Oscillations and Waves Chapter - Physics JEE Main

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 physics” and 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.

JEE Main Physics Chapters 2024

Important Topics of Oscillations and Waves Chapter

• 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 Main

List of Important formulas for Oscillations and Waves Chapter

JEE Main Oscillations and Waves 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,

$F=-{\displaystyle \frac{\partial U}{\partial x}}$

After partially derivative of $U$, we get;

$F=-{\displaystyle \frac{\partial (k{x}^4)}{\partial x}}$

$F=-4k{x}^3$ ……….(1)

Here we can write force as,

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

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

$m{\displaystyle \frac{{\text{d}}^{2}x}{\text{d}{t}^2}}=-4k{x}^3$ …………(3)

The standard equation of SHM is,

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

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

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

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

$-m{\omega }^{2}x=-4k{x}^3$

Om simplification,

$m{\omega }^2=4k{x}^2$

$m({\displaystyle \frac{2\pi }{T}}{)}^2=4k{x}^2$.........(As, $\omega ={\displaystyle \frac{2\pi }{T}}$)

${\displaystyle \frac{2\pi }{T}}=\sqrt{{\displaystyle \frac{4k{x}^2}{m}}}$

$T=2\pi \sqrt{{\displaystyle \frac{m}{4k{x}^2}}}$

Hence, the time period of the oscillation is $2\pi \sqrt{{\displaystyle \frac{m}{4k{x}^2}}}$.

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.

2. A sonometer wire of length $1.5m$ is made of steel. The tension in it produces an elastic strain $1\mathrm{\%}$. 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, $l=1.5m$

Elastic strain, ${\displaystyle \frac{\mathrm{\Delta }L}{L}}=1\mathrm{\%}$

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,

$Y={\displaystyle \frac{F}{A(\mathrm{\Delta }L/L)}}$

After rearranging we can write;

$F=YA({\displaystyle \frac{\mathrm{\Delta }L}{L}})$

Putting the values of known quantities, we get;

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

The expression of fundamental frequency is,

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

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

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

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

Upon simplification, we get;

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

$\nu =178.2Hz$

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

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, $L=1m$

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

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

Tension in the string, $T=8.0N$

Frequency, $\nu =100Hz$

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,

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

And the wavelength-frequency relationship is given as,

$\nu ={\displaystyle \frac{v}{\lambda }}$

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

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

$\nu \lambda =\sqrt{{\displaystyle \frac{T}{m}}}$

$\lambda ={\displaystyle \frac{1}{\nu }}\sqrt{{\displaystyle \frac{T}{m}}}$

After putting the values of all quantities, we get;

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

$\lambda ={\displaystyle \frac{40}{100}}=0.40m=40cm$

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

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.

2. A tuning fork vibrates with frequency $256Hz$ 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 $340m/s$) (JEE Main 2018)

a.  190 cm                     

b.  180 cm

c.  220 cm                     

d.  200 cm

Sol:

Given that,

Frequency of oscillation of tuning fork =$256Hz$

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,

$\nu ={\displaystyle \frac{3v}{2l}}$

After rearrangement, we can write;

$l={\displaystyle \frac{3v}{2\nu }}$

Noe putting the values of known quantities, we get;

$l={\displaystyle \frac{3\times 340}{2\times 255}}$ 

$l=2.00m=200cm$

Therefore, the length of the open pipe is $200m$. 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 $360s^{-1}$, then calculate the frequency of beats produced?

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

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: $6\tau$)

JEE Main Physics Oscillations and Waves Study Materials

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Conclusion

In this article, we're diving into the fascinating world of Oscillations and Waves in physics, specifically tailored for JEE Main students. We'll unravel the secrets of how things move back and forth and create waves. You'll learn about oscillations, wave properties, and practical problem-solving. Whether you're a beginner or need a quick refresher, our free PDFs provide you with clear explanations and questions to sharpen your skills. Master this chapter, and you'll breeze through your exams with confidence!