after which the splash is heard, $t = {t_1} + {t_2}$

$ = 7.82 + 0.88 = 8.7\;{\text{s}}$

6.  You have learnt that a traveling wave in one dimension is represented by a function $y = f(x$, there $x$ and $t$ must appear in the combination $x - vt$ or $x + vt$, 1.e. $y = f$ $(x \pm vt).$ Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:

(a) ${(x - vt)^2}({\mathbf{b}})\log \left[ {\dfrac{{x + vt}}{{{x_0}}}} \right]$ (c) $\dfrac{1}{{(x + vt)}}$

Ans: No;

(a) Does not represent a wave

(b) Represents a wave

(c) Does not represent a wave

The converse of the given statement is not true. The essential requirement for a function to represent a travelling wave ${\text{k}}$ : that it should remain finite for all values of $x$and $t$.

Explanation:

(a) Far $x = 0$ and $t = 0$, the function ${(x - v)^2}$ becomes 0 .

Hence, for $x = 0$ and $t = 0$, the function represents a point and not a wave.

(b) For $x = 0$ and $t = 0$, the function

$\log \left( {\dfrac{{x + vt}}{{{x_0}}}} \right) = \log 0 = x$

Since the function does not converge to a finite value for $x = 0$ and $t = 0$, it represents a travelling wave.

(c) For $x = 0$ and $t = 0$, the function

$\dfrac{1}{{x + vt}} = \dfrac{1}{0} = x$

Since the function does not converge to a finite value for $x = 0$ and $t = 0$, it does not represent a travelling wave.

7: For the travelling harmonic wave

$y(x,t) = 2.0\cos 2\pi (10t - 0.0080x + 0.35)$

Where $x$ and $y$ are in cm and $t$ in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of

(a) $4\;{\text{m}}$, (b) $0.5\;{\text{m}},({\text{c}}),\dfrac{\lambda }{2}$ (d) $\dfrac{{3.\lambda }}{4}$

Ans: Equation for a travelling harmonic wave is given as:

$y(x,t) = 20\cos 2\pi (10t - 0.0050x + 0.35)$

$ = 2.0\cos (20\pi t - 0.016 - x + 0.70\pi )$

Where,

Propagation constant, $k = 0.0160\pi $

Amplitude, $a = 2\;{\text{cm}}$

Angular frequency, $\emptyset  = 20\pi $ rad/s

Phase difference is given by the relation:

$\phi  = kx = \dfrac{{2\pi }}{\lambda }$

(a) For $x = 4\;{\text{m}} = 400\;{\text{cm}}$

$\Phi  = 0.016\pi  \times 400 = 6.4\pi {\text{rad}}$

(b) For $0.5\;{\text{m}} = 50\;{\text{cm}}$

$\Phi  = 0.016\pi  \times 50 = 0.8{\text{mrad}}$

$(c){\text{For}}x = \dfrac{\lambda }{2}$

$\phi  = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{2} = \pi rad$

(d) For $x = \dfrac{{3\lambda }}{4}$

$\phi  = \dfrac{{2\pi }}{2} \times \dfrac{{3\lambda }}{4} = 1.5\pi rad$

8. A meter-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency $340\;{\text{Hz}}$ ) when the tube length is $25.5\;{\text{cm}}$ or $79.3\;{\text{cm}}$. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Ans:  Given data,

Frequency of the turning fork, ${\text{v}} = 340\;{\text{Hz}}$

Because one end of the supplied pipe is connected to a piston, it will behave as a pipe with one end closed and the other open, as illustrated in the diagram.

Such a system produces odd harmonics. The fundamental note in a closed pipe is given by the relation:

${l_1} = \dfrac{\lambda }{4}$

Where,

Length at the pipe, ${l_1} = 25.5\;{\text{cm}} = 0.255\;{\text{m}}$

$\therefore \lambda  = 4{l_1} = 4 \times 0.255\;{\text{m}} = 1.02\;{\text{m}}$

The speed of sound is given by the relation:

$v = v\lambda  = 340 \times 1.02 = 346.8\;{\text{m}}/{\text{s}}$

9. A steel rod $100\;{\text{cm}}$ long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be $2.53{\text{kHz}}$. What is the speed of sound in steel?

Ans: Given data,

 Length of the steel rod, $l = 100\;{\text{cm}} = 1\;{\text{m}}$

Fundamental frequency of vibration, $v = 2.53{\text{kHz}} = 2.53 \times {10^{{\text{iHz}}}}$

When the rod is plucked at its middle, an antinode (A) is formed at its center, and nodes (N) are formed at its two ends, as shown in the given figure.

The distance between two successive nodes is $\dfrac{\lambda }{2}$.

$\therefore l = \dfrac{\lambda }{2}$

$\lambda  = 2l = 2 \times 1 = 2m$

The speed of sound in steel is given by the relation:

$v = {v_\lambda }$

$ = 2.53 \times {10^3} \times 2$

$ = 5.06 \times {10^3}\;{\text{m}}/{\text{s}}$

$ = 5.06\;{\text{km}}/{\text{s}}$

(e) A pulse is made up of many waves with varying wavelengths. Depending on the nature of the medium, these waves travel at varying speeds in a dispersive medium. The shape of a wave pulse is distorted as a result of this.

10. A train, standing at the outer signal of a railway station blows a whistle of frequency $400\;{\text{Hz}}$ in still air. (i) What is the frequency of the whistle for a platform observer when the train $(a)$ approaches the platform with a speed of $10\;{\text{m}}_5^{ - 1},(\;{\text{b}})$ recedes from the platform with a speed of $10\;{\text{m}} = 1?$ (ii) What is the speed of sound in each ease? The speed of sound in still air can be taken as $340\;{\text{m}}\;{\text{s}} - 1$.

Ans: (i) Given that,

(a) Frequency of the whistle, ${\text{v}} = 400\;{\text{Hz}}$

Speed of the train, ${v_T} = 10\;{\text{m}}/{\text{s}}$

Speed of sound, $v = 340\;{\text{m}}/{\text{s}}$

The apparent frequency $\left( {{v^7}} \right)$ of the whistle as the train approaches the platform is given by

the relation:

${v^\prime } = \left( {\dfrac{v}{{v - {v_r}}}} \right)v$

$ = \left( {\dfrac{{340}}{{340 - 10}}} \right) \times 400 = 412.12\;{\text{Hz}}$

(b) The apparent frequency $\left( {{v^{\prime \prime }}} \right)$ ot the whistle as the train recedes from the platform is given by the relation:

${v^{\prime \prime }} = \left( {\dfrac{v}{{v - {v_r}}}} \right)v$

$ = \left( {\dfrac{{340}}{{340 + 10}}} \right) \times 400 = 388.57\;{\text{Hz}}$

(ii) The apparent shift in sound frequency is produced by the source and observer's relative motions. The speed of sound is unaffected by these relative motions. As a result, the speed of sound in air is constant in both circumstances. i.e, $340\;{\text{m}}$/s

11. A SONAR system fixed in a submarine operates at a frequency $40.0{\text{kH}}$. An enemy submarine moves towards the SONAR with a speed of $360\;{\text{km}}\;{{\text{h}}^{ - 1}}$ - What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be $1450\;{\text{m}}\;{{\text{s}}^{ - 1}}$

Ans: Given that,

Operating frequency of the $50{\text{NAR}}5{\text{ssem}},{\text{V}} = 40{\text{kHz}}$

Speed of the enemy submarine, ${v_4} = 360\;{\text{km}}/{\text{h}} = 100\;{\text{m}}/{\text{s}}$

Speed of sound in water, $v = 1450\;{\text{m}}/{\text{s}}$

The source is at rest and the observer (enemy submarine) is moving toward it. Hence, the apparent frequency $\left( {{v^\prime }} \right)$ received and reflected by the submarine is given by the relation:

${v^\prime } = \left( {\dfrac{{v + {v_A}}}{v}} \right)v$

$ = \left( {\dfrac{{1450 + 100}}{{1450}}} \right) \times 40 = 42.76{\text{ldH}}$

The frequency (v") received by the enemy submarine is given by the relation:

${{\text{v}}^{\prime \prime }} = \left( {\dfrac{v}{{v + {v_2}}}} \right){v^\prime }$

where, ${v_2} = 100\;{\text{m}}/{\text{s}}$

$\therefore {V^{\prime \prime }} = \left( {\dfrac{{1450}}{{1450 - 100}}} \right) \times 42.76 = 45.93{\text{kHz}}$

4 Marks Questions

1. A pipe $20\;{\text{cm}}$ long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a $430\;{\text{Hz}}$ source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is $340\;{\text{m}}_5^{ - 1}$ )

Ans: Given data ,

 First (Fundamental); No

Length of the pipe, ${\text{I}} = 20\;{\text{cm}} = 0.2\;{\text{m}}$

Source frequency $ = n$ the normal mode of frequency, ${v_n} = 430\;{\text{Hz}}$

Speed of sound, $v = 340\;{\text{m}}\;{\text{s}}$

In a closed pipe, the $n$the normal mode of frequency is given by the relation:

${v_n} = (2n - 1)\dfrac{v}{4};{\text{n}}$ is an integer $ = 0,1,23 \ldots  \ldots $

$430 = (2n - 1)\dfrac{{340}}{{4 \times 0.2}}$

$2n - 1 = \dfrac{{430 \times 4 \times 0.2}}{{340}} = 1.01$

$2n = 2.01$

$n = 1$

Hence, the first mode of vibration frequency is resonantly excited by the given source.

In a pipe open at both ends, the nth mode of vibration frequency is given by the relation:

${v_n} = \dfrac{{Nv}}{{2l}}$

${v_n} = \dfrac{{2{{\sqrt V }_A}}}{v}$

$ = \dfrac{{2 \times 0.2 \times 430}}{{340}} = 0.5$

Since the number of the mode of vibration $(n)$ has to be an integer, the given source does not produce a resonant vibration in an open pipe.

2. Explain why (or how):

(a) In a sound wave, a displacement node is a pressure antinode and vice versa,

(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any "eyes",

(e) A violin note and sitar note may have the same frequency. yet we can distinguish between the two notes.

(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and

(e) The shape of a pulse gets distorted during propagation in a dispersive medium.

Ans: (a) A node is a position where the vibration amplitude is the smallest and the pressure is the highest. An antinode, on the other hand, is a place where the vibration amplitude is greatest and the pressure is lowest.

As a result, a pressure antinode is nothing more than a displacement node, and vice versa.

(b) Bats produce ultrasonic sound waves at a very high frequency. Obstacles re-direct these waves back toward them. With the help of its cerebral sensors, a bat receives a reflected wave (frequency) and calculates the distance, direction, nature, and size of an obstacle.

(c) A sitar's and a violin's overtones, as well as the strength of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a Violin even if they have the same frequency of vibration.

(d) Shear modulus is a property of solids. They are able to withstand shearing force. Fluids yield to shearing stress because they have no defined shape. A transverse wave propagates in such a way that it causes shearing stress in a medium. A wave of this nature can only propagate in solids, not in gases.

The bulk moduli of solids and fluids are different. They have the ability to withstand compressive stress.

3. One end of a long string of linear mass density $8.0 \times {10^{ - 3}}\;{\text{kg}}\;{{\text{m}}^{ - 1}}$ is connected to an electrically driven tuning fork of frequency $256{\text{H}}$. The other end passes over a pulley and is tied to a pan containing a mass of $90\;{\text{kg}}$. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At $t = 0$, the left end (fork end) of the string $x = 0$ has zero transverse displacement $(y = 0)$ and is moving along positive $y$ -direction. The amplitude of the wave is $5.0{\text{em}}$. Write down the transverse displacement $y$ as a function of $x$ and $t$ that describes the wave on the string.

Ans: The equation of a travelling wave propagating along the positive $y$ -direction is given by the displacement equation:

$y(x,t) = a\sin (wt - kx) - (0)$

Linear mass density, $\lambda  = 8.0 \times {10^{ - 3}}$ kg ${{\text{m}}^{ - 1}}$

Frequency of the tuning fork, ${\text{v}} = 256{\text{Hz}}$

Amplitude of the wave, $a = 5.0\;{\text{cm}} = 0.05\;{\text{m}} \ldots ({\text{m}}$

Mass of the pan, $m = 90\;{\text{kg}}$

Tension in the string. ${\text{T}} = {\text{mg}} = 90 \times 9.8 = {\text{BRZN}}$

The velocity of the transverse wave $v$, is given by the relation:

$v = \sqrt {\dfrac{T}{\mu }} $

$ = \sqrt {\dfrac{{882}}{{8.0 \times {{10}^{ - 1}}}}}  = 332\;{\text{m}}/{\text{s}}$

Angular frequency $ \cdot \omega  = 2\pi v$

$ = 2 \times 3.14 \times 256$

$ = 1608.5 = 1.6 \times {10^3}{\text{rad}}/{\text{s}} - ({\text{ui}})$

Wavelength, $\lambda  = \dfrac{v}{v} = \dfrac{{332}}{{256}}m$

$\therefore $ Propagation constant, $k = \dfrac{{2\pi }}{\lambda }$

$ = \dfrac{{2 \times 3.14}}{{\dfrac{{332}}{{256}}}} = 4.84{w^{ - 1}}\quad  \ldots  \ldots ..0$

displacement equation:

\[y(x,t) = 0.05\sin \left( {1.6 \times {{10}^3}t - 4.84x} \right)m\]

4.  Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse $(5)$ and longitudinal $(P)$ sound waves. Typically the speed of S wave is about $4.0\;{\text{km}}\;{\text{s}} - 1$, and that of ${\text{P}}$ wave is $80\;{\text{l}}{{\text{m}}_5}\;{{\text{s}}^{ - 1}}$ - A seismograph records $P$ and $S$ waves from an earthquake. The first $P$ wave arrives 4 min before the first $S$ wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?

Ans: Assume ,

Let ${v_s}$ and ${v_F}$ be the velocities of $S$ and $P$ waves respectively.

Let $L$ be the distance between the epicenter and the seismograph.

We have:

L. $ = {v_g}{t_3}(a)$

$1. = {v_P}{l_P}(i)$

Where,

${t_5}$ and ${t_p}$ are the respective times taken by the S and P waves to reach the seismograph from the epicenter

It is given that:

${v_p} = 8\;{\text{km}}/{\text{s}}$

${v_s} = 4\;{\text{km}}/{\text{s}}$

From equations (i) and (ii), we have:

${v_3}{t_S} = {v_k}{t_p}4{t_3} = 8{t_p}$

${t_s} = 2{t_P}(iii)$

It is also given that

${t_y} - {t_P} = 4\;{\text{min}} = 240s$

$2t - {t_p} = 240$

${t_P} = 240$

And ${t_3} = 2 \times 240 = 480\;{\text{s}}$

From equation (ii), we get

$I = 8 \times 240$

$ = 1920\;{\text{km}}$

Hence, the earthquake occurs at a distance of $1920\;{\text{km}}$ from the seismograph.

5. A train, standing in a station-yard, blows a whistle ot frequency $400\;{\text{Hz}}$ in still air. The wind starts blowing in the direction from the yard to the station with at a speed of $10\;{\text{m}}$ ${5^{ - 1}}$. What are the frequency, wavelength, and speed of sound for an observer standing on the station's platform? is the situation exactly identical to the ease when the air is still and the observer runs towards the yard at a speed of $10\;{\text{m}}\;{{\text{s}}^{ - 1}}$ ? The speed of sound in still air can be taken as $340\;{\text{m}}\;{{\text{s}}^{ - 1}}$ -

Ans: Given that,

 For the stationary observer: $400\;{\text{Hz}},0.875\;{\text{m}};350\;{\text{m}}/{\text{s}}$

For the running observer Not exactly identical

For the stationary observer:

Frequency of the sound produced by the whistle, ${\text{v}} = 400\;{\text{Hz}}$

Speed of sound $ = 340\;{\text{m}}/{\text{s}}$

Velocity of the wind, $v = 10\;{\text{m}}/{\text{s}}$

As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer wit be the same as that produced by the source, Le. $400\;{\text{Hz}}$.

The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units, i.e.,

Effective speed of the sound, ${v_4} = 340 + 10 = 350\;{\text{m}}/{\text{s}}$

The wavelength ( $\lambda $ ) of the sound heard by the observer is given by the relation:

$\lambda  = \dfrac{{{v_4}}}{v} -  = \dfrac{{350}}{{400}} = 0.875\;{\text{m}}$

For the running observer:

Velocity of the observer, ${v_s} = 10\;{\text{m}}/{\text{s}}$

The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency $\left( {{v^n}} \right)$.

This is given by the relation:

${v^\prime } = \left( {\dfrac{v}{{v - {v_r}}}} \right)v$

$ = \left( {\dfrac{{340 + 10}}{{340}}} \right) \times 400 = 411.76Hz$

since the air is still, the effective speed of sound $ = 340 + 0 = 340\;{\text{m}}/{\text{s}}$

The source is at rest. Hence, the wavelength of the sound will not change, Le., $\lambda $ remains

$0.875\;{\text{m}}$.

Hence, the given two situations are not exactly identical

5 Marks Questions

1. Use the formula $v = \sqrt {\dfrac{{yP}}{\rho }} $ to explain why the speed of sound in air

(a) is independent of pressure,

(b) increases with temperature,

(e) Increases with humidity-

Ans: Given that,

(a) Take the relation:

$v = \sqrt {\dfrac{{{y^P}}}{\rho }} ........(i)$

Where,

Density, $\rho  = \dfrac{{{\text{ Moss }}}}{{{\text{ Volume }}}} = \dfrac{N}{V}$

${\text{M}} = $ molecular weight of the ${\text{gas}}$

${\text{V}} = $ Volume of the ${\text{gas}}$

Hence, equation(i) reduces to

${\text{v}} = \sqrt {\dfrac{{yP\;{\text{V}}}}{{\text{M}}}}  \cdots (ii)$

Now from the ideal gas equation for $n = 1:$

${\text{PV}} = {\text{m}}T$

For constant $T,PV = $

Since both $M$ and ${\text{Y}}$ are constants, $v$ = Constant

Hence, at a constant temperature, the speed of sound in a gaseous medium is independent at the change in the pressure of the gas.

(b) Now,

Tale the relation:

$v = \sqrt {\dfrac{{yP}}{\rho }}  \ldots  \ldots (i)$

For one mole at an ideal gas, the gas equation can be written as:

${\text{PV}} = {\text{FT}}$

$P = \dfrac{{RT}}{V} \ldots (ii)$

Substituting equation (ii) in equation (i), we get:

$v = \sqrt {\dfrac{{yRT}}{{V\rho }}}  = \sqrt {\dfrac{{yRT}}{M}} ........(iv)$

Where,

$M = pv$ is a constant

${\text{Y}}$ and ${\text{F}}$ are also constants

We conclude from equation (iv) that $v\alpha \sqrt T $ 

As a result, the speed of sound in a gas is proportional to the square root of the temperature of the gaseous medium, Le. In other words, the speed of sound increases as the temperature of the gaseous medium rises and vice versa. 

(e) Now assume that,

${v_m}$ and ${V_d}$ be the speeds of sound in moist air and dry air respectively.

And ${\rho _n}$ and ${\rho _A}$ be the densities of moist air and dry air respectively.

Take the relation:

$v = \sqrt {\dfrac{{yP}}{\rho }} $

Hence, the speed of sound in moist air is:

${v_n} = \sqrt {\dfrac{{yP}}{{{\rho _m}}}} $

And the speed of sound in dry air is:

${v_f} = \sqrt {\dfrac{{yP}}{{{\rho _d}}}} $

On dividing equations $(1)$ and (ii), we get:

$\dfrac{{{v_m}}}{{{v_A}}} = \sqrt {\dfrac{{1P}}{{{\rho _x}}} \times \dfrac{\rho }{{y\rho }}}  = \sqrt {\dfrac{{{\rho _A}}}{{{P_m}}}} $

However, the presence of water vapor reduces the density of air, 1e.,

${\rho _d} < {\rho _m}$

$\therefore {V_k} > {v_6}$

As a result, sound travels faster in moist air than in dry air. As a result, the speed of sound increases with humidity in a gaseous medium.

2. A transverse harmonic wave on a string is described by

$y(x,t) = 3.0$ sin $\left( {36t + 0.018x + \dfrac{\pi }{4}} \right) \ldots .......(i)$

Where $x$ and $y$ are in $c m$ and $t$ in s. The positive direction of $x$ is from left to right.

(a) Is this a travelling wave or a stationary wave?

If it is travelling, what are the speed and direction of its propagation?

(b) What are its amplitude and frequency?

(c) What is the initial phase at the origin?

(d) What is the least distance between two successive crests in the wave?

Ans:  (a) Yes; Speed = $20\;{\text{m}}4$, Direction = Right to left

(b) $3\;{\text{cm}},5.73{\text{Hz}}$

(c) $\dfrac{\pi }{4}$

(d) $3.49\;{\text{m}}$

Explanation:

(a) The equation of a progressive wave travelling from right to Left is given by the displacement function:

$y(x,t) = a\sin (\omega t + kx + \theta ) \ldots (2)$

The given equation is:

$y(x,t) = 3.0$ sin $\left( {36t + 0.018x + \dfrac{\pi }{4}} \right).... \ldots (ii)$

On comparing both the equations, we find that equation $(i)$ represents a travelling wave, propagating from right to left.

Now, using equations $(i)$ and (ii), we can write:

$\phi  = 36{\text{rad}}/{\text{s}}$ and $k = 0.018\;{{\text{m}}^{ - 1}}$

We know that:

$v = \dfrac{\omega }{{2\pi }}$ and $\lambda  = \dfrac{{2\pi }}{k}$

Also,

$v = vh$

$\therefore v = \left( {\dfrac{\omega }{{2\pi }}} \right) \times \left( {\dfrac{{2\pi }}{k}} \right) = \dfrac{\omega }{k}$

$ = \dfrac{{36}}{{0.018}} = 2000\;{\text{cm}}/{\text{s}} = 20\;{\text{m}}/{\text{s}}$

Hence, the speed of the given travelling wave is $20\;{\text{m}}$ k.

(b) Amplitude of the given wave, $a = 3\;{\text{cm}}$

Frequency of the given wave:

$v = \dfrac{\omega }{{2\pi }} = \dfrac{{36}}{{2x3.14}} = 5.73\;{\text{Hz}}$

(c) On comparing equations $(1)$ and (ii), we find that the initial phase angle, $\phi  = \dfrac{\pi }{4}$

(d) The distance between two successive crests ơ troughs is equal to the wavelength of the wave.

Wavelength is given by the relation:

$k = \dfrac{{2\pi }}{\lambda }$

$\therefore k = \dfrac{{2\pi }}{\lambda } = \dfrac{{2 \times 3.14}}{{0.018}} = 348.89\;{\text{cm}} = 3.49\;{\text{m}}$

3.  For the wave described in Exercise 15.8, plot the displacement $(y)$ versus $(t)$ graphs for $x = 0,2$and $4\;{\text{cm}}$. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

Ans: All the waves have different phases.

The given transverse harmonic wave is

$y(x,t) = 3.0\sin \left( {36t + 0.018x + \dfrac{\pi }{4}} \right)(1)$

For $x = 0$, the equation reduces to:

$y(0,t) = 3.0\sin \left( {36t + \dfrac{\pi }{4}} \right)$

Also, $\omega  = \dfrac{{2\pi }}{T} = 36{\text{rad}}/{{\text{s}}^{ - 1}}$

$\therefore T = \dfrac{\pi }{8}s$

Now, plotting $y$ vs. t graphs using the different values of $t$, as listed in the given table. 

For $x = 0,x = 2$, and $x = 4$, the phases of the three waves wit get changed. This is because amplitude and frequency are invariant for any change in $x$. The $y$ -t plots of the three waves are shown in the given figure.

4. The transverse displacement of a string (clamped at its both ends) is given by

$y(x,t) = 0.06\sin \dfrac{2}{3}x\cos (120\pi t)$

Where $x$ and $y$ are in $m$ and $t$ in $s$. The length of the string is $1.5\;{\text{m}}$ and its mass is $3.0 \times {10^{ - 2}}\;{\text{kg}}$

Answer the following:

(a) Does the function represent a travelling wave or a stationary wave?

(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

(c) Determine the tension in the string.

Ans: (a)The general equation representing a stationary wave is given by the displacement function:

$y(x,t) = 2a\sin kxdb{\cos ^n}{\text{r}}$

This equation is similar to the given equation:

$y(x,t) = 0.06\sin \left( {\dfrac{2}{3}x} \right)\cos (120\pi t)$

Hence, the given function represents a stationary wave.

(b) A wave travelling along the positive $x$ -direction is given as:

${y_1} = a\sin \left( {{\alpha ^2} - kx} \right)$

The wave travelling along the negative $x$ -direction is given as:

${y_2} = a\sin (ct - kx)$

The superposition of these two waves yields:

$y = {y_1} + {y_2} = a\sin (\omega t - kx) = a\sin (\omega t + kx)$

$ = a\sin (\omega r)\cos (kx) = a\sin (kx)\cos (ar) = a\sin (\omega t)\cos (kx) = a\sin (kx)\cos (\omega t)$

$ =  - 2a\sin (kx)\cos (at)$

$ =  - 2a\sin \left( {\dfrac{{2\pi }}{\lambda }x} \right)\cos (2\pi v) \ldots  \ldots  \ldots (1)$

The transverse displacement of the string is given as:

$y(x,t) = 0.06\sin \left( {\dfrac{{2\pi }}{3}x} \right)\cos (120\pi )$

Comparing equations $(1)$ and $(10)$, we have:

$\dfrac{{2\pi }}{\lambda } = \dfrac{{2\pi }}{3}$

Wavelength, ${\text{h}} = 3\;{\text{m}}$

It is given that:

$120\pi  = 2\pi h$

Frequency, ${\text{v}} = 60{{\text{H}}_2}$

Wave speed, $v = $ vh

$ = 60 \times 3 = 180\;{\text{m}}/{\text{s}}$

(e) The velocity of a transverse wave travelling in a string is given by the relation:

$v = \sqrt {\dfrac{T}{\lambda }}  \cdots  \ldots  \ldots (1)$

Where,

Velocity of the transverse wave, $v = 180\;{\text{m}}/{\text{s}}$

Mass of the string, $m = 3.0 \times {10^{ - 1}}\;{\text{kg}}$

Length of the string, $l = 1.5\;{\text{m}}$

Mass per unit length of the string $\lambda  = \dfrac{m}{l}$

$ = \dfrac{{3.0}}{{1.5}} \times {10^{ - 2}}$

$ = 2 \times {10^{ - 1}}\;{\text{kg}}\;{{\text{m}}^{ - 1}}$

Tension in the string $ = T$

From equation (i), tension can be obtained as:

${\text{T}} = {{\text{v}}^2}\mu $

$ = {(180)^2} \times 2 \times {10^{ - 2}}$

$ = 648\;{\text{N}}$

5. A travelling harmonic wave on a string is described by

$y(x,t) = 7.5{\text{sin}}\left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)$

(a) What are the displacement and velocity of oscillation of a point at $x = 1\;{\text{cm}}$, and $t = 1$ s? Is this velocity equal to the velocity of wave propagation?

(b) Locate the points of the string which have the same transverse displacements and velocity as the $x = 1$cm point at $t = 2s,5s$ and 11 s.

Ans: (a) The given harmonic wave is:

$y\left( {{x_1}t} \right) = 7,5\sin \left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)$

Far $x = 1\;{\text{cm}}$ and $t = 15$,

$y(1,1) = 7.5\sin \left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)$

$ = 7.5\sin \left( {12.0050 + \dfrac{\pi }{4}} \right)$

Where, $\theta  = 12.0050 + \dfrac{\pi }{4} = 12.0050 + \dfrac{{3.14}}{4} = 12.79{\text{rad}}$

$ = \dfrac{{180}}{{3.14}} \times 12.79 = {732.81^{\prime \prime }}$

$\therefore y = (1,1) = 7.5\sin \left( {{{732.81}^*}} \right)$

$ = 7.5\sin (90 \times 8 + 12.81) = 7.5\sin {{12.81}^{0}}$

$ = 7.5 \times 0.2217$

$ = 1.6629 \times 1.663\;{\text{cm}}$

The velocity of the oscillation at a given point and time is given as

$v = \dfrac{d}{{dt}}y\left( {{x_1}t} \right) = \dfrac{d}{{dt}}\left[ {7.5\sin \left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)} \right]$

$7.5 \times 12\cos \left( {0.0050x + 12t + \dfrac{\pi }{4}} \right)$

A ${t_x} = 1$ cm and t=1 1 :

$v = y(1,1) = 90\cos \left( {12.005 + \dfrac{\pi }{4}} \right)$

\[=90\cos 73281{}^\circ =90\cos \left( 90\cos \text{ }\!\!~\!\!\text{ }\times 8+{{12.81}^{*}} \right)\]

$ = 90\cos \left( {{{12.81}^{\prime \prime }}} \right)$

$ = 90 \times 0.975 = 87.75\;{\text{cm}}/{\text{s}}$

Now, the equation of a propagating wave ss given by:

$y\left( {{x_n}t} \right) = a\sin (kx + wt + \phi )$

Where,

$k = \dfrac{{2\pi }}{\lambda }$

$\therefore k = \dfrac{{2\pi }}{\lambda }$

And $\theta  = 2\pi v$

$\therefore v = \dfrac{\omega }{{2\pi }}$

speed, $v = v\lambda  = \dfrac{\omega }{k}$

Where.

$\omega  = 12{\text{rad}}/{\text{s}}$

${\text{K}} = 0.0050\;{{\text{m}}^{ - 1}}$

$\therefore v = \dfrac{{12}}{{0.0050}} = 2400\;{\text{cm}}/{\text{s}}$

Hence, the velocity of the wave oscillation at $x = 1\;{\text{cm}}$ and $t = 1\;{\text{s}}$ is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:

$k = \dfrac{{2\pi }}{\lambda }$

$\therefore A = \dfrac{{2\pi }}{k} = \dfrac{{2 \times 3.14}}{{0.0050}}$

$ = 1253\;{\text{cm}} = 12.56\;{\text{m}}$

Therefore, all the points at distances $\lambda .(n =  \pm 1, \pm 2, \ldots $ and $s00n)$, Le. $ \pm 12.56\;{\text{m}}, \pm 25.12$ ${\text{m}}$, . and $s$ on for $x = 1\;{\text{cm}}$, will have the same displacement as the $x = 1\;{\text{cm}}$ points at $t = 2\;{\text{s}}$, 5 s, and $11\;{\text{s}}$.

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