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📘 Oscillations — JEE Mains PYQs
Step-by-Step Solutions with Detailed Explanations
Master Oscillations for JEE Main with previous year questions on SHM, pendulums, and spring-mass systems. Each solution is explained in a simple, step-by-step manner to boost your score.
Oscillations — JEE Mains PYQs
JEE Main 2018–2025Q1. JEE Main 2025 (8 April Shift 2)
The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves,
\(y_1(x, t) = 4 \sin(kx - \omega t)\) and \(y_2(x, t) = 2 \sin(kx - \omega t + \frac{2\pi}{3})\), are :
(Take the angular frequency of initial waves same as \(\omega\))
(1) \([6, \frac{2\pi}{3}]\) (2) \([6, \frac{\pi}{3}]\)
(3) \([\sqrt{3}, \frac{\pi}{6}]\) (4) \([2\sqrt{3}, \frac{\pi}{6}]\)
(1) \([6, \frac{2\pi}{3}]\) (2) \([6, \frac{\pi}{3}]\)
(3) \([\sqrt{3}, \frac{\pi}{6}]\) (4) \([2\sqrt{3}, \frac{\pi}{6}]\)
✅ Answer & Explanation:
Correct Option: (4)
- Resultant Amplitude \(A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos\phi}\)
- \(A = \sqrt{4^2 + 2^2 + 2(4)(2) \cos(120^\circ)} = \sqrt{16 + 4 - 8} = \sqrt{12} = 2\sqrt{3}\).
- Resultant Phase \(\alpha\): \(\tan\alpha = \frac{A_2 \sin\phi}{A_1 + A_2 \cos\phi} = \frac{2 \sin(120^\circ)}{4 + 2 \cos(120^\circ)} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}\).
- Thus, \(\alpha = \pi/6\).
Q2. JEE Main 2025 (28 Jan Shift 2)
Assertion (A): Knowing initial position \(x_0\) and initial momentum \(p_0\) is enough to determine the position and momentum at any time \(t\) for a simple harmonic motion with a given angular frequency \(\omega\).
Reason (R): The amplitude and phase can be expressed in terms of \(x_0\) and \(p_0\).
Choose the correct answer from the options given below :
(1) (A) is false but (R) is true
(2) (A) is true but (R) is false
(3) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(4) Both (A) and (R) are true and (R) is the correct explanation of (A)
Reason (R): The amplitude and phase can be expressed in terms of \(x_0\) and \(p_0\).
Choose the correct answer from the options given below :
(1) (A) is false but (R) is true
(2) (A) is true but (R) is false
(3) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(4) Both (A) and (R) are true and (R) is the correct explanation of (A)
✅ Answer & Explanation:
Correct Option: (4)
The general equation of SHM is \(x(t) = A \sin(\omega t + \phi)\). The constants \(A\) and \(\phi\) are uniquely determined by the initial conditions \(x(0) = x_0\) and \(v(0) = p_0/m\). Thus, knowing the initial state completely defines the future motion.
The general equation of SHM is \(x(t) = A \sin(\omega t + \phi)\). The constants \(A\) and \(\phi\) are uniquely determined by the initial conditions \(x(0) = x_0\) and \(v(0) = p_0/m\). Thus, knowing the initial state completely defines the future motion.
Q3. JEE Main 2025 (24 Jan Shift 1)
A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s , then \(D/d\) is :
(1) \(16/5\) (2) 10
(3) \(15/4\) (4) 25
(1) \(16/5\) (2) 10
(3) \(15/4\) (4) 25
✅ Answer & Explanation:
Correct Option: (4)
- \(12.5 \, \text{s} = 6T + \frac{T}{4}\).
- Total distance \(D = 6 \times (4A) + A = 25A = 25 \, \text{cm}\).
- Total displacement \(d\): Particle moves from 0 to \(A\), so \(d = A = 1 \, \text{cm}\).
- Ratio \(\frac{D}{d} = 25\).
Q4. JEE Main 2024 (09 Apr Shift 1)
The position, velocity and acceleration of a particle executing SHM are found to have magnitudes of 4 m, \(2 \, \text{m s}^{-1}\) and \(16 \, \text{m s}^{-2}\) at a certain instant. The amplitude of the motion is \(\sqrt{x} \, \text{m}\) where \(x\) is ________.
✅ Answer & Explanation:
Correct Answer: 17
- Magnitude of acceleration \(|a| = \omega^2 x \implies 16 = \omega^2 (4) \implies \omega = 2\).
- Velocity \(v = \omega \sqrt{A^2 - x^2} \implies 2 = 2 \sqrt{A^2 - 16} \implies 1 = A^2 - 16 \implies A^2 = 17\).
- Thus, \(x = 17\).
Q5. JEE Main 2021 (27 Aug Shift 2)
Two simple harmonic motion are represented by the equations:
\(y_1 = 10 \sin(3\pi t + \frac{\pi}{3})\)
\(y_2 = 5 (\sin 3\pi t + \sqrt{3} \cos 3\pi t)\)
Ratio of amplitude of \(y_1\) to \(y_2\) is \(x : 1\). The value of \(x\) is :
\(y_1 = 10 \sin(3\pi t + \frac{\pi}{3})\)
\(y_2 = 5 (\sin 3\pi t + \sqrt{3} \cos 3\pi t)\)
Ratio of amplitude of \(y_1\) to \(y_2\) is \(x : 1\). The value of \(x\) is :
✅ Answer & Explanation:
Correct Answer: 1
- Amplitude of \(y_1\) is \(A_1 = 10\).
- For \(y_2\): \(A_2 = \sqrt{5^2 + (5\sqrt{3})^2} = \sqrt{25 + 75} = 10\).
- Ratio \(A_1 : A_2 = 10 : 10 = 1 : 1\). Thus, \(x = 1\).
Q6. JEE Main 2021 (20 Jul Shift 2)
A particle is making SHM along the X-axis. If at distances \(x_1\) and \(x_2\) from the mean position the velocities of the particle are \(v_1\) and \(v_2\), respectively. The time period of its oscillation is given as:
(1) \(T = 2\pi \sqrt{\frac{x_2^2 + x_1^2}{v_1^2 - v_2^2}}\) (2) \(T = 2\pi \sqrt{\frac{x_2^2 + x_1^2}{v_1^2 + v_2^2}}\)
(3) \(T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 + v_2^2}}\) (4) \(T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}\)
(1) \(T = 2\pi \sqrt{\frac{x_2^2 + x_1^2}{v_1^2 - v_2^2}}\) (2) \(T = 2\pi \sqrt{\frac{x_2^2 + x_1^2}{v_1^2 + v_2^2}}\)
(3) \(T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 + v_2^2}}\) (4) \(T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{v_1^2 - v_2^2}}\)
✅ Answer & Explanation:
Correct Option: (4)
- Using \(v^2 = \omega^2(A^2 - x^2)\):
- \(v_1^2 = \omega^2 A^2 - \omega^2 x_1^2\) and \(v_2^2 = \omega^2 A^2 - \omega^2 x_2^2\).
- Subtracting: \(v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2) \implies \omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}\).
- Since \(T = 2\pi/\omega\), Option (4) is correct.
Q7. JEE Main 2025 (4 April Shift 1)
Two simple pendulums having lengths \(l_1\) and \(l_2\) with negligible string mass undergo angular displacements \(\theta_1\) and \(\theta_2\), from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?
(1) \(\theta_1 l_2^2 = \theta_2 l_1^2\) (2) \(\theta_1 l_1 = \theta_2 l_2\)
(3) \(\theta_1 l_1^2 = \theta_2 l_2^2\) (4) \(\theta_1 l_2 = \theta_2 l_1\)
(1) \(\theta_1 l_2^2 = \theta_2 l_1^2\) (2) \(\theta_1 l_1 = \theta_2 l_2\)
(3) \(\theta_1 l_1^2 = \theta_2 l_2^2\) (4) \(\theta_1 l_2 = \theta_2 l_1\)
✅ Answer & Explanation:
Correct Option: (4)
- Angular acceleration \(\alpha = -\frac{g}{l} \theta\).
- Given \(\alpha_1 = \alpha_2 \implies \frac{g}{l_1} \theta_1 = \frac{g}{l_2} \theta_2 \implies \frac{\theta_1}{l_1} = \frac{\theta_2}{l_2}\).
- Rearranging gives \(\theta_1 l_2 = \theta_2 l_1\).
Q8. JEE Main 2025 (3 April Shift 1)
Two blocks of masses \(m\) and \(M, (M > m)\), are placed on a frictionless table as shown in figure. A massless spring with spring constant \(k\) is attached with the lower block. If the system is slightly displaced and released then (\(\mu\) = coefficient of friction between the two blocks)
(A) The time period of small oscillation is \(T = 2\pi \sqrt{\frac{m+M}{k}}\)
(B) The acceleration of the blocks is \(a = \frac{kx}{M+m}\)
(C) Magnitude of frictional force on upper block is \(\frac{m \mu |x|}{M+m}\)
(D) Max amplitude of upper block, if it does not slip, is \(\frac{\mu(M+m)g}{k}\)
Choose the correct answer from the options given below:
(1) A, B, D Only (2) B, C, D Only
(3) C, D, E Only (4) A, B, C Only
(B) The acceleration of the blocks is \(a = \frac{kx}{M+m}\)
(C) Magnitude of frictional force on upper block is \(\frac{m \mu |x|}{M+m}\)
(D) Max amplitude of upper block, if it does not slip, is \(\frac{\mu(M+m)g}{k}\)
Choose the correct answer from the options given below:
(1) A, B, D Only (2) B, C, D Only
(3) C, D, E Only (4) A, B, C Only
✅ Answer & Explanation:
Correct Option: (1)
- (A) True: Total mass is \((m+M)\).
- (B) True: \(F = kx = (M+m)a \implies a = kx/(M+m)\).
- (D) True: For no slipping, \(ma_{max} \le \mu mg \implies m(\frac{kA}{M+m}) \le \mu mg \implies A \le \frac{\mu(M+m)g}{k}\).
Q9. JEE Main 2025 (23 Jan Shift 1)
A light hollow cube of side length 10 cm and mass 10 g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is \(y\pi \times 10^{-2} \, \text{s}\), where the value of \(y\) is : [density of water \(= 10^3 \, \text{kg/m}^3\)]
(1) 6 (2) 2
(3) 4 (4) 1
(1) 6 (2) 2
(3) 4 (4) 1
✅ Answer & Explanation:
Correct Option: (2)
- For a floating body pushed down: \(T = 2\pi \sqrt{\frac{m}{\rho A g}}\).
- \(T = 2\pi \sqrt{\frac{0.01}{1000 \times 0.01 \times 10}} = 2\pi \sqrt{\frac{0.01}{100}} = 2\pi \times 0.01 = 2\pi \times 10^{-2} \, \text{s}\).
- Thus, \(y = 2\).
Q10. JEE Main 2025 (22 Jan Shift 2)
Assertion (A): A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
Reason (R): The mass of the pendulum remains unchanged at Earth and the other planet.
(1) (A) is false but (R) is true
(2) (A) is true but (R) is false
(3) Both (A) and (R) are true and (R) is correct explanation of (A)
(4) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
Reason (R): The mass of the pendulum remains unchanged at Earth and the other planet.
(1) (A) is false but (R) is true
(2) (A) is true but (R) is false
(3) Both (A) and (R) are true and (R) is correct explanation of (A)
(4) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
✅ Answer & Explanation:
Correct Option: (4)
- \(g_{planet} = G(4M)/(2R)^2 = g_{earth}\). Since \(g\) is same, \(T\) is same. Assertion is true.
- Mass of pendulum bob is constant. Reason is true.
- However, the time period is independent of the bob's mass; it's the equality of \(g\) that makes \(T\) same.
Q11. JEE Main 2024 (30 Jan Shift 2)
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be _________ s. [take \(g = \pi^2 \, \text{m s}^{-2}\)]
✅ Answer & Explanation:
Correct Answer: 8
- At distance \(2R\) from center, \(g' = g \frac{R^2}{(2R)^2} = \frac{g}{4} = \pi^2/4\).
- \(T = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{4}{\pi^2/4}} = 2\pi \times \frac{4}{\pi} = 8 \, \text{s}\).
Q12. JEE Main 2023 (31 Jan Shift 1)
In the figure given below, a block of mass \(M = 490 \, \text{g}\) placed on a frictionless table is connected with two springs having same spring constant (\(K = 2 \, \text{N m}^{-1}\)). If the block is horizontally displaced, then the number of complete oscillations it will make in \(14\pi\) seconds will be ______.
✅ Answer & Explanation:
Correct Answer: 20
- Springs are in parallel: \(K_{eff} = 2K = 4 \, \text{N/m}\).
- \(T = 2\pi \sqrt{\frac{M}{K_{eff}}} = 2\pi \sqrt{\frac{0.49}{4}} = 2\pi \times \frac{0.7}{2} = 0.7\pi \, \text{s}\).
- Total oscillations \(n = \frac{\text{Total time}}{T} = \frac{14\pi}{0.7\pi} = 20\).
Q13. JEE Main 2021 (24 Feb Shift 2)
In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. If each spring has spring constant k, the frequency of oscillation of given body is:
(1) \(\frac{1}{2\pi} \sqrt{\frac{2k}{Mg \sin\alpha}}\) (2) \(\frac{1}{2\pi} \sqrt{\frac{k}{Mg} \sin\alpha}\)
(3) \(\frac{1}{2\pi} \sqrt{\frac{k}{2M}}\) (4) \(\frac{1}{2\pi} \sqrt{\frac{2k}{M}}\)
(3) \(\frac{1}{2\pi} \sqrt{\frac{k}{2M}}\) (4) \(\frac{1}{2\pi} \sqrt{\frac{2k}{M}}\)
✅ Answer & Explanation:
Correct Option: (4)
The two springs act in parallel for restoring force. The effective spring constant is \(K_{eff} = 2k\). The frequency is independent of the inclination \(\alpha\): \(f = \frac{1}{2\pi} \sqrt{\frac{2k}{M}}\).
The two springs act in parallel for restoring force. The effective spring constant is \(K_{eff} = 2k\). The frequency is independent of the inclination \(\alpha\): \(f = \frac{1}{2\pi} \sqrt{\frac{2k}{M}}\).
Q14. JEE Main 2021 (17 Mar Shift 1)
Consider two identical springs each of spring constant k and negligible mass compared to mass M. Fig (a) shows one of them and Fig (b) shows their series combination. The ratio of time period of oscillation \(T_b / T_a = \sqrt{x}\), where value of x is _____.
✅ Answer & Explanation:
Correct Answer: 2
- \(T_a = 2\pi \sqrt{\frac{M}{k}}\).
- For series combination: \(k_{eq} = \frac{k}{2}\).
- \(T_b = 2\pi \sqrt{M/(\frac{k}{2})} = 2\pi \sqrt{2M/k}\).
- Ratio \(T_b/T_a = \sqrt{2}\). Thus, \(x = 2\).
Q15. JEE Main 2025 (24 Jan Shift 2)
A particle oscillates along the x-axis according to the law, \(x(t) = x_0 \sin^2(\frac{t}{2})\) where \(x_0 = 1 \, \text{m}\). The kinetic energy (K) of the particle as a function of x is correctly represented by the graph :
Choose correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (4)
- \(x = \sin^2(\frac{t}{2}) = \frac{1 - \cos t}{2}\).
- Velocity \(v = \frac{dx}{dt} = \frac{1}{2} \sin t\).
- Kinetic Energy \(K \propto v^2 \propto \sin^2 t = 1 - \cos^2 t = 1 - (1 - 2x)^2 = 4x(1-x)\).
- The graph of \(K\) vs \(x\) is a downward opening parabola passing through \(x=0\) and \(x=1\).
Q16. JEE Main 2022 (28 Jul Shift 2)
The potential energy of a particle of mass 4 kg in motion along the x-axis is given by \(U = 4(1 - \cos 4x) \, \text{J}\). The time period of the particle for small oscillation \((\sin\theta \simeq \theta)\) is \((\frac{\pi}{K}) \, \text{s}\). The value of K is _____.
✅ Answer & Explanation:
Correct Answer: 2
- For small \(x\), \(\cos 4x \approx 1 - \frac{(4x)^2}{2} = 1 - 8x^2\).
- \(U = 4(1 - (1 - 8x^2)) = 32x^2\).
- Force \(F = -dU/dx = -64x\). Thus, effective spring constant \(k = 64\).
- \(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{4}{64}} = 2\pi \times \frac{1}{4} = \frac{\pi}{2} \, \text{s}\).
- Thus, \(K = 2\).
Q17. JEE Main 2021 (27 Aug Shift 1)
The variation of displacement with time of a particle executing free SHM is shown in the figure. The potential energy \(U(x)\) versus time (t) plot of the particle is correctly shown in figure:
Choose correct graph (1, 2, 3, or 4).
✅ Answer & Explanation:
Correct Option: (4)
Potential energy \(U = \frac{1}{2} k x^2\). Since displacement is sinusoidal, \(U\) follows a \(\sin^2\) curve, which is always non-negative and reaches zero twice per oscillation period. Graph (4) correctly depicts this.
Potential energy \(U = \frac{1}{2} k x^2\). Since displacement is sinusoidal, \(U\) follows a \(\sin^2\) curve, which is always non-negative and reaches zero twice per oscillation period. Graph (4) correctly depicts this.
Q18. JEE Main 2021 (25 Feb Shift 2)
The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers \(30^\circ\) in 0.1 s. The perpendicular projection P from A on the diameter MN represents the SHM of P. The restoration force per unit mass when P touches M will be :
(1) 0.49 N (2) 9.87 N
(3) 50 N (4) 100 N
(3) 50 N (4) 100 N
✅ Answer & Explanation:
Correct Option: (2)
- Angular velocity \(\omega = \frac{\Delta \theta}{\Delta t} = \frac{\pi/6}{0.1} = \frac{5\pi}{3} \, \text{rad/s}\).
- Restoration force per unit mass (acceleration) \(|a| = \omega^2 x\).
- At point M (extreme position), \(x = R = 0.36 \, \text{m}\).
- \(|a| = (\frac{5\pi}{3})^2 \times 0.36 = \frac{25 \pi^2}{9} \times 0.36 = \pi^2 \approx 9.87 \, \text{m/s}^2\).
📘 Exam Preparation Tip for Oscillations
- SHM Equations: \(x = A \sin(\omega t + \phi)\), \(v = A\omega \cos(\omega t + \phi)\), \(a = -\omega^2 x\). Understand the relationships between displacement, velocity, and acceleration.
- Time Period: For spring-mass: \(T = 2\pi \sqrt{m/k}\). For simple pendulum: \(T = 2\pi \sqrt{L/g}\). For compound pendulum: \(T = 2\pi \sqrt{I/(mgd)}\).
- Energy in SHM: Total energy \(E = \frac{1}{2}kA^2\) is constant. \(K = \frac{1}{2}k(A^2 - x^2)\), \(U = \frac{1}{2}kx^2\). Energy is exchanged between kinetic and potential forms.
- Spring Combinations: In parallel: \(k_{eff} = k_1 + k_2\). In series: \(1/k_{eff} = 1/k_1 + 1/k_2\).
- Superposition of SHMs: When two SHMs of same frequency are superimposed, the resultant amplitude and phase are given by vector addition in phasor diagram.
- Variable Acceleration: For problems with \(a = -\omega^2 x\), recognize SHM. For small oscillations, use approximations like \(\sin\theta \approx \theta\).