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JEE Important Chapter - Oscillations and Waves

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Important Concept of Oscillations and Waves for JEE

Important Concept of Oscillations and Waves for JEE

The chapter Waves and Oscillations includes the concepts like periodic motion, oscillatory motion, simple harmonic motion and also gives the answer about “what is oscillating motion”. Oscillation is the process of occurring again and again; the variations in any quantity or measure around its equilibrium value through time.


Therefore, we can define the oscillatory motion as the motion in which a body moves to and fro or back and forth repeatedly about a fixed point i.e, mean position in a fixed interval of time. Some of the oscillation examples are the motion of a pendulum of a wall clock and a girl swinging on a swing.


In simple harmonic motion, it tells us about how the energy, time period and equation of motion of a particle can be given. Also, explain how the expression of the time period of a simple pendulum can be given using this concept. 


This chapter also explains about “what is a wave in physicsand various wave types like transverse and longitudinal waves. It also includes concepts like the velocity of longitudinal waves, sound waves and how the stationary waves are formed.


Now, let's move on to the important concepts and formulae related to JEE and JEE Main exams along with a few solved examples.


Important Topics for Oscillations and Waves

  • Simple harmonic motion

  • Total energy in SHM

  • Simple pendulum

  • Oscillation in a loaded spring

  • Velocity of sound in gases

  • Standing waves in string and normal mode

  • Beats

  • Doppler’s effect in sound


Oscillations and Waves Important Concept for JEE

Name of the Concept

Key Points of the Concepts

1. What is an oscillation in physics or what is the basic principle of oscillation? 

  • Oscillation is described as the process of recurring changes of any quantity or measure in time about its equilibrium value. Oscillation may also be described as a periodic fluctuation in a matter's value between two values or near its centre value.

2. Simple harmonic motion

  • Simple harmonic motion is a type of periodic motion in which a particle moves back and forth around a mean position under the influence of a restoring force that is always directed towards the mean position and whose magnitude is proportional to the particle's displacement from the mean position at any given instant.

  • The expression of displacement of a particle is given as,

$y=a\sin(\omega t+\phi_o)$

  • Expression of velocity of a particle is given as,

$V=\omega\sqrt{a^2-y^2}$

  • Expression of acceleration of a particle at extreme position is,

$A=-\omega^2 a$

3. Total energy in SHM

  • Total energy in SHM is expressed as the sum of kinetic energy and potential energy of a particle. It is given as,

$E=\dfrac{1}{2}m \omega^2 a^2$

4. Simple pendulum

  • Pendulum's time period is

$T=2\pi\sqrt{\dfrac{l}{g}}$

  • If the effective length ($l$) of the pendulum is much greater than radius of earth($R_e$) then time period of the simple pendulum is,

$T=2\pi\sqrt{\dfrac{R_e}{(1+R_e/l)g}}$

  • If the bob of the simple pendulum is suspended by a wire then effective length of the it will increase with rise in temperature so in this case the time period of simple pendulum is,

$T=2\pi\sqrt{\dfrac{l}{g}}~[1+\dfrac{1}{2}\dfrac{Mg}{\pi r^2 Y}]$

  • If a pendulum is mounted on a trolley which is moving down on an inclined plane of inclination $\theta$ then in this case the time period of a simple pendulum is,

$T=2\pi \sqrt{\dfrac{l}{g\cos\theta}}$

  • If a simple pendulum is in a carriage which is accelerating with acceleration $a$ then its time period is,

  1. In case of upward acceleration

$T=2\pi \sqrt{\dfrac{l}{g+a}}$

  1. In case of downward acceleration

$T=2\pi \sqrt{\dfrac{l}{g-a}}$

  1. Incase accelerating horizontally,

$T=2\pi \sqrt{\dfrac{l}{\sqrt{g^2+a^2}}}$

5. Oscillation in a loaded spring

  • Time period of a spring vibrating horizontally with mass $m$ attached to it is,

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

  • Time period of a spring vibrating vertically with mass $m$ attached to it is,

$T=2\pi\sqrt{\dfrac{l}{g}}$

  • When two spring having spring constant $k_1$ and $k_2$ are in parallel combination then the expression of time period is given as,

$T=2\pi\sqrt{\dfrac{m}{k_1+k_2}}$

  • When two spring having spring constant $k_1$ and $k_2$ are in series combination then the expression of time period is given as,

$T=2\pi\sqrt{\dfrac{m(k_1+k_2)}{k_1k_2}}$

6. The velocity of sound in gases

  • The speed of longitudinal waves in a fluid is given as,

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

Here, $B$ is bulk modulus and $\rho$ is density.

  • Newton's formula for sound velocity in gases is,

$v=\sqrt{\dfrac{B_i}{\rho}}=\sqrt{\dfrac{P}{\rho}}$

Here, $B_i$ is isothermal elasticity and $P$ is the pressure of a gas.

  • After Laplace’s correction it is given as,

$v=\sqrt{\dfrac{B_a}{\rho}}=\sqrt{\dfrac{\gamma P}{\rho}}$

Here, $B_a$ is adiabatic elasticity and $\gamma$ is atomicity of a gas. 

7. What is a wave in physics?

  • Wave is known as the regular and ordered transmission of disturbances from one location to another. Surface waves on the water are the most well-known example, but sound, light, and the motion of subatomic particles also display wavelike qualities.

8. Standing waves in string and normal mode

  • When two waves of opposite phases interfere then standing waves are formed i.e, a wave in which the waveform doesn’t move.

  • For a normal mode of vibration in a string,

  1. Wavelength is given as,

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

Here, $n$ corresponds to normal modes of vibrations in a string and $L$ is the length of the string.

  1. Frequency is given as,

$\nu=\dfrac{n}{2L}\sqrt{\dfrac{T}{m}}$

9. Beats

  • Beats occur when two sound waves with almost similar frequencies and amplitudes travelling in the same direction superimpose on each other.

  • Mathematically, it is given as;

$\nu_2-\nu_1=m$

Here, $\nu_2-\nu_1$ is the difference in frequencies and $m$ is the beats per second.

10. Doppler’s effect in sound

  • This effect tells us that whenever there is a relative motion between a source of sound, the listener and the intervening medium, the apparent frequency of the sound heard by the listener differs from the frequency produced by the source.

  • It’s expression is given as,

$\nu'=\dfrac{v\pm v_L}{v\pm v_S}\nu$

Here, $v_L$ is the velocity of the listener, $v_S$ is the velocity of source and $\nu$ is the original frequency of source.

  • All velocities along the direction source to the listener are treated as positive and vice versa, according to the sign convention.



List of Important formulas for Oscillations and Waves

S.No.

Name of the Concept

Formula


Simple harmonic motion

  • Expression of displacement of a particle,

$y=a\sin(\omega t+\phi_o)$

  • Expression of velocity of a particle,

$V=\omega\sqrt{a^2-y^2}$

  • Expression of acceleration of a particle,

$A=-\omega^2 a$


Total energy in SHM

  • $E=\dfrac{1}{2}m \omega^2 a^2$


Simple pendulum

  • Time period of simple pendulum,

$T=2\pi\sqrt{\dfrac{l}{g}}$

  • When $l > > R_e$

$T=2\pi\sqrt{\dfrac{R_e}{(1+R_e/l)g}}$


Oscillation in a loaded spring

  • Time period of a spring vibrating horizontally,

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

  • Time period of a spring vibrating vertically,

$T=2\pi\sqrt{\dfrac{l}{g}}$

  • When two spring having spring constant $k_1$ and $k_2$ are in parallel combination,

$T=2\pi\sqrt{\dfrac{m}{k_1+k_2}}$

  • When two spring having spring constant $k_1$ and $k_2$ are in series combination,

$T=2\pi\sqrt{\dfrac{m(k_1+k_2)}{k_1k_2}}$


Velocity of sound in gases

  • Newton's formula for sound velocity in gases,

$v=\sqrt{\dfrac{B_i}{\rho}}=\sqrt{\dfrac{P}{\rho}}$

  • After Laplace’s correction,

$v=\sqrt{\dfrac{B_a}{\rho}}=\sqrt{\dfrac{\gamma P}{\rho}}$



Speed of transverse waves in string 

  • $v=\sqrt{\dfrac{T}{m}}$

Here, $m$= mass per unit lenght


Standing waves in string and normal mode

  • Wavelength for normal modes of oscillations in string,

 $\lambda=\dfrac{2L}{n}$

  • Frequency expression,

$\nu=\dfrac{n}{2L}\sqrt{\dfrac{T}{m}}$


Beats

  • $\nu_2-\nu_1=m$


Doppler’s effect in sound

  • $\nu'=\dfrac{v\pm v_L}{v\pm v_S}\nu$


Solved Examples 

  1. A particle with mass $m$ oscillates about the origin on the x-axis. Its potential energy is calculated as $U(x)=k\,x^4$, where $k$ is a positive constant. If the amplitude of the oscillation is $a$, then calculate its time period of the oscillation.

Sol:

Given that,

Potential energy, $U(x)=k\,x^4$

In order to solve this first we have to use the relation between force and potential energy of a particle and after converting it to standard simple harmonic motion we can obtain the time period of the oscillation.

The relation of force between potential energy is,

$F=-\dfrac{\partial U}{\partial x}$

After partially derivative of $U$, we get;

$F=-\dfrac{\partial (k\,x^4)}{\partial x}$

$F=-4kx^3$ ……….(1)

Here we can write force as,

$F=ma=m \dfrac{\text{d}^2x}{\text{d}t^2}$.......(2)

After putting the value of $F$ from equation (2) into equation (1), we get;

$m \dfrac{\text{d}^2x}{\text{d}t^2} =-4kx^3$ …………(3)

The standard equation of SHM is,

$\dfrac{\text{d}^2x}{\text{d}t^2}+\omega^2x=0$ 

$\dfrac{\text{d}^2x}{\text{d}t^2}=-\omega^2x$.......(4)

Here, $x=a\sin\omega t$

Now putting the value of $\dfrac{\text{d}^2x}{\text{d}t^2}$ from eq.(4) to eq.(3), we get;

$-m \omega^2x=-4kx^3$

Om simplification,

$m\omega^2 = 4kx^2$

$m(\dfrac{2\pi}{T})^2= 4kx^2$.........(As, $\omega=\dfrac{2\pi}{T}$)

$\dfrac{2\pi}{T}=\sqrt{\dfrac{4kx^2}{m}}$

$T=2\pi\sqrt{\dfrac{m}{4kx^2}}$

Hence, the time period of the oscillation is $2\pi\sqrt{\dfrac{m}{4kx^2}}$.


Key Point: The relation between force and potential energy of the particle and the standard equation of simple harmonic motion is crucial for solving such problems.


  1. A sonometer wire of length $1.5\,m$ is made of steel. The tension in it produces an elastic strain $1\%$. What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times 10^3\,kg/m^3$ and $2.2\times 10^{11}\,N/m^2$ respectively ?

Sol: 

Given that,

Length of the wire, $l=1.5\,m$

Elastic strain, $\dfrac{\Delta L}{L}= 1\%$

Density, $\rho= 7.7 \times 10^3\,kg/m^3$

Young’s modulus of elasticity, $Y=2.2\times 10^{11}\,N/m^2$

In order to solve this problem first we have to use the relation of Young’s modulus of elasticity to find the value of force and after that we can put it in the expression of fundamental frequency.

The expression of Young’s modulus is given as,

$Y=\dfrac{F}{A (\Delta L/L)}$

After rearranging we can write;

$F=YA(\dfrac{\Delta L}{L})$

Putting the values of known quantities, we get;

$F=2.2\times 10^{11}\times A \times \dfrac{1}{100}=2.2A \times 10^9$

The expression of fundamental frequency is,

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

$\nu=\dfrac{1}{2l}\sqrt{\dfrac{F}{A \times 1 \times \rho}}$

After putting the values of $F$ and $\rho$, we get;

$\nu=\dfrac{1}{2 \times 1.5}\sqrt{\dfrac{2.2A \times 10^9}{A \times 1 \times \rho}}$

Upon simplification, we get;

$\nu=\dfrac{10^3}{3}\sqrt{\dfrac{2}{7}}$

$\nu=178.2\,Hz$

Hence, the fundamental frequency of steel is $178.2\,Hz$.


Key Point: The expression of Young’s modulus and fundamental frequency is essential to solve this problem.


Previous Year Questions from JEE Paper

  1. A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to- (JEE Main 2019)

a. 16.6 cm

b. 10.0 cm

c. 20.0 cm

d. 33.3 cm

Sol:

Given that,

Length of the string, $L=1\,m$

Mass of the string, $M=5\times 10^{-3}\,kg$

Hence, mass per unit length, $m=\dfrac{M}{L}=\dfrac{5\times 10^{-3}}{1}\,kg/m$

Tension in the string, $T=8.0\,N$

Frequency, $\nu=100\,Hz$

In order to solve this problem, we first use the relation of speed of the transverse wave in the string and after that using the relation between the frequency and wavelength we are able to find the distance between the successive nodes.

Now the expression for speed of transverse waves in string is given as,

$v=\sqrt{\dfrac{T}{m}}$...........(i)

And the wavelength-frequency relationship is given as,

$\nu=\dfrac{v}{\lambda}$

$v = \nu \lambda$............(ii)

Now, putting the value of $v$ using eq.(ii) in eq.(ii),

$\nu \lambda = \sqrt{\dfrac{T}{m}}$

$\lambda =\dfrac{1}{\nu} \sqrt{\dfrac{T}{m}}$

After putting the values of all quantities, we get;

$\lambda =\dfrac{1}{100} \sqrt{\dfrac{8.0}{5\times 10^{-3}}}$

$\lambda = \dfrac{40}{100} = 0.40\,m=40\,cm$

The distance between the two successive nodes is $\dfrac{\lambda}{2}$, therefore it is $\dfrac{40}{2}\,cm=20\,cm$. 

Hence, option c is correct.


Key Point: The relation of speed of transverse wave and frequency of the wave is important to solve such a problem.


  1. A tuning fork vibrates with frequency $256\,Hz$ and gives one best per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound of air is $340\,m/s$) (JEE Main 2018)

a.  190 cm                     

b.  180 cm

c.  220 cm                     

d.  200 cm

Sol:

Given that,

Frequency of oscillation of tuning fork =$256\,Hz$

Therefore, the frequency of oscillation of open pipe = $(256\pm 1)\,Hz$

To solve this problem we have to use the relation of the third normal mode of vibration of an open pipe to find the length of the open pipe.

The relation of third normal mode of vibration is given as,

$\nu=\dfrac{3v}{2l}$

After rearrangement, we can write;

$l=\dfrac{3v}{2\nu}$

Noe putting the values of known quantities, we get;

$l=\dfrac{3\times 340}{2\times 255}$ 

$l= 2.00\,m=200\,cm$

Therefore, the length of the open pipe is $200\,m$. Hence option d is correct.


Key Point: The relation of normal modes of vibration in open pipe and beats per second is important to solve this problem.


Practice Questions

1. Two wires are fixed on a sonometer. Their tensions are in the ratio $8:1$, their lengths are in the ratio $36:35$, the diameters are in the ratio $4:1$ and densities are in the ratio $1:2$. If the note of the higher pitch has a frequency $360\,s^{-1}$, then calculate the frequency of beats produced?

(Ans: $10\,s^{-1}$)

 

2. A particle moves with simple harmonic motion in a straight line. In the first $\tau\,s$, after starting from rest it travels a distance $a$, and in next $\tau\,s$ it travels distance $2a$ in the same direction, then calculates the time period of oscillation.

(Ans: $6\tau$)


Conclusion

In this article, we have discussed the concepts like simple harmonic motion, simple pendulum, standing waves and velocity of sound in gases medium etc. We have also discussed “what is the basic principle of oscillation” and also “what is wave”. Mention various important formulas and concepts of the chapter.

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JEE Main 2022 Book Solutions and PDF Download

JEE Main 2022 Book Solutions and PDF Download

View all JEE Main Important Books
In order to prepare for JEE Main 2022, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2022 exam so that they can grab the top rank in the all India entrance exam.
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books
Maths
NCERT Book for Class 12 Maths
books
Physics
NCERT Book for Class 12 Physics
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JEE Main Mock Tests

JEE Main Mock Tests

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JEE Main 2022 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2022 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.
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JEE MAIN MOCK TEST - 1
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 3
3 hr  • 75 questions • OBJECTIVE
JEE MAIN MOCK TEST - 2
3 hr  • 75 questions • OBJECTIVE
Toppers

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JEE Main 2022 Cut-Off

JEE Main 2022 Cut-Off

JEE Main Cut Off
NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
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JEE Main 2022 Results

JEE Main 2022 Results

JEE Main 2022 June and July Session Result - NTA has announced JEE Main result on their website. To download the Scorecard for JEE Main 2022 June and July Session, visit the official website of JEE Main NTA.
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Rank List
Counselling
Cutoff
JEE Main 2022 state rank lists will be released by the state counselling committees for admissions to the 85% state quota and to all seats in NITs and CFTIs colleges. JEE Main 2022 state rank lists are based on the marks obtained in entrance exams. Candidates can check the JEE Main 2022 state rank list on the official website or on our site.
The NTA will conduct JEE Main 2022 counselling at https://josaa.nic.in/. There will be two rounds of counselling for admission under All India Quota (AIQ), deemed and central universities, NITs and CFTIs. A mop-up round of JEE Main counselling will be conducted excluding 15% AIQ seats, while the dates of JEE Main 2022 June and July session counselling for 85% state quota seats will be announced by the respective state authorities.
NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.
Want to know which Engineering colleges in India accept the JEE Main 2022 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.
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Counselling

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FAQs on JEE Important Chapter - Oscillations and Waves

FAQ

1. What is the weightage of the Waves and Oscillations in JEE?

This chapter almost includes 2-3 questions every year and contributes nearly up to 3-4% of the weightage. Therefore, it is important to cover this chapter while preparing for JEE.

2. Is it necessary to study oscillations and waves for the JEE main?

Oscillations and Waves for JEE Main is a critical topic that must be thoroughly studied. Every year, 2-3 questions from this topic appear in the test. These issues are relevant to our daily lives, but we are too distracted to recognise or comprehend them. Therefore, it is necessary to study it.

3. Is it really good to prepare JEE test questions from previous years?

Practicing past year questions from any chapter helps us learn which topics are significant for the exam. It also offers us a sense of the difficulty level of the subjects requested from the chapter. As a result, it is important to practice the previous year's questions for a better comprehension of the chapter's main issues, as well as to try to build your own oscillations and waves notes for the last revision before the test.