NCERT Solutions for Class 11 Physics Chapter 14  Waves
Wave is a quivering disruption that passes through a medium because of repetition of regular motion of particles of any medium. Ch 14 Physics Class 11 helps create a solid base for future courses like engineering and medical science.
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Chapter Name:  Chapter 14  Waves 
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NCERT solutions for Class 11 Physics Chapter 14 introduce students to many new concepts like effects of Doppler, types of waves and their interrelationships. Students in class 11 have to gain indepth knowledge about different topics on this subject to score better in their exams, as well as rank higher in competitive tests.
Vedantu’s PDF on Physics Class 11 Waves NCERT solutions will help students get familiar with the increasing importance of waves in various mediums.
Mechanical Properties Of Solids Chapter at a Glance  Class 11 NCERT Solutions
A wave is a disturbance that propagates in space, transport energy and momentum from one point to another without the transport of matter.
Mechanical transverse waves are produced in such type of medium which have shearing property, so they are known as shear wave or Swave
A crest is a portion of the medium, which is raised temporarily above the normal position of rest of particles of the medium, when a transverse wave passes.
A trough is a portion of the medium, which is depressed temporarily below the normal position of rest of particles of the medium, when a transverse wave passes.
Longitudinal Wave Motion: Longitudinal waves have oscillatory motion of the medium particles that produce regions of compression (high pressure) and rarefaction (low pressure) which propagate in space with time (see figure).
The regions of high particle density are called compressions and regions of low particle density are called rarefactions.
Wavelength $\lambda $ (length of one wave): Distance traveled by the wave during the time interval in which any one particle of the medium completes one cycle about its mean position. We may also define wavelength as the distance between any two nearest particle of the medium, vibrating in the same phase
Phase: Phase is a quantity which contains all information related to any vibrating particle in a wave. For equation $y= A\;sin\left ( \omega tkx \right ); \left ( \omega tkx \right )=phase$
Wave number $(v\bar{})$: it is defined as $(v\bar{})=\frac{1}{\gamma \lambda }=\frac{k}{2\pi }=$ number of waves in unit length of the wave pattern.
Differential equation of Harmonic Progressive Waves:
Wave velocity: The velocity with which the disturbance, or planes of equal (wave front), travel through the medium is called wave (or phase) velocity
Transverse wave: A transverse wave is a moving wave whose oscillations are perpendicular to the direction of the wave
The speed of a wave on a string is given by
$v=\sqrt{\frac{T}{\mu }}$
where T is tension in the string (in Newtons) and $\mu$ is mass per unit length of the string (kg/m).
When a traveling wave s established on a string, energy is transmitted along the direction of propagation of the wave, in form of potential energy and kinetic energy
Intensity of Sound Waves: The amount of energy carried per unit time by a wave is called its power and power per unit area held perpendicular to the direction of energy flow is called intensity.
Loudness: Audible intensity range for humans: The ability of human to perceive intensity at different frequencies are different. The perception of intensity is maximum at 1000 Hz and perception of intensity decreases as the frequency decreases or increases from 1000Hz.
Decibel Scale: The logarithmic scale which is used for comparing two sound intensity is called decibel scale. The intensity level $\beta$
described in terms of decibels is defined as $\beta = 10 \;log\left ( \frac{I}{I_{0}} \right )(dB)$
Superposition of Waves: The phenomenon of intermixing of two or more waves to produce a new wave is called Superposition of waves. Therefore, according to the superposition principle.
The resultant displacement of a particle at any point of the medium, at any instant of time is the vector sum of the displacement caused to the particle by the individual waves.
Coherence: Two sources are said to be coherent if the phase difference between them does not change with time. In this case their resultant intensity at any point in space remains constant with time. Two independent sources of sound are generally incoherent in nature, i.e. phase difference between them changes with time and hence the resultant intensity due to them at any point in space changes with time.
Standing Waves: Standing waves can be transverse or longitudinal, e.g., in strings (under tension) if reflected wave exists, the waves are transversestationary, while in organ pipes waves are longitudinalstationary.
Closed Organ Pipe: The tube which is closed at one end and open at the other end is called closed organ pipe. Open Organ Pipe: The tube which is open at both ends is called an open organ pipe.
Beats: When two sound waves of same amplitude and different frequency superimpose, then intensity at any point in space varies periodically with time. This effect is called beats. Beat phenomenon can be used for determining an unknown frequency by sounding it together with a source of known frequency.
Doppler’s Effect: The apparent change in frequency or pitch due to relative motion of source an observer along the line of sight is called the Doppler Effect.
Assumptions: (i) The velocity of the source, the observer and the medium are along the line joining the positions of the source and the observer.
(ii) The velocity of the source and the observer is less than velocity of sound.
Access NCERT Solutions for Class 11 Physics Chapter 14 – Waves
1. A string of mass \[\mathbf{2}.\mathbf{50}\text{ }\mathbf{kg}\] is under a tension of \[\mathbf{200}\text{ }\mathbf{N}\]. The length of the stretched string is \[\mathbf{20}.\mathbf{0m}\]. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Ans: It is provided that,
Mass of the string, \[M=2.50\text{ }kg\]
Tension in the string, \[T=200\text{ }N\]
String length, \[\text{l}=20.0\text{ }m\]
Mass per unit length, \[\mu =\frac{M}{l}=\frac{2.50}{20}=0.125kg{{m}^{1}}\]
The transverse wave’s velocity in the string is given by:
\[v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=40m{{s}^{1}}\]
Clearly, the time taken by disturbance to reach the other end is, \[t=\frac{l}{v}=\frac{20}{40}=0.50s\]
2. A stone dropped from the top of a tower of height \[\mathbf{300}\text{ }\mathbf{m}\]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 \[\mathbf{340}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{1}}}\]? (\[\mathbf{g}=\mathbf{9}.\mathbf{8}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{2}}}\])
Ans: It is provided that,
Tower height, \[s=300\text{ }m\]
Stone’s initial velocity, \[u=0\]
Acceleration, \[a=\mathbf{g}=\mathbf{9}.\mathbf{8}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{2}}}\]
Sound speed in air \[=340\text{ }m/s\]
The time that stone takes to strike the water in the pond can be estimated using the motion’s second equation, as:
\[s=u{{t}_{1}}+\frac{1}{2}gt_{1}^{2}\]
\[\Rightarrow 300=0+\frac{1}{2}\times 9.8\times t_{1}^{2}\]
We get,
\[{{t}_{1}}=\sqrt{\frac{300\times 2}{9.8}}=7.82s\]
Time taken by the sound to reach the tower top, \[{{t}_{2}}=\frac{300}{340}=0.88s\]
Therefore, the time after which the sound of splash is heard, \[t={{t}_{1}}+{{t}_{2}}\]
\[\Rightarrow t=7.82+0.88=8.7s\]
The time after which the sound of splash is heard is $8.7s$.
3. A steel wire has a length of \[\mathbf{12}.\mathbf{0}\text{ }\mathbf{m}\] and a mass of \[\mathbf{2}.\mathbf{10}\text{ }\mathbf{kg}\]. 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 \[20{}^{0}C=343m{{s}^{1}}\].
Ans: It is provided that,
Steel wire’s length, \[l=12\text{ }m\]
Steel wire’s mass, \[m=2.10\text{ }kg\]
Velocity of the transverse wave, \[v=343\text{ }m{{s}^{1}}\]
Mass per unit length, \[\mu =\frac{m}{l}=\frac{2.10}{12}=0.175kg{{m}^{1}}\]
For tension T, transverse wave’s velocity can be calculated using the relation:
\[v=\sqrt{\frac{T}{\mu }}\]
\[\Rightarrow T={{v}^{2}}\mu \]
\[\Rightarrow T={{(343)}^{2}}\times 0.175=20588.575\approx 2.06\times {{10}^{4}}N\]
Tension in the wire is \[2.06\times {{10}^{4}}N\].
4. Use the formula \[v=\sqrt{\frac{\gamma P}{\rho }}\] to explain why the speed of sound in air
(a) is independent of pressure,
Ans: We have,
\[v=\sqrt{\frac{\gamma P}{\rho }}\] …...(i)
Where,
Density, \[\rho =\frac{Mass}{Volume}=\frac{M}{V}\]
M = Molecular weight of gas
V = Volume of gas
Hence, equation (i) becomes:
\[v=\sqrt{\frac{\gamma PV}{M}}\] ……(ii)
Now ideal gas equation for \[n=1\] is:
\[PV=RT\]
For constant T, \[PV=Constant\]
Both \[M\] and \[\gamma \] are constants, \[v=Constant\]
Hence, the speed of sound is independent of the change in the pressure of the gas at a constant temperature.
(b) increases with temperature,
Ans: We have,
\[v=\sqrt{\frac{\gamma P}{\rho }}\] ……(i)
Now ideal gas equation for \[n=1\] is:
\[PV=RT\]
\[P=\frac{RT}{V}\] ……(ii)
Substituting (ii) in (i), we get:
\[v=\sqrt{\frac{\gamma RT}{V\rho }}=\sqrt{\frac{\gamma RT}{M}}\] ……(iii)
Where,
Mass, \[M=\rho V\]is a constant
\[\gamma \]and \[R\]are also constants.
We get from equation (iii),
\[v\propto \sqrt{T}\].
Hence, the sound speed in a gas is directly proportional to the square root of the gaseous medium’s temperature, i.e., the sound speed increases with rise in the gaseous medium’s temperature and vice versa.
(c) increases with humidity.
Ans: Let \[{{v}_{m}}\] and \[{{v}_{d}}\] are the sound speed in moist air and dry air respectively and \[{{\rho }_{m}}\] and \[{{\rho }_{d}}\] are the densities of moist air and dry air respectively.
We have,
\[v=\sqrt{\frac{\gamma P}{\rho }}\]
The speed of sound in moist air is:
\[{{v}_{m}}=\sqrt{\frac{\gamma P}{{{\rho }_{m}}}}\] ……(i)
The speed of sound in dry air is:
\[{{v}_{d}}=\sqrt{\frac{\gamma P}{{{\rho }_{d}}}}\] ……(ii)
On dividing equations (i) and (ii), we get:
\[\frac{{{v}_{m}}}{{{v}_{d}}}=\sqrt{\frac{\gamma P}{{{\rho }_{m}}}\times \frac{{{\rho }_{d}}}{\gamma P}}=\sqrt{\frac{{{\rho }_{d}}}{{{\rho }_{m}}}}\] ……(iii)
However, the presence of water vapour decreases the density of air, i.e.,
\[{{\rho }_{d}}<{{\rho }_{m}}\]
\[\Rightarrow {{v}_{m}}>{{v}_{d}}\]
Hence, the speed of sound in moist air is higher than it is in dry air. Thus, in a gaseous medium, the sound speed increases with humidity.
5. You have learnt that a travelling wave in one dimension is represented by a function \[y=f(x,t)\] where x and t must appear in the combination \[xvt\] or \[x+vt\], i.e. \[y=f(x\pm \nu t)\]. Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:
(a) \[{{(xvt)}^{2}}\]
Ans: No.
For \[x=0\] and \[t=0\], the function \[{{(x+vt)}^{2}}\] becomes \[0\].
Hence, for \[x=0\] and \[t=0\], the function represents a point.
(b) \[log\left[ \frac{x+vt}{{{x}_{0}}} \right]\]
Ans: Yes.
For \[x=0\] and \[t=0\], the function \[\log \left( \frac{x+vt}{{{x}_{0}}} \right)=\log 0=\infty \]
Since the function does not converge to a finite value for \[x=0\] and \[t=0\], it does not represent a travelling wave.
(c) \[\frac{1}{(x+vt)}\]
Ans: No.
For \[x=0\] and \[t=0\], the function
\[\frac{1}{x+vt}=\frac{1}{0}=\infty \]
Since the function does not converge to a finite value for \[x=0\] and \[t=0\], it does not represent a travelling wave.
The converse is not true. The requirement for a wave function of a travelling wave is that for all x and t values, wave function should have a finite value. Therefore, none can represent a travelling wave.
6. A bat emits ultrasonic sound of frequency \[\mathbf{1000}\text{ }\mathbf{kHz}\] in air. If the sound meets a water surface, what is the wavelength of
(a) the reflected sound,
Ans: We have,
Frequency of the ultrasonic sound, \[\nu =1000kHz={{10}^{6}}Hz\]
Speed of sound in air, \[{{v}_{a}}=340m{{s}^{1}}\]
The wavelength \[({{\lambda }_{r}})\]of the reflected sound is given by:
\[{{\lambda }_{r}}=\frac{{{v}_{a}}}{\nu }=\frac{340}{{{10}^{6}}}=3.4\times {{10}^{4}}m\]
(b) the transmitted sound? Speed of sound in air is \[340m{{s}^{1}}\] and in water \[1486m{{s}^{1}}\].
Ans: We have,
Ultrasonic sound’s frequency, \[\nu =1000kHz={{10}^{6}}Hz\]
Sound speed in water, \[{{v}_{w}}=1486m/s\]
The wavelength is given as: \[{{\lambda }_{t}}=\frac{1486}{{{10}^{6}}}=1.49\times {{10}^{3}}m\]
7. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is \[\mathbf{1}.\mathbf{7}\text{ }\mathbf{km}{{\mathbf{s}}^{\mathbf{1}}}\]? The operating frequency of the scanner is \[\mathbf{4}.\mathbf{2}\text{ }\mathbf{MHz}.\]
Ans: It is provided that,
Sound speed in the tissue, \[v=1.7Km{{s}^{1}}=1.7\times {{10}^{3}}m{{s}^{1}}\]
Scanner’s operating frequency, \[\nu =4.2MHz=4.2\times {{10}^{6}}Hz\]
The wavelength of sound wave in the tissue is given by:
\[\lambda =\frac{v}{\nu }=\frac{1.7\times {{10}^{3}}}{4.2\times {{10}^{6}}}=4.1\times {{10}^{4}}m\]
The wavelength of sound in the tissue is \[4.1\times {{10}^{4}}m\].
8. A transverse harmonic wave on a string is described by \[y(x,t)=3.0\sin \left( 36t+0.018x+\frac{\pi }{4} \right)\]
Where x and y are in cm 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?
Ans: The given equation is the equation of a travelling wave, moving from right to left because it is an equation of the type
\[y(x,t)=A\sin (\omega t+kx+\phi )\]
Here, \[A=3.0cm\], \[\omega =36ra{{d}^{1}},k=0.018cm\]and \[\phi =\frac{\pi }{4}\]
\[\therefore \]Speed of wave propagation is given by,
\[v=\frac{\omega }{k}=\frac{36rad{{s}^{1}}}{0.018c{{m}^{1}}}=\frac{36rad{{s}^{1}}}{0.018\times {{10}^{2}}{{m}^{1}}}=20m{{s}^{1}}\]
The speed of wave propagation is \[20m{{s}^{1}}\].
(b) What are its amplitude and frequency?
Ans: Amplitude of wave, \[A=3.0cm=0.03m\]
Frequency of wave, \[\nu =\frac{\omega }{\,2\pi }=\frac{36}{2\pi }=5.7Hz\]
(c) What is the initial phase at the origin?
Ans: Initial phase at origin, \[\phi =\frac{\pi }{4}\]rad
(d) What is the least distance between two successive crests in the wave?
Ans: Least distance between two successive crests in the wave,\[\Rightarrow \lambda =\frac{2\pi }{k}=\frac{2\pi }{0.018}=349cm=3.49m\]
9. For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for \[\mathbf{x}=\mathbf{0},\text{ }\mathbf{2}\text{ }\mathbf{and}\text{ }\mathbf{4}\text{ }\mathbf{cm}\]. 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:
\[y(x,t)=3.0\sin \left( 36t+0.018x+\frac{\pi }{4} \right)\] ……(i)
For \[x=0\], the equation becomes:
\[y(0,t)=3.0\sin \left( 36t+\frac{\pi }{4} \right)\]
Also, \[\omega =\frac{2\pi }{T}=36rad/s\]
\[\therefore T=\frac{2\pi }{36}s\]
Now, plotting y vs t graphs using the different values of t, as listed in the given table.
t(s)  0  \[\frac{T}{8}\]  \[\frac{2T}{8}\]  \[\frac{3T}{8}\]  \[\frac{4T}{8}\]  \[\frac{5T}{8}\]  \[\frac{6T}{8}\]  \[\frac{7T}{8}\] 
y(cm)  \[\frac{3\sqrt{2}}{2}\]  3  \[\frac{3\sqrt{2}}{2}\]  0  \[\frac{3\sqrt{2}}{2}\]  3  \[\frac{3\sqrt{2}}{2}\]  0 
For \[x=0,\text{ }x=2,\text{ }and\text{ }x=4\], the phases of the three waves will get altered. This is because amplitude and frequency are same for any change in x. The \[yt\] plots of the three waves are shown in the given figure.
10. For the travelling harmonic wave \[y(x,t)=2.0\cos 2\pi \left( 10t0.0080x+0.35 \right)\]
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 \[\mathbf{4}\text{ }\mathbf{m},\text{ }\mathbf{0}.\mathbf{5}\text{ }\mathbf{m},\] \[\frac{\lambda }{2}\], \[\frac{3\lambda }{4}\].
(a) \[\mathbf{4}\text{ }\mathbf{m}\]
Ans: Equation for a travelling harmonic wave is given by:
\[y(x,t)=2.0\cos 2\pi \left( 10t0.0080x+0.35 \right)\]
\[\Rightarrow y(x,t)=2.0\cos (20\pi t0.016\pi x+0.70\pi )\]
Where, Propagation constant, \[k=0.0160\pi \]
Amplitude, \[a=2cm\]
Angular frequency, \[\omega =20\pi rad/s\]
Phase difference is given by:
\[\phi =kx=\frac{2\pi }{\lambda }\]
for \[x=4m=400\text{ }cm\]
\[\Rightarrow \phi =0.016\pi \times 400=6.4\pi \] rad
(b) \[\mathbf{0}.\mathbf{5}\text{ }\mathbf{m}\]
Ans: Phase difference is given by:
\[\phi =kr=\frac{2\pi }{\lambda }\]
For \[x=0.5\text{ }m=50\text{ }cm\]
\[\Rightarrow \phi =0.016\pi \times 50=0.8\pi \] rad
(c) \[\frac{\lambda }{2}\]
Ans: Phase difference is given by:
\[\phi =kr=\frac{2\pi }{\lambda }\]
For \[x=\frac{\lambda }{2}\]
\[\Rightarrow \phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{2}=\pi \] rad
(d) \[\frac{3\lambda }{4}\]
Ans: Phase difference is given by:
\[\phi =kr=\frac{2\pi }{\lambda }\]
For \[x=\frac{3\lambda }{4}\]
\[\Rightarrow \phi =\frac{2\pi }{\lambda }\times \frac{3\lambda }{4}=1.5\pi rad\]
11. The transverse displacement of a string (clamped at its both ends) is given by \[y(x,t)=0.06\sin \frac{2\pi x}{3}\cos (120\pi t)\] Where x and y are in m and t in s. The length of the string is \[\mathbf{1}.\mathbf{5}\text{ }\mathbf{m}\] and its mass is \[\mathbf{3}.\mathbf{0}\text{ }\times \text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{2}}}\mathbf{kg}\].
Answer the following:
(a) Does the function represent a travelling wave or a stationary wave?
Ans: The general equation of stationary wave is given by:
\[y(x,t)=Asin(kx)cos(\omega t)\]
This given equation is similar to the equation of stationary wave:
\[y(x,t)=0.06\sin \frac{2\pi x}{3}\cos (120\pi t)\]
Hence, the given function is 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?
Ans: The transverse displacement of the string is given as:
\[y(x,t)=0.06\sin \left( \frac{2\pi }{3}x \right)\cos (120\pi t)\]
From above equation, $k=\frac{2\pi }{3}$
\[\therefore \] Wavelength, $\lambda =\frac{2\pi }{k}=\frac{2\pi }{\frac{2\pi }{3}}=3m$
It is given that:
\[120\pi =2\pi \nu \]
Frequency, \[\nu =60\]Hz
Wave speed, \[v=\nu \lambda =60\times 3=180m{{s}^{1}}\]
(c) Determine the tension in the string.
Ans: The transverse wave’s velocity travelling in a string is given by the relation:
\[v=\sqrt{\frac{T}{\mu }}\] ……(i)
Where, Velocity of the transverse wave, \[v=180m{{s}^{1}}\]
Mass of the string, \[m=3.0\times {{10}^{2}}kg\]
Length of the string, \[l\text{ }=\text{ }1.5\text{ }m\]
Mass per unit length of the string, \[\mu =\frac{m}{l}=\frac{3.0}{1.5}\times {{10}^{2}}=2\times {{10}^{2}}kg{{m}^{1}}\]
Tension in the string \[=\text{ }T\]
From equation (i), tension can be obtained as:
\[T={{v}^{2}}\mu \]
\[\Rightarrow T={{(180)}^{2}}\times 2\times {{10}^{2}}\]
\[\Rightarrow T=648N\]
The tension in string is $648N$.
12.(i) For the wave on a string described in Question 11, do all the points on the string oscillate with the same
(a) frequency,
Ans: Yes, all the points on the string vibrate with the same frequency, except at the nodes which are having zero frequency.
(b) phase,
Ans: Yes, all the points in any oscillating loop have the same phase, except at the nodes.
(c) amplitude? Explain your answers.
Ans: No, all the points in any oscillating loop have different vibration amplitudes.
(ii) What is the amplitude of a point \[\mathbf{0}.\mathbf{375}\text{ }\mathbf{m}\] away from one end?
Ans: The given equation is:
\[y(x,t)=0.06\sin \left( \frac{2\pi }{3}x \right)\cos \left( 120\pi t \right)\]
For x = 0.375 m and t = 0
Amplitude\[=0.06\sin \left( \frac{2\pi }{3}x \right)\cos 0\]
\[\Rightarrow a=0.06\sin \left( \frac{2\pi }{3}\times 0.375 \right)\times 1\]
\[\Rightarrow a=0.06\sin \left( 0.25\pi \right)=0.06\sin \left( \frac{\pi }{4} \right)\]
\[\Rightarrow a=0.06\times \frac{1}{\sqrt{2}}=0.042m\]
The value of amplitude is \[0.042m\].
13. 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 travelling wave, (ii) a stationary wave or (iii) none at all:
(a) \[y=2\cos (3x)sin(10t)\]
Ans: This equation demonstrates a stationary wave because the harmonic terms \[kx\] and \[wt\] seem separately in the equation.
(b) \[y=2\sqrt{xvt}\]
Ans: This equation is not having any harmonic term. Therefore, it is not representing either a stationary wave or travelling wave.
(c) \[y=3\sin (5x0.5t)+4\cos (5x0.5t)\]
Ans: This equation demonstrates a travelling wave as it is having harmonic terms \[kx\] and \[wt\] are in \[kx~wt\] combination.
(d) \[y=\cos x\sin t+\cos 2x\sin 2t\]
Ans: This equation demonstrates a stationary wave because it is having harmonic terms \[kx\] and \[wt\] separately in the equation. It is the superposition of two stationary waves.
14. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of \[\mathbf{45}\text{ }\mathbf{Hz}\]. The mass of the wire is \[\mathbf{3}.\mathbf{5}\text{ }\times \text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{2}}}\mathbf{kg}\] and its linear mass density is \[\mathbf{4}.\mathbf{0}\text{ }\mathbf{x}\text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{2}}}\mathbf{kg}\text{ }{{\mathbf{m}}^{\mathbf{1}}}\]. What is
(a) the speed of a transverse wave on the string, and
Ans: Provided that,
Mass of the wire, \[m=3.5\times {{10}^{2}}kg\]
Linear mass density, \[\mu =\frac{m}{l}=4.0\times {{10}^{2}}kg{{m}^{1}}\]
Frequency of vibration, \[\nu =45Hz\]
Length of the wire, \[l=\frac{m}{\mu }=\frac{3.5\times {{10}^{2}}}{4.0\times {{10}^{2}}}=0.875m\]
The wavelength of the stationary wave \[(\lambda )\] is given by:
\[\lambda =\frac{2l}{n}\] where, \[n=\]number of nodes
For fundamental node, \[n=1\]:
\[\lambda =2l\]
\[\lambda =2\times 0.875=1.75m\]
The speed of the transverse wave in the string is given as:
\[v=\nu \lambda =45\times 1.75=78.75m{{s}^{1}}\]
The speed of transverse wave is \[78.75m{{s}^{1}}\].
(b) the tension in the string?
Ans: The tension produced in the string is given by the relation:
\[T={{v}^{2}}\mu \]
\[\Rightarrow T={{(78.75)}^{2}}\times 4.0\times {{10}^{2}}=248.06N\]
The tension in the string is $248.06N$.
15. A metrelong 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 \[\mathbf{340}\text{ }\mathbf{Hz}\]) when the tube length is \[\mathbf{25}.\mathbf{5}\text{ }\mathbf{cm}\text{ }\mathbf{or}\text{ }\mathbf{79}.\mathbf{3}\text{ }\mathbf{cm}\]. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Ans: It is provided that,
Frequency of the turning fork, \[\nu =340\text{ }Hz\]
Since the given pipe is attached with a movable piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.
Such a system gives odd harmonics. The relation of fundamental note in a closed pipe is given by:
\[{{l}_{1}}=\frac{\lambda }{4}\]
Where, length of the pipe \[{{l}_{1}}=25.5cm=0.255m\]
\[\Rightarrow \lambda =4{{l}_{1}}=4\times 0.255=1.02m\]
The relation of sound speed is given by:
\[v=\nu \lambda =340\times 1.02=346.8m{{s}^{1}}\]
16. A steel rod \[\mathbf{100}\text{ }\mathbf{cm}\] long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be \[\mathbf{2}.\mathbf{53}\text{ }\mathbf{kHz}.\] What is the speed of sound in steel?
Ans: We have,
Length of the steel rod, \[l\text{ }=\text{ }100\text{ }cm\text{ }=\text{ }1\text{ }m\]
Fundamental frequency of vibration, \[\nu =2.53kHz=2.53\times {{10}^{3}}Hz\]
An antinode (A) is formed at its centre, and nodes (N) are formed at its two ends when the rod is plucked at its middle, as shown in the given figure.
The distance between two successive nodes is \[\frac{\lambda }{2}\]
\[\therefore l=\frac{\lambda }{2}\]
\[\lambda =2l=2\times 1=2m\]
The sound speed in steel is given by:
\[v=\nu \lambda \]
\[\Rightarrow v=2.53\times {{10}^{3}}\times 2\]
\[\Rightarrow v=5.06\times {{10}^{3}}m{{s}^{1}}\]
\[\Rightarrow v=5.06km{{s}^{1}}\]
The speed of sound in steel is \[5.06km{{s}^{1}}\].
17. A pipe \[\mathbf{20}\text{ }\mathbf{cm}\] long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a \[\mathbf{430}\text{ }\mathbf{Hz}\] source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is \[\mathbf{340}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{1}}}\]).
Ans: Provided that,
Length of the pipe, \[l=20\text{ }cm=0.2\text{ }m\]
Source frequency \[=\text{ }{{n}^{th}}\] normal mode of frequency, \[{{\nu }_{n}}=\text{ }430\text{ }Hz\]
Speed of sound, \[v=340\text{ }m/s\]
In a closed pipe, the nth normal mode of frequency is given by the relation:\[{{\nu }_{n}}=(2n1)\frac{v}{4l}\]
Where, n is an integer \[=\text{ }0,1,2,3\ldots ..\]
\[\Rightarrow 430=\frac{(2n1)340}{4\times 0.2}\]
\[\Rightarrow 2n1=\frac{430\times 4\times 0.2}{340}\]
\[\Rightarrow 2n1=1.01\]
\[\Rightarrow n\approx 1\]
Clearly, the vibration frequency’s first mode is resonantly excited by the given source. In a pipe open at both ends, the \[{{n}^{th}}\] mode of vibration frequency is given by:
\[{{\nu }_{n}}=\frac{nv}{2l}\]
\[\Rightarrow n=\frac{2l{{v}_{n}}}{v}\]
\[\Rightarrow n=\frac{2\times 0.2\times 430}{340}=0.5\]
Since the number of the vibration mode has to be an integer, the given source does not generate a resonant vibration in an open pipe.
18. Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency \[\mathbf{6}\text{ }\mathbf{Hz}\]. The tension in the string A is slightly reduced and the beat frequency is found to reduce to \[\mathbf{3}\text{ }\mathbf{Hz}\]. If the original frequency of A is \[\mathbf{324}\text{ }\mathbf{Hz}\], what is the frequency of B?
Ans: It is provided that,
Frequency of string A, \[{{f}_{A}}=324Hz\]
Frequency of string B \[={{f}_{B}}\]
Beat’s frequency, \[n=6\text{ }Hz\]
Beat’s frequency is given as:
\[n=\left {{f}_{A}}\pm {{f}_{B}} \right\]
\[\Rightarrow 6=324\pm {{f}_{B}}\]
\[\Rightarrow {{f}_{n}}=\text{ }330\text{ }Hz\text{ }or\text{ }318\text{ }Hz\]
Frequency decreases with a reduction in the tension in a string because frequency is directly proportional to the tension’s square root. It is given as:
\[\nu \propto \sqrt{T}\]
Hence, the beat frequency cannot be \[330\text{ }Hz\]
\[\therefore {{f}_{B}}=318Hz\]
19. Explain why (or how):
(a) In a sound wave, a displacement node is a pressure antinode and vice versa,
Ans: A node is a point where the vibration amplitude is the minimum and pressure is the greatest. An antinode is a point where the maximum vibration amplitude and pressure are the lowest. Therefore, 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”,
Ans: Bats emit powerful highfrequency ultrasonic sound waves. These waves get reflected toward them by obstructions. A bat takes a reflected wave and measures the distance, size, direction, and nature of an obstacle with the aid of its brain senses.
(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
Ans: The overtones generated by a sitar and a violin, and the powers of these overtones, are different. Hence, one can differentiate between the notes produced by a sitar and a violin even if they have a similar vibration frequency.
(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
Ans: Solids have a shear modulus. They can provide shearing stress. Since fluids do not hold any definite shape, they produce shearing stress. The transverse wave propagation is such that it provides shearing stress in a medium. The propagation of such a wave is probable only in solids and not in gases. Both solids and fluids have their respective bulk moduli. They can provide compressive stress. Hence, longitudinal waves can pass through solids and fluids.
(e) the shape of a pulse gets distorted during propagation in a dispersive medium.
Ans: A pulse is a combination of waves having various wavelengths. These waves move in a dispersive medium with varying velocities, depending on the medium's nature. This results in the shape distortion of a wave pulse.
20. A train, standing at the outer signal of a railway station blows a whistle of frequency \[\mathbf{400}\text{ }\mathbf{Hz}\] 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 \[\mathbf{10}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{1}}}\],
Ans: Provided that,
Frequency of the whistle, \[\nu =400\text{ }Hz\]
Speed of the train, \[{{v}_{T}}=10\text{ }m/s\]
Speed of sound, \[v=340\text{ }m/s\]
The whistle’s apparent frequency \[\left( \nu \right)\] as the train approaches the platform is given by:
\[\nu '=\left( \frac{v}{v{{v}_{T}}} \right)\nu \]
\[\Rightarrow \nu '=\left( \frac{340}{34010} \right)\times 400=412.12Hz\]
(b) recedes from the platform with a speed of \[\mathbf{10}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{1}}}\]?
Ans: The apparent frequency (\[\nu ''\]) of the whistle as the train recedes from the platform is given by the relation:
\[\nu ''=\left( \frac{v}{v+{{v}_{T}}} \right)\nu \]
\[\Rightarrow \nu ''=\left( \frac{340}{340+10} \right)\times 400=388.57Hz\]
The apparent frequency of the whistle is $388.57Hz$.
(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as \[\mathbf{340}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{1}}}\].
Ans: The apparent change in sound frequency is caused by the relative motions of the source and the observer. These relative motions generate no effect on the sound speed. Therefore, the sound speed in the air in both cases remains the same, i.e., \[340\text{ }m/s.\]
21. A train, standing in a stationyard, blows a whistle of frequency \[\mathbf{400}\text{ }\mathbf{Hz}\] in still air. The wind starts blowing in the direction from the yard to the station with at a speed of \[\mathbf{10}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{1}}}\]. What is the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of \[\mathbf{10}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{1}}}\]? The speed of sound in still air can be taken as \[\mathbf{340}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{1}}}.\]
Ans: For the stationary observer:
Frequency of the sound generated by the whistle, \[\nu =400\text{ }Hz\]
Speed of sound \[=\text{ }340\text{ }m/s\]
Velocity of the wind, \[v=10\text{ }m/s\]
As there is no relative motion between the observer and the source, the sound’s frequency heard by the observer will be similar to that produced by the source, i.e., \[400\text{ }Hz\].
The wind is blowing towards the observer. Hence, the sound’s effective speed increases by \[10\] units, i.e., \[{{v}_{e}}=340+10=350m/s\]
The wavelength \[(\lambda )\] of the sound heard by the observer is given by:
\[\lambda =\frac{{{v}_{e}}}{\nu }=\frac{350}{400}=0.875m\]
For the running observer:
Velocity of the observer, \[{{v}_{o}}=10\text{ }m/s\]
The observer is moving towards the source. As a result, the relative motions of the source and the observer, there is a frequency change (\[\nu '\])
This is given by the relation:
\[\nu '=\left( \frac{v+{{v}_{o}}}{v} \right)\nu \]
\[\Rightarrow \nu '=\left( \frac{340+10}{340} \right)\times 400=411.76Hz\]
Since the air is still so, the effective speed of sound \[=340+0=340\text{ }m/s\]
The source is at rest. Hence, the sound’s wavelength will not change, i.e., $\lambda $remains \[0.875\text{ }m.\]
Hence, the two given situations are not exactly identical.
Additional Exercise:
22. A travelling harmonic wave on a string is described by
\[y(x,t)=7.5\sin \left( 0.0050x+12t+\frac{\pi }{4} \right)\]
(a) What are the displacement and velocity of oscillation of a point at \[\mathbf{x}\text{ }=\text{ }\mathbf{1}\text{ }\mathbf{cm},\text{ }\mathbf{and}\text{ }\mathbf{t}\text{ }=\text{ }\mathbf{1}\text{ }\mathbf{s}\]? Is this velocity equal to the velocity of wave propagation?
Ans: The given harmonic wave is:
\[y(x,t)=7.5\sin \left( 0.0050x+12t+\frac{\pi }{4} \right)\]
For \[x\text{ }=\text{ }1\text{ }cm\] and \[t\text{ }=\text{ }1\text{ }s,\]
\[y(1,1)=7.5\sin \left( 0.0050+12+\frac{\pi }{4} \right)\]
\[\Rightarrow y(1,1)=7.5\sin \left( 12.0050+\frac{\pi }{4} \right)\]
\[\Rightarrow y(1,1)=7.5\sin \theta \]
\[Where,\theta =12.0050+\frac{\pi }{4}=12.0050+\frac{3.14}{4}=12.78rad\]
\[\Rightarrow \theta =\frac{180}{3.14}\times 12.78={{732.81}^{o}}\]
\[\therefore y(1,1)=7.5\sin ({{732.81}^{o}})\]
\[\Rightarrow y(1,1)=7.5\sin (90\times 80+{{12.81}^{o}})=7.5\sin {{12.81}^{o}}\]
\[\Rightarrow y(1,1)=7.5\times 0.2217\]
\[\Rightarrow y(1,1)=1.6629\approx 1.663cm\]
The velocity of the oscillation at a given point and time is given as:
\[v=\frac{d}{dt}y(x,t)=\frac{d}{dt}\left[ 7.5\sin \left( 0.0050x+12t+\frac{\pi }{4} \right) \right]\]
\[\Rightarrow v=7.5\times 12\cos \left( 0.0050x+12t+\frac{\pi }{4} \right)\]
At \[x\text{ }=\text{ }1\text{ }cm\] and \[t\text{ }=\text{ }1s\],
\[v=y(1,1)=90\cos \left( 12.005+\frac{\pi }{4} \right)\]
\[\Rightarrow v=90\cos ({{732.81}^{o}})=90\cos \left( 90\times 8+{{12.81}^{o}} \right)\]
\[\Rightarrow v=90\cos ({{12.81}^{o}})\]
\[\Rightarrow v=90\times 0.975=87.75cm/s\]
Now, the equation of a propagating wave is given by:
\[y(x,t)=asin(kx+\omega t+\phi )\]
Where, \[k=\frac{2\pi }{\lambda }\]
\[\lambda =\frac{2\pi }{k}\] and \[\omega =2\pi \nu \]
\[\Rightarrow \nu =\frac{\omega }{2\pi }\]
\[Speed,v=\nu \lambda =\frac{\omega }{k}\]
Where, \[\omega =12rad/s\]
\[k=0.0050{{m}^{1}}\]
\[\Rightarrow v=\frac{12}{0.0050}=2400cm/s\]
Hence, the velocity of the wave oscillation at \[x=1\text{ }cm\] and \[t=1\text{ }s\] is not equal to the wave propagation’s velocity.
(b) Locate the points of the string which have the same transverse displacements and velocity as the \[\mathbf{x}=\mathbf{1}\text{ }\mathbf{cm}\] point at \[\mathbf{t}=\mathbf{2}\text{ }\mathbf{s},\text{ }\mathbf{5}\text{ }\mathbf{s}\text{ }\mathbf{and}\text{ }\mathbf{11}\text{ }\mathbf{s}.\]
Ans: The relation of propagation constant with wavelength is given by:
\[k=\frac{2\pi }{\lambda }\]
\[\Rightarrow \lambda =\frac{2\pi }{k}=\frac{2\times 3.14}{0.0050}=1256cm=12.56m\]
Therefore, all the points at distances \[n\lambda (n=\pm 1,\pm 2,...and\text{ }so\text{ }on)\], i.e., \[\pm 12.56m,\pm 25.12m,..\ldots \] and so on for \[x=1\text{ }cm\], will have the same displacement as the \[x=1\text{ }cm\] points at \[t=2\text{ }s,\text{ }5\text{ }s,\text{ }and\text{ }11\text{ }s\].
23. A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
(a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation?
Ans: (a) The sound pulse does not have a constant wavelength or frequency. However, the sound speed vibration remains the same, equal to the sound speed in that medium.
(b) If the pulse rate is \[\mathbf{1}\] after every \[\mathbf{20}\text{ }\mathbf{s}\], (that is the whistle is blown for a split of second after every \[\mathbf{20}\text{ }\mathbf{s}\]), is the frequency of the note produced by the whistle equal to \[\frac{1}{20}\] or \[\mathbf{0}.\mathbf{05}\text{ }\mathbf{Hz}\]?
Ans: The short pip produced after every \[20\text{ }s\] does not implies that the frequency of the whistle is \[\frac{1}{20}\] or \[0.05\text{ }Hz\]. It means that \[0.05\text{ }Hz\] is the frequency of the repetition of the pip of the whistle.
24. One end of a long string of linear mass density \[\mathbf{8}.\mathbf{0}\text{ }\mathbf{x}\text{ }\mathbf{1}{{\mathbf{0}}^{\mathbf{3}}}\mathbf{kg}\text{ }{{\mathbf{m}}^{\mathbf{1}}}\]is connected to an electrically driven tuning fork of frequency \[\mathbf{256}\text{ }\mathbf{Hz}\]. The other end passes over a pulley and is tied to a pan containing a mass of \[\mathbf{90}\text{ }\mathbf{kg}\]. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At \[\mathbf{t}=\mathbf{0}\], the left end (fork end) of the string \[\mathbf{x}=\mathbf{0}\] has zero transverse displacement (\[\mathbf{y}=\mathbf{0}\]) and is moving along positive ydirection. The amplitude of the wave is \[\mathbf{5}.\mathbf{0}\text{ }\mathbf{cm}\]. Write down the transverse displacement y as function of x and t that describes the wave on the string.
Ans: The equation of a travelling wave propagating along the positive ydirection is given by the displacement equation:
\[y(x,t)=a\sin (\omega tkx)\] ……(i)
Linear mass density, \[\mu =8.0\times {{10}^{3}}kg{{m}^{1}}\]
Frequency of the tuning fork, \[\nu =256\text{ }Hz\]
Amplitude of the wave, \[a=5.0\text{ }cm=0.05\text{ }m\]……(ii)
Mass of the pan, \[\text{m}=90\text{ }kg\]
Tension in the string, \[T=mg=90x9.8=882\text{ }N\]
The transverse wave’s velocity is given by:
\[v=\sqrt{\frac{T}{\mu }}\]
\[\Rightarrow v=\sqrt{\frac{882}{8.0\times {{10}^{3}}}}=332m/s\]
Angular frequency, \[\omega =2\pi \nu \]
\[\Rightarrow w=2\times 3.14\times 256\]
\[\Rightarrow w=1608.5=16\times {{10}^{3}}rad/s\]……(iii)
Wavelength, \[\lambda =\frac{v}{\nu }=\frac{332}{256}m\]
\[\text{Propagation constant},k=\frac{2\pi }{\lambda }\]
\[\Rightarrow k=\frac{2\times 3.14}{{}^{332}/{}_{256}}=4.84{{m}^{1}}\]……(iv)
Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get,
\[y(x,t)=0.05\sin (1.6\times {{10}^{3}}t4.84x)m\]
This describes the wave of the string.
25. A SONAR system fixed in a submarine operates at a frequency \[\mathbf{40}\text{ }\mathbf{kHz}\]. An enemy submarine moves towards the SONAR with a speed of \[\mathbf{360}\text{ }\mathbf{km}{{\mathbf{h}}^{\mathbf{1}}}\]. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be \[\mathbf{1450}\text{ }\mathbf{m}{{\mathbf{s}}^{\mathbf{1}}}\].
Ans: It is provided that,
SONAR system’s operating frequency, \[\nu =40\text{ }kHz\]
Speed of the enemy submarine, \[{{v}_{e}}=360km/h=100\text{ }m/s\]
Sound speed in water, \[v=1450\text{ }m/s\]
The source is at rest and the observer is moving towards it. Hence, the apparent frequency (\[\nu '\]) received and reflected by the submarine is given by:
\[\nu '=\left( \frac{v+{{v}_{e}}}{v} \right)\nu \]
\[\Rightarrow v'=\left( \frac{1450+100}{1450} \right)\times 40=42.76kHz\]
The frequency (\[\nu ''\]) received by the enemy submarine is given by:
\[\nu ''=\left( \frac{v}{v{{v}_{s}}} \right)\nu '\]
Where, \[{{v}_{s}}=\text{ }100\text{ }m/s\]
\[\Rightarrow \nu ''=\left( \frac{1450}{1450100} \right)42.72=45.93kHz\]
26. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically, the speed of S wave is about \[\mathbf{4}.\mathbf{0}\text{ }\mathbf{km}{{\mathbf{s}}^{\mathbf{1}}}\], and that of P wave is \[\mathbf{8}.\mathbf{0}\text{ }\mathbf{km}{{\mathbf{s}}^{\mathbf{1}}}\]. A seismograph records P and S waves from an earthquake. The first P wave arrives \[\mathbf{4}\] min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
Ans: Let \[{{v}_{s}}\] and \[{{v}_{p}}\] be the velocities of S and P waves respectively.
Let L be the distance between the seismograph and the epicentre.
We have, \[L={{v}_{s}}{{t}_{s}}\] ……(i)
\[L={{v}_{p}}{{t}_{p}}\] ……(ii)
Where, \[{{t}_{s}}\] and \[{{t}_{p}}\] are the respective times taken by the S and P waves to reach the seismograph from the epicentre.
It is given that, \[{{v}_{p}}=8km/s\]
\[{{v}_{s}}=4km/s\]
From equations (i) and (ii), we have:
\[{{v}_{s}}{{t}_{s}}={{v}_{p}}{{t}_{p}}\]
\[4{{t}_{s}}=8{{t}_{p}}\]
\[{{t}_{s}}=2{{t}_{p}}\] ……(iii)
It is also given that:
\[{{t}_{s}}{{t}_{p}}=4\min =240s\]
\[\Rightarrow 2{{t}_{p}}{{t}_{p}}=240\]
\[\Rightarrow {{t}_{p}}=240\]
And \[{{t}_{s}}=2\times 240=480s\]
From equation (ii), we get:
\[L=8\times 240=1920\text{ }km\]
Clearly, the earthquake occurs at a distance of \[1920\text{ }km\] from the seismograph.
27. A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is \[\mathbf{40}\text{ }\mathbf{kHz}\]. During one fast swoop directly toward a flat wall surface, the bat is moving at \[\mathbf{0}.\mathbf{03}\] times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Ans: Bat’s ultrasonic beep frequency, \[\nu =40\text{ }kHz\]
Velocity of the bat, \[{{v}_{b}}=0.03v\]
Where, v is the velocity of sound in air
The apparent frequency of the sound striking the wall is given by:
\[\nu '=\left( \frac{v}{v{{v}_{b}}} \right)\nu \]
\[\Rightarrow \nu '=\left( \frac{v}{v0.03v} \right)\times 40\]
\[\Rightarrow \nu '=\frac{40}{0.97}kHz\]
This frequency is reflected by the stationary wall (vs =0) toward the bat.
The frequency (\[\nu ''\]) of the received sound is given by the relation:
\[\nu ''=\left( \frac{v+{{v}_{b}}}{v} \right)\nu '\]
\[\Rightarrow \nu ''=\left( \frac{v+0.03v}{v} \right)\times \frac{40}{0.97}\]
\[\Rightarrow \nu ''=\frac{1.03\times 40}{0.97}=42.47kHz\]
The frequency of the received sound is \[42.47kHz\].
NCERT Solutions for Class 11 Physics Chapter 14 Free PDF Download
The PDF of NCERT Class 11 Physics Chapter 14 solutions have all the essential topics, subtopics with comprehensive explanations which assist students of class 11 to grasp the concepts better.
Waves Class 11 NCERT solutions PDF provides students with many practise exercises as well as a detailed explanation of the questions and solutions provided in the chapter.
Class 11 Physics Ch 14 NCERT solutions lay the foundation for class 12 CBSE boards, which determines the basis of their admission in higher education institutes. The focus on concepts and theoretical aspects of this chapter is essential to build a proper foundation for the Class 12 CBSE board exam. It has been designed with the latest guidelines and syllabus of CBSE and NCERT.
Physics is a subject that has many numerical, formulas and graphical representations. Students often require easy references before the final exam. The PDF contains many forms of practise questions like shortquestions, long questions, MCQ’s, diagrammatic and illustrative representations.
Students can quickly recap and revise different formulas and numerical valuations before class 11 final exams without any rush or stress.
NCERT Solutions for Class 11 Physics Chapter 14
The Waves Chapter of Class 11 PDF covers many topics such as:
Waves.
Nature of waves in different mediums.
Graphical representation of waves and its properties.
Comparison between waves.
Speed of sound.
Harmonic motion.
Resonance among waves and calculation.
Displace and velocity of oscillations.
Waves are connected with rotating electrons, protons, neutrons and other fundamental particles. Waves are also associated with molecules and atoms. The chapter describes many features of a wave, and it’s characteristics like frequency, amplitude, wavelength, phase, resonance and wave displacement.
The waves are in various forms like sounds, lights, radio signals and microwaves. This chapter of Waves Class 11 NCERT solutions has many detailed sections as well, which are also crucial for engineering and college admission entrance tests. The chapter will further teach students about the idea of wave motion.
Vedantu’s experienced team of teachers has designed the syllabus according to the latest guidelines of NCERT and CBSE.
Benefits of NCERT Solutions for Class 11 Physics Chapter 14 PDF
There are many benefits to using PDF that can be downloaded from Vedantu’s website.
The benefits are:
The PDF will help students approach their revision for the final class 11 exam in a more efficient way.
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PDF of NCERT solutions for Class 11 Physics Chapter 14 is available for free from Vedantu’s website.
NCERT solutions For Class 11 Physics Chapter Waves is designed by an expert team of teachers, who have many years of experience in teaching science and engineering students around the country.
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Define Overtone
The term ‘overtone’ is applied to standing waves and is an essential part of this chapter. An overtone is a frequency that is more than the primary frequency of sound. Overtone is put in an application to higherfrequency standing waves.
It is a type of frequency that is usually provided by instruments. Overtone is also called second harmonic. One of the main advantages is that overtone can adapt any valuation of the fundamental frequency.
FAQs on NCERT Solutions for Class 11 Physics Chapter 14  Waves
1. What is the Fundamental Property of Waves?
Every wave has several different types of fundamental properties. The basic properties of waves can be described as reflection, refraction, diffraction and interference. The waves also consist of wavelength, frequency, amplitude and speed. Its amplitude, length and frequency generally explain a wave.
2. What are Some Characteristics of a Wave?
Some of the characteristics of a wave includes frequency, amplitude, wavelength, phase, resonance and wave displacement.
3. What are the Topics Covered Under Chapter 14, Waves?
The topics covered under this chapter includes:
Waves.
Nature of waves in different mediums.
Graphical representation of waves and its properties.
Comparison between waves.
Speed of sound.
Harmonic motion.
Resonance among waves and calculation.
Displace and velocity of oscillations.
4. What is a transverse wave motion?
A transverse wave motion is where the particles of the medium are seen to vibrate at right angles towards the direction that the wave propagates. Some examples of these are waves on a string, electromagnetic waves and surface water waves. It is to be noted that in the case of the electromagnetic wave, the disturbance caused is not the result of the vibration that the particles produce but is the result of the oscillation that occurs in the electric and magnetic fields taking place at right angles to the direction in which the wave travels.
5. What are the factors that influence the velocity of the sound?
Some factors affecting the velocity of the sound are as follows:
The velocity of the gas is always inversely proportional to the square root of the density of the respective gas.
The velocity of the sound is always directly proportional to the square of its absolute temperature in a gas
It is usually independent of the change in the pressure of the gas but on the condition that the temperature remains constant.
In the case of moist air, the velocity of the sound is greater as compared to the dry air.
6. What is the Doppler effect?
The Doppler effect propagates the philosophy that in the case of the relative motion between the source of the sound and the listener, the frequency that the source emits is different from the frequency of the sound that the listener encounters.
For more such insight into the chapter, it is important that the students refer to the NCERT solutions and practice the exercises that it provides. It will help the student to dwell deeper into the topics and get a better look at the subject matter.
7. Is Class 11 Physics Chapter 14 hard?
No, Class 11 Physics Chapter 14 is not hard. But to be able to do well in the exam, good preparation is required. To avail this, it is important that the student read all the chapters thoroughly and mark all the important areas so that it gets easier to refer to them in the time of need. They should also cultivate the habit of making their own notes so that they are able to retain all the concepts and ideas that the chapter presents. Lastly, it goes without saying that the student should get their hands on the NCERT solutions and practice all the given exercises. With the practice of these simple steps, they will definitely be able to score and secure good grades in the exam.
8. From where you can download the NCERT Solutions for Class 11 Physics Chapter 14?
The students can download the NCERT Solutions easily from the website of Vedantu for free. They can also be downloaded from the Vedantu app. These NCERT solutions will prove to be the best source of guidance for the students as they present the exercises with the aim of preparing the student for the exam. With the regular practice of the contents of the NCERT solutions, the student will develop the idea of how to answer any question in the examination.