Oscillations JEE Main PYQs (2020–2026)

Solved Oscillations PYQs

3 recent questions with answers. The full set of 122 is free to practise.

1. JEE Main 2026EasySpring-Mass Systems
   Q1

   A spring of force constant $15N/m$ is cut into two pieces. If the ratio of their length is $1:3$ , then the force constant of smaller piece is $\_\_\_\_N/m$ .

   • A20
   • B45
   • C60
   • D15

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   Correct answer: (C)

   The spring constant k of a spring is inversely proportional to the length L of the spring.
   
   $$k\propto \frac{1}{L}$$
   
   This means that the product of the spring constant and the length is a constant for a given spring:
   
   $$k\cdot L=\text{ Constant }$$
   
   The force constant of original spring is $k_o=15N/m$ .
   
   Let the original length be L .
   
   The spring is cut into two pieces with a length ratio of $1:3$ .
   
   $$x+3x=L\Rightarrow x=\frac{L}{4}$$
   
   Hence, the length of smaller spring is $L_1=x=\frac{L}{4}$ and the length of larger spring is $L_2=3x=\frac{3L}{4}$ .
   
   Let $k_1$ be the force constant of the smaller piece.
   
   Using the inverse proportionality relationship:
   
   $$k_o{L}_o=k_1{L}_1$$
   
   $$\Rightarrow 15\times L=k_1\times \left(\frac{L}{4}\right)\Rightarrow k_1=60N/m$$
   
   Therefore, the force constant of smaller piece is $60N/m$ .
   
   Hence, the correct option is (C).
2. JEE Main 2026MediumSimple Pendulum and Other Oscillating Systems
   Q2

   A simple pendulum of string length 30 cm performs 20 oscillations in 10 s . The length of the string required for the pendulum to perform 40 oscillations in the same time duration is
   
   $\_\_\_\_$ cm . [Assume that the mass of the pendulum remains same.]

   • A0.75
   • B7.5
   • C15
   • D120

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   Correct answer: (B)

   The time period (T) of a simple pendulum is the time taken to complete one full oscillation.
   
   For a simple pendulum of the length of the string $(L)$ and the acceleration due to gravity $(g)$ the formula for the time period is,
   
   $$T=2\pi \sqrt{\frac{L}{g}}$$
   
   So, the time period is directly proportional to the square root of the length $(T\propto \sqrt{L})$ .
   
   $$T^2\propto L\Rightarrow \propto T^2$$
   
   Case - 1
   
   The initial length of the string is $L_1=30cm$
   
   $N_1=20$ is the number of oscillations completed in total time $t_1=10s$
   
   So, the initial time period of oscillation is
   
   $$T_1=\frac{\text{ Total Time }}{\text{ Number of Oscillations }}=\frac{10}{20}=0.5s$$
   
   Case - 2
   
   $N_2=40$ is the number of oscillations completed in the same time $t_2=t_1=10s$
   
   So, the final time period of oscillation is,
   
   $$T_2=\frac{\text{ Total Time }}{\text{ Number of Oscillations }}=\frac{10}{40}=0.25s$$
   
   We need to find the required length $\left(L_2\right)$ corresponding to the final length of the string, Using the proportionality $L\propto T^2$
   
   $$\frac{L_2}{L_1}={\left(\frac{T_2}{T_1}\right)}^2$$
   
   $$\Rightarrow \frac{L_2}{30}={\left(\frac{0.25}{0.50}\right)}^2=\frac{1}{4}$$
   
   $$\Rightarrow \frac{L_2}{30}=\frac{1}{4}\Rightarrow L_2=\frac{30}{4}cm=7.5cm$$
   
   Therefore, the required length of the string is 7.5 cm .
   
   Hence, the correct option is (B).
3. JEE Main 2026MediumEnergy in Simple Harmonic Motion
   Q3

   The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s. The frequency of this simple harmonic oscillator is ______ Hz. [ take $\pi =\frac{22}{7}$ ]

   • A88
   • B14
   • C28
   • D176

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   Correct answer: (B)

   In simple harmonic motion, the kinetic energy oscillates with angular frequency $2\omega$ , where $\omega$ is the angular frequency of the oscillator.
   
   Given:
   
   $$2\omega =176\text{rad/s}$$
   
   So,
   
   $$\omega =\frac{176}{2}=88\text{rad/s}$$
   
   Now, frequency is related to angular frequency by:
   
   $$\omega =2\pi f$$
   
   Therefore,
   
   $$f=\frac{\omega }{2\pi }=\frac{88}{2\times \frac{22}{7}}$$
   
   $$f=\frac{88}{\frac{44}{7}}=88\times \frac{7}{44}$$
   
   $$f=2\times 7=14\text{Hz}$$
   
   So, the correct answer is:
   
   $$\boxed{14}$$
   
   Option B.

Topic-wise question count

Where JEE Main has asked from Oscillations, most frequent first.

• Kinematics and Equation of Motion of SHM32
• Simple Pendulum and Other Oscillating Systems31
• Spring-Mass Systems28
• Energy in Simple Harmonic Motion25
• Superposition of SHMs4
• Damped and Forced Oscillations2

Year-wise Oscillations questions in JEE Main

Number of questions from this chapter in each year's paper(s).