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Oscillations — JEE Previous Year Questions

Every Oscillations question asked in JEE Main and JEE Advanced across the last 186 papers — 141 questions, each with its correct answer. Free to read, no account needed.

Questions

141

Papers it appeared in

117/186

Appearance rate

63%

All 141 Oscillations questions

Most recent papers first.

Q1·PhysicsNumericalJEE Advanced 2026
A tank contains two immiscible liquids of densities 6ρ6\rho6ρ and 2ρ2\rho2ρ. The higher density liquid is filled up to a height L/2L/2L/2 from the bottom. A thin rod of density ρ\rhoρ and length LLL is fully immersed and hinged at the bottom so that it can oscillate freely, as shown in the figure. If the rod is slightly disturbed from its equilibrium, the time period of small oscillations is 2πnLg\frac{2\pi}{n}\sqrt{\frac{L}{g}}n2π​gL​​, where ggg is the acceleration due to gravity. The value of nnn is:

Correct answer: 1.73

Step-by-step solution →
Q2·PhysicsSingle correctJEE Main 2026
The frequency of oscillation of a mass mmm suspended by a spring is v1v_1v1​. If the length of the spring is cut to half, the same mass oscillates with frequency v2v_2v2​. The value of v2/v1v_2/v_1v2​/v1​ is ________.
  1. (A)1
  2. (B)2
  3. (C)2\sqrt{2}2​
  4. (D)3\sqrt{3}3​

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correctJEE Main 2026
A spring stretches by 2 mm when it is loaded with a mass of 200 g. From equilibrium position the mass is further pulled down by 2 mm and released. The frequency associated with the system and maximum energy in the spring are __________ Hz and __________ J, respectively. (Take g =10= 10=10 m/s2^{2}2)
  1. (A)550π\frac{5\sqrt{50}}{\pi}π550​​ and 8×10−38 \times 10^{-3}8×10−3
  2. (B)550π\frac{5\sqrt{50}}{\pi}π550​​ and 8
  3. (C)105010\sqrt{50}1050​ and 2×10−32 \times 10^{-3}2×10−3
  4. (D)550π\frac{5\sqrt{50}}{\pi}π550​​ and 16×10−316 \times 10^{-3}16×10−3

Correct answer: (A)

Step-by-step solution →
Q4·PhysicsSingle correctJEE Main 2026
A particle is executing simple harmonic motion. Its amplitude is A and time period is 5 sec. The time required by it to move from x = A to x=A2x = \frac{A}{\sqrt{2}}x=2​A​ is ________ sec.
  1. (A)1/4
  2. (B)5/4
  3. (C)5/8
  4. (D)3/8

Correct answer: (C)

Step-by-step solution →
Q5·PhysicsSingle correctJEE Main 2026
Match List-I with List-II. Choose the correct answer from the options given below :
List-IList-II
A.sin² ωtI.Periodic with time period T=πωT = \dfrac{\pi}{\omega}T=ωπ​ but not simple harmonic motion (SHM)
B.sin³(2ωt)II.Periodic with time period T=2πωT = \dfrac{2\pi}{\omega}T=ω2π​ but Not SHM
C.sin(ωt) + cos(πωt)III.Periodic with time period T=πωT = \dfrac{\pi}{\omega}T=ωπ​ and SHM
D.cos ωt + cos 2ωtIV.Non-periodic
  1. (A)A-III, B-I, C-IV, D-II
  2. (B)A-II, B-I, C-III, D-IV
  3. (C)A-III, B-II, C-IV, D-I
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (A)

Step-by-step solution →
Q6·PhysicsNumericalJEE Main 2026
The velocity of a particle executing simple harmonic motion along xxx-axis is described as v2=50−x2v^2 = 50 - x^2v2=50−x2, where xxx represents displacement. If the time period of motion is x7\frac{x}{7}7x​ s, the value of xxx is ______.

Correct answer: 44

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
A uniform disc of radius RRR and mass MMM is free to oscillate about the axis AAA as shown in the figure. For small oscillations the time period is ______. (ggg is acceleration due to gravity)
  1. (A)2π5R4g2\pi\sqrt{\frac{5R}{4g}}2π4g5R​​
  2. (B)2π2R3g2\pi\sqrt{\frac{2R}{3g}}2π3g2R​​
  3. (C)2π3R2g2\pi\sqrt{\frac{3R}{2g}}2π2g3R​​
  4. (D)2π3Rg2\pi\sqrt{\frac{3R}{g}}2πg3R​​

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
The equation of motion of a particle is given by x=asin⁡(50t+π3)x = a \sin\left(50t + \frac{\pi}{3}\right)x=asin(50t+3π​) cm. The particle will come to rest at time t1t_1t1​ and it will have zero acceleration at time t2t_2t2​. The t1t_1t1​ and t2t_2t2​ respectively are ________.
  1. (A)π300 s\frac{\pi}{300}\,s300π​s, π75 s\frac{\pi}{75}\,s75π​s
  2. (B)π75 s\frac{\pi}{75}\,s75π​s, π300 s\frac{\pi}{300}\,s300π​s
  3. (C)π300 s\frac{\pi}{300}\,s300π​s, π25 s\frac{\pi}{25}\,s25π​s
  4. (D)π50 s\frac{\pi}{50}\,s50π​s, π100 s\frac{\pi}{100}\,s100π​s

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsNumericalJEE Main 2026
The displacement of a particle, executing simple harmonic motion with time period TTT, is expressed as x(t)x(t)x(t) = A sin ωt, where AAA is the amplitude. The maximum value of potential energy of this oscillator is found at ttt = TTT/2β. The value of β is __________.

Correct answer: 2

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses 1 kg and 0.2 kg with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is :
  1. (A)30
  2. (B)27
  3. (C)20
  4. (D)5

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correctJEE Main 2026
A cylindrical block of mass M and area of cross section A is floating in a liquid of density ρ and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is________.
  1. (A)2πMρAg2\pi\sqrt{\dfrac{M}{\rho Ag}}2πρAgM​​
  2. (B)π2MρAg\pi\sqrt{\dfrac{2M}{\rho Ag}}πρAg2M​​
  3. (C)πρAMg\pi\sqrt{\dfrac{\rho A}{Mg}}πMgρA​​
  4. (D)2πρAMg2\pi\sqrt{\dfrac{\rho A}{Mg}}2πMgρA​​

Correct answer: (A)

Step-by-step solution →
Q12·PhysicsSingle correctJEE Main 2026
A spring of force constant 15 N/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
  1. (A)15
  2. (B)20
  3. (C)60
  4. (D)45

Correct answer: (C)

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2026
A simple pendulum of string length 30 cm performs 20 oscillations in 10s. 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.]
  1. (A)120
  2. (B)0.75
  3. (C)7.5
  4. (D)15

Correct answer: (C)

Step-by-step solution →
Q14·PhysicsSingle correctJEE Main 2026
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 π=227]\left[\text{take } \pi = \frac{22}{7}\right][take π=722​]
  1. (A)14
  2. (B)88
  3. (C)28
  4. (D)176

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correctJEE Advanced 2025
As shown in the figures, a uniform rod OO′ of length ℓ\ellℓ is hinged at the point O and held in place vertically between two walls using two massless springs of same spring constant. The springs are connected at the midpoint and at the top-end (O′) of the rod, as shown in Fig.1 and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is f1f_1f1​. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2 and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is f2f_2f2​. Ignoring gravity and assuming motion only in the plane of the diagram, the value of f1f2\frac{f_1}{f_2}f2​f1​​ is:
  1. (A)2
  2. (B)2\sqrt{2}2​
  3. (C)52\sqrt{\frac{5}{2}}25​​
  4. (D)25\sqrt{\frac{2}{5}}52​​

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correctJEE Advanced 2025
The center of a disk of radius r and mass m is attached to a spring of spring constant k, inside a ring of radius R > r as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as T=2πωT = \frac{2\pi}{\omega}T=ω2π​. The correct expression for ω\omegaω is ( g is the acceleration due to gravity):
  1. (A)23(gR−r+km)\sqrt{\frac{2}{3}\left(\frac{g}{R-r}+\frac{k}{m}\right)}32​(R−rg​+mk​)​
  2. (B)2g3(R−r)+km\sqrt{\frac{2g}{3(R-r)}+\frac{k}{m}}3(R−r)2g​+mk​​
  3. (C)16(gR−r+km)\sqrt{\frac{1}{6}\left(\frac{g}{R-r}+\frac{k}{m}\right)}61​(R−rg​+mk​)​
  4. (D)14(gR−r+km)\sqrt{\frac{1}{4}\left(\frac{g}{R-r}+\frac{k}{m}\right)}41​(R−rg​+mk​)​

Correct answer: (A)

Step-by-step solution →
Q17·PhysicsSingle correctJEE Main 2025
A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring’s natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance xxx m (x<2x<2x<2) from the wall, the speed of the block will be:
  1. (A)10[1−(2−x)]3/2 m/s10[1-(2-x)]^{3/2}\ m/s10[1−(2−x)]3/2 m/s
  2. (B)10[1−(2−x)2]1/2 m/s10[1-(2-x)^2]^{1/2}\ m/s10[1−(2−x)2]1/2 m/s
  3. (C)10[1−(2−x)2] m/s10[1-(2-x)^2]\ m/s10[1−(2−x)2] m/s
  4. (D)10[1−(2−x)2]2 m/s10[1-(2-x)^2]^{2}\ m/s10[1−(2−x)2]2 m/s

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correctJEE Main 2025
The equation of a wave travelling on a string is y=sin⁡[20πx+10πt]y=\sin[20\pi x+10\pi t]y=sin[20πx+10πt], where xxx and ttt are distance and time in SI units. The minimum distance between two points having the same oscillating speed is:
  1. (A)5.0 cm
  2. (B)20 cm
  3. (C)10 cm
  4. (D)2.5 cm

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsSingle correctJEE Main 2025
Two simple pendulums having lengths l1l_1l1​ and l2l_2l2​ with negligible string mass undergo angular displacements θ1\theta_1θ1​ and θ2\theta_2θ2​ from their mean positions, respectively. If the angular accelerations of both pendulums are same, then which expression is correct?
  1. (A)θ1l12=θ2l22\theta_1 l_1^2=\theta_2 l_2^2θ1​l12​=θ2​l22​
  2. (B)θ1l1=θ2l2\theta_1 l_1=\theta_2 l_2θ1​l1​=θ2​l2​
  3. (C)θ1l1=θ2l22\theta_1 l_1=\theta_2 l_2^2θ1​l1​=θ2​l22​
  4. (D)θ1l2=θ2l1\theta_1 l_2=\theta_2 l_1θ1​l2​=θ2​l1​

Correct answer: (D)

Step-by-step solution →
Q20·PhysicsSingle correctJEE Main 2025
Two blocks of masses mmm and MMM (M>m)(M>m)(M>m) are placed on a frictionless table, the upper block (mass mmm) resting on the lower block (mass MMM); a massless spring of spring constant kkk is attached to the lower block, and μ\muμ is the coefficient of friction between the two blocks. If the system is slightly displaced and released then: (A) The time period of small oscillation of the two blocks is T=2πm+MkT=2\pi\sqrt{\dfrac{m+M}{k}}T=2πkm+M​​. (B) The acceleration of the blocks is a=kxM+ma=\dfrac{kx}{M+m}a=M+mkx​ (x=x=x= displacement from the mean position). (C) The magnitude of the frictional force on the upper block is mμ∣x∣M+m\dfrac{m\mu|x|}{M+m}M+mmμ∣x∣​. (D) The maximum amplitude of the upper block, if it does not slip, is μ(M+m)gk\dfrac{\mu(M+m)g}{k}kμ(M+m)g​. (E) Maximum frictional force can be μ(M+m)g\mu(M+m)gμ(M+m)g. Choose the correct answer from the options given below:
  1. (A)(A), (B), (D) Only
  2. (B)(B), (C), (D) Only
  3. (C)(C), (D), (E) Only
  4. (D)(A), (B), (C) Only

Correct answer: (A)

Step-by-step solution →
Q21·PhysicsSingle correctJEE Main 2025
A particle is subjected to two simple harmonic motions as: x1=7sin⁡5tx_1=\sqrt{7}\sin 5tx1​=7​sin5t cm and x2=27sin⁡(5t+π3)x_2=2\sqrt{7}\sin\left(5t+\dfrac{\pi}{3}\right)x2​=27​sin(5t+3π​) cm, where xxx is displacement and ttt is time in seconds. The maximum acceleration of the particle is x×10−2x\times10^{-2}x×10−2 ms−2^{-2}−2. The value of xxx is:
  1. (A)175
  2. (B)25725\sqrt{7}257​
  3. (C)575\sqrt{7}57​
  4. (D)125

Correct answer: (A)

Step-by-step solution →
Q22·PhysicsSingle correctJEE Main 2025
Two bodies A and B of equal mass are suspended from two massless springs of spring constant k1k_1k1​ and k2k_2k2​, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is
  1. (A)k1k2\sqrt{\dfrac{k_1}{k_2}}k2​k1​​​
  2. (B)k1k2\dfrac{k_1}{k_2}k2​k1​​
  3. (C)k2k1\dfrac{k_2}{k_1}k1​k2​​
  4. (D)k2k1\sqrt{\dfrac{k_2}{k_1}}k1​k2​​​

Correct answer: (A)

Step-by-step solution →
Q23·PhysicsSingle correctJEE Main 2025
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Knowing initial position x0x_0x0​ and initial momentum p0p_0p0​ is enough to determine the position and momentum at any time t for a simple harmonic motion with a given angular frequency ω\omegaω. Reason (R): The amplitude and phase can be expressed in terms of x0x_0x0​ and p0p_0p0​. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both (A) and (R) are true and (R) is NOT the correct explanation of (A)
  2. (B)(A) is false but (R) is true
  3. (C)(A) is true but (R) is false
  4. (D)Both (A) and (R) are true and (R) is the correct explanation of (A)

Correct answer: (D)

Step-by-step solution →
Q24·PhysicsSingle correctJEE Main 2025
A particle is executing simple harmonic motion with time period 2s and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then Dd\dfrac{D}{d}dD​ is :-
  1. (A)154\dfrac{15}{4}415​
  2. (B)25
  3. (C)10
  4. (D)165\dfrac{16}{5}516​

Correct answer: (B)

Step-by-step solution →
Q25·PhysicsSingle correctJEE Main 2025
A particle oscillates along the x-axis according to the law, x(t)=x0sin⁡2(t2)x(t)=x_0\sin^2\left(\dfrac{t}{2}\right)x(t)=x0​sin2(2t​) where x0=1x_0=1x0​=1 m. The kinetic energy (K) of the particle as a function of x is correctly represented by the graph:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q26·PhysicsSingle correctJEE Main 2025
A light hollow cube of side length 10 cm and mass 10g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is yπ×10−2y\pi\times10^{-2}yπ×10−2 s, where the value of yyy is (Acceleration due to gravity, g=10g=10g=10 m/s2^22, density of water =103=10^3=103 kg/m3^33)
  1. (A)2
  2. (B)6
  3. (C)4
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q27·PhysicsNumericalJEE Advanced 2024
Two particles, 1 and 2, each of mass mmm, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x0x_0x0​, are oscillating with amplitude aaa and angular frequency ω\omegaω. Thus, their positions at time ttt are given by x1(t)=(x0+d)+asin⁡ωtx_1(t) = (x_0 + d) + a \sin \omega tx1​(t)=(x0​+d)+asinωt and x2(t)=(x0−d)−asin⁡ωtx_2(t) = (x_0 - d) - a \sin \omega tx2​(t)=(x0​−d)−asinωt, respectively, where d>2ad > 2ad>2a. Particle 3 of mass mmm moves towards this system with speed u0=aω/2u_0 = a\omega/2u0​=aω/2, and undergoes instantaneous elastic collision with particle 2, at time t0t_0t0​. Finally, particles 1 and 2 acquire a center of mass speed vcmv_{\text{cm}}vcm​ and oscillate with amplitude bbb and the same angular frequency ω\omegaω. If the collision occurs at time t0t_0t0​ = π/(2ω)\pi/(2\omega)π/(2ω), then the value of 4b2/a24b^2/a^24b2/a2 will be

Correct answer: 4.25

Step-by-step solution →
Q28·PhysicsNumericalJEE Advanced 2024
Two particles, 1 and 2, each of mass mmm, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x0x_0x0​, are oscillating with amplitude aaa and angular frequency ω\omegaω. Thus, their positions at time ttt are given by x1(t)=(x0+d)+asin⁡ωtx_1(t) = (x_0 + d) + a \sin \omega tx1​(t)=(x0​+d)+asinωt and x2(t)=(x0−d)−asin⁡ωtx_2(t) = (x_0 - d) - a \sin \omega tx2​(t)=(x0​−d)−asinωt, respectively, where d>2ad > 2ad>2a. Particle 3 of mass mmm moves towards this system with speed u0=aω/2u_0 = a\omega/2u0​=aω/2, and undergoes instantaneous elastic collision with particle 2, at time t0t_0t0​. Finally, particles 1 and 2 acquire a center of mass speed vcmv_{\text{cm}}vcm​ and oscillate with amplitude bbb and the same angular frequency ω\omegaω. If the collision occurs at time t0t_0t0​ = 0, the value of vcm/(aω)v_{\text{cm}}/(a\omega)vcm​/(aω) will be

Correct answer: 0.75

Step-by-step solution →
Q29·PhysicsSingle correctJEE Advanced 2024
Two beads, each with charge qqq and mass mmm, are on a horizontal, frictionless, non-conducting, circular hoop of radius RRR. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [ε0\varepsilon_{0}ε0​ is the permittivity of free space.]
  1. (A)q2/(4πε0R3m)q^{2} / \left(4\pi\varepsilon_{0}R^{3}m\right)q2/(4πε0​R3m)
  2. (B)q2/(32πε0R3m)q^{2} / \left(32\pi\varepsilon_{0}R^{3}m\right)q2/(32πε0​R3m)
  3. (C)q2/(8πε0R3m)q^{2} / \left(8\pi\varepsilon_{0}R^{3}m\right)q2/(8πε0​R3m)
  4. (D)q2/(16πε0R3m)q^{2} / \left(16\pi\varepsilon_{0}R^{3}m\right)q2/(16πε0​R3m)

Correct answer: (B)

Step-by-step solution →
Q30·PhysicsSingle correctJEE Advanced 2024
A block of mass 5 kg moves along the x-direction subject to the force F=(−20x+10)F = (-20x + 10)F=(−20x+10) N, with the value of xxx in metre. At time t=0t = 0t=0 s, it is at rest at position x=1x = 1x=1 m. The position and momentum of the block at t=(π/4)t = (\pi/4)t=(π/4) s are
  1. (A)−0.5-0.5−0.5 m, 5 kg m/s
  2. (B)0.5 m, 0 kg m/s
  3. (C)0.5 m, −5-5−5 kg m/s
  4. (D)−1-1−1 m, 5 kg m/s

Correct answer: (C)

Step-by-step solution →
Q31·PhysicsNumericalJEE Main 2024
A particle of mass 0.50 kg executes simple harmonic motion under force F=−50 (N/m) xF=-50\,(\text{N/m})\,xF=−50(N/m)x. The time period of oscillation is x35\dfrac{x}{35}35x​ s. The value of x is _______. (Given π=227\pi=\dfrac{22}{7}π=722​)

Correct answer: 22

Step-by-step solution →
Q32·PhysicsNumericalJEE Main 2024
The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of 4 m, 2 ms−12\ \mathrm{ms^{-1}}2 ms−1 and 16 ms−216\ \mathrm{ms^{-2}}16 ms−2 at a certain instant. The amplitude of the motion is x\sqrt{x}x​ m where xxx is ______.

Correct answer: 17

Step-by-step solution →
Q33·PhysicsNumericalJEE Main 2024
An object of mass 0.2 kg executes simple harmonic motion along the x axis with frequency of 25π\dfrac{25}{\pi}π25​ Hz. At the position x = 0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is ________ cm.

Correct answer: 6

Step-by-step solution →
Q34·PhysicsNumericalJEE Main 2024
A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is _______ cm/s.

Correct answer: 12

Step-by-step solution →
Q35·PhysicsNumericalJEE Main 2024
An elastic spring under tension of 3 N3\,N3N has a length aaa. Its length is bbb under tension 2 N2\,N2N. For its length (3a−2b)(3a-2b)(3a−2b), the value of tension will be ___ N.

Correct answer: 5

Step-by-step solution →
Q36·PhysicsNumericalJEE Main 2024
The displacement of a particle executing SHM is given by x=10sin⁡(ωt+π3)x=10\sin\left(\omega t+\tfrac{\pi}{3}\right)x=10sin(ωt+3π​) m. The time period of motion is 3.14 s. The velocity of the particle at t=0t=0t=0 is ______ m/s.

Correct answer: 10

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Q37·PhysicsSingle correctJEE Main 2024
In a simple harmonic motion, the total mechanical energy of given system is E. If mass of the oscillating particle P is doubled then the new energy of the system for same amplitude is
  1. (A)E2\tfrac{E}{\sqrt{2}}2​E​
  2. (B)E
  3. (C)E2E\sqrt{2}E2​
  4. (D)2E2E2E

Correct answer: (B)

Step-by-step solution →
Q38·PhysicsNumericalJEE Main 2024
A mass m is suspended from a spring of negligible mass and the system oscillates with a frequency f1f_1f1​. The frequency of oscillations if a mass 9m is suspended from the same spring is f2f_2f2​. The value of f1f2\dfrac{f_1}{f_2}f2​f1​​ is __________.

Correct answer: 3

Step-by-step solution →
Q39·PhysicsNumericalJEE Main 2024
A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is 2A3\dfrac{2A}{3}32A​. The new amplitude of motion is nA3\dfrac{nA}{3}3nA​. The value of n is ______.

Correct answer: 7

Step-by-step solution →
Q40·PhysicsNumericalJEE Main 2024
The time period of simple harmonic motion of mass M in the given figure is παM5K\pi\sqrt{\dfrac{\alpha M}{5K}}π5KαM​​, where the value of α\alphaα is ______.

Correct answer: 12

Step-by-step solution →
Q41·PhysicsNumericalJEE Main 2024
A simple harmonic oscillator has an amplitude A and time period 6π6\pi6π second. Assuming the oscillation starts from its mean position, the time required by it to travel from x=Ax=Ax=A to x=32Ax=\dfrac{\sqrt{3}}{2}Ax=23​​A will be πx\dfrac{\pi}{x}xπ​ s, where x=x=x= ___.

Correct answer: 2

Step-by-step solution →
Q42·PhysicsNumericalJEE Main 2024
When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is x8\dfrac{x}{8}8x​, where x=x=x= ______ .

Correct answer: 9

Step-by-step solution →
Q43·PhysicsNumericalJEE Main 2024
A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is α\sqrt{\alpha}α​ cm, where α=\alpha=α= __________.

Correct answer: 12

Step-by-step solution →
Q44·PhysicsSingle correctJEE Main 2024
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle θ\thetaθ of thread deflection in the extreme position will be :
  1. (A)tan⁡−1(2)\tan^{-1}(\sqrt2)tan−1(2​)
  2. (B)2tan⁡−1(12)2\tan^{-1}\left(\dfrac12\right)2tan−1(21​)
  3. (C)tan⁡−1(12)\tan^{-1}\left(\dfrac12\right)tan−1(21​)
  4. (D)2tan⁡−1(15)2\tan^{-1}\left(\dfrac{1}{\sqrt5}\right)2tan−1(5​1​)

Correct answer: (B)

Step-by-step solution →
Q45·PhysicsNumericalJEE Advanced 2023
Two point-like objects of masses 20 gm and 30 gm are fixed at the two ends of a rigid massless rod of length 10 cm. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is 1.2×10−81.2 \times 10^{-8}1.2×10−8 Nm rad−1rad^{-1}rad−1 . The angular frequency of the oscillations in n×10−3n \times 10^{-3}n×10−3 rad s−1s^{-1}s−1 . The value of n is ______

Correct answer: 10

Step-by-step solution →
Q46·PhysicsSingle correctJEE Main 2023
In a linear simple harmonic motion (SHM) (A) Restoring force is directly proportional to the displacement. (B) The acceleration and displacement are opposite in direction. (C) The velocity is maximum at mean position. (D) The acceleration is minimum at extreme points. Choose the correct answer from the options given below :
  1. (A)(A), (B) and (C) only
  2. (B)(C) and (D) only
  3. (C)(A), (B) and (D) only
  4. (D)(A), (C) and (D) only

Correct answer: (A)

Step-by-step solution →
Q47·PhysicsSingle correctJEE Main 2023
A particle executes SHM of amplitude A. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:
  1. (A)2A\sqrt{2}A2​A
  2. (B)2A2A2A
  3. (C)12A\dfrac{1}{\sqrt{2}}A2​1​A
  4. (D)12A\dfrac{1}{2}A21​A

Correct answer: (C)

Step-by-step solution →
Q48·PhysicsSingle correctJEE Main 2023
Which graph represents the difference between total energy and potential energy of a particle executing SHM Vs its distance from mean position:
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q49·PhysicsNumericalJEE Main 2023
At a given point of time the value of displacement of a simple harmonic oscillator is given as y=Acos⁡(30∘)y=A\cos(30^\circ)y=Acos(30∘). If amplitude is 40 cm40\,cm40cm and kinetic energy at that time is 200 J200\,J200J, the value of force constant is 1.0×10x Nm−11.0\times 10^{x}\,Nm^{-1}1.0×10xNm−1. The value of xxx is _____.

Correct answer: 4

Step-by-step solution →
Q50·PhysicsSingle correctJEE Main 2023
A particle is executing Simple Harmonic Motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be:
  1. (A)1 : 1
  2. (B)2 : 1
  3. (C)1 : 4
  4. (D)1 : 3

Correct answer: (D)

Step-by-step solution →
Q51·PhysicsSingle correctJEE Main 2023
The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q52·PhysicsSingle correctJEE Main 2023
A particle executes S.H.M. of amplitude A along x-axis. At t=0t=0t=0, the position of the particle is x=A2x=\dfrac{A}{2}x=2A​ and it moves along positive x-axis. The displacement of particle in time t i.e. x=Asin⁡(ωt+δ)x=A\sin(\omega t+\delta)x=Asin(ωt+δ), then the value of δ\deltaδ will be:
  1. (A)π6\dfrac{\pi}{6}6π​
  2. (B)π3\dfrac{\pi}{3}3π​
  3. (C)π4\dfrac{\pi}{4}4π​
  4. (D)π2\dfrac{\pi}{2}2π​

Correct answer: (A)

Step-by-step solution →
Q53·PhysicsNumericalJEE Main 2023
A rectangular block of mass 555 kg attached to a horizontal spiral spring executes simple harmonic motion of amplitude 111 m and time period 3.143.143.14 s. The maximum force exerted by the spring on the block is _______ N.

Correct answer: 20

Step-by-step solution →
Q54·PhysicsSingle correctJEE Main 2023
For a periodic motion represented by the equation Y=sin⁡ωt+cos⁡ωtY=\sin\omega t+\cos\omega tY=sinωt+cosωt, the amplitude of the motion is:
  1. (A)0.50.50.5
  2. (B)2\sqrt{2}2​
  3. (C)111
  4. (D)222

Correct answer: (B)

Step-by-step solution →
Q55·PhysicsSingle correctJEE Main 2023
For a particle P revolving round the centre O with radius of circular path rrr and angular velocity ω\omegaω, as shown in the figure, the projection of OP on the x-axis at time t is
  1. (A)x(t)=rcos⁡(ωt+π6)x(t)=r\cos\left(\omega t+\dfrac{\pi}{6}\right)x(t)=rcos(ωt+6π​)
  2. (B)x(t)=rcos⁡(ωt)x(t)=r\cos(\omega t)x(t)=rcos(ωt)
  3. (C)x(t)=rsin⁡(ωt+π6)x(t)=r\sin\left(\omega t+\dfrac{\pi}{6}\right)x(t)=rsin(ωt+6π​)
  4. (D)x(t)=rcos⁡(ωt−π6)x(t)=r\cos\left(\omega t-\dfrac{\pi}{6}\right)x(t)=rcos(ωt−6π​)

Correct answer: (A)

Step-by-step solution →
Q56·PhysicsNumericalJEE Main 2023
A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10 cm. The maximum tension in the string is found to be x40\dfrac{x}{40}40x​ N. The value of xxx is

Correct answer: 99

Step-by-step solution →
Q57·PhysicsSingle correctJEE Main 2023
A small block of mass 100 g100\,g100g is tied to a spring of spring constant 7.5 N/m7.5\,N/m7.5N/m and length 20 cm20\,cm20cm. The other end of spring is fixed at a particular point AAA. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad/s5\,rad/s5rad/s about point AAA, then tension in the spring is:
  1. (A)1.5 N1.5\,N1.5N
  2. (B)0.75 N0.75\,N0.75N
  3. (C)0.25 N0.25\,N0.25N
  4. (D)0.50 N0.50\,N0.50N

Correct answer: (B)

Step-by-step solution →
Q58·PhysicsSingle correctJEE Main 2023
A mass mmm is attached to two springs as shown in figure. The spring constants of two springs are K1K_1K1​ and K2K_2K2​. For the frictionless surface, the time period of oscillation of mass is:
  1. (A)12πK1+K2m\dfrac{1}{2\pi}\sqrt{\dfrac{K_1+K_2}{m}}2π1​mK1​+K2​​​
  2. (B)12πK1−K2m\dfrac{1}{2\pi}\sqrt{\dfrac{K_1-K_2}{m}}2π1​mK1​−K2​​​
  3. (C)2πmK1+K22\pi\sqrt{\dfrac{m}{K_1+K_2}}2πK1​+K2​m​​
  4. (D)2πmK1−K22\pi\sqrt{\dfrac{m}{K_1-K_2}}2πK1​−K2​m​​

Correct answer: (C)

Step-by-step solution →
Q59·PhysicsSingle correctJEE Main 2023
Choose the correct length (L) versus square of time period (T2^22) graph for a simple pendulum executing simple harmonic motion.
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q60·PhysicsNumericalJEE Main 2023
The amplitude of a particle executing SHM is 333 cm. The displacement at which its kinetic energy will be 25%25\%25% more than the potential energy is _______ cm.

Correct answer: 2

Step-by-step solution →
Q61·PhysicsNumericalJEE Main 2023
A block is fastened to a horizontal spring. The block is pulled to a distance x=10x=10x=10 cm from its equilibrium position (at x=0x=0x=0) on a frictionless surface from rest. The energy of the block at x=5x=5x=5 cm is 0.250.250.25 J. The spring constant of the spring is _________ Nm−1^{-1}−1.

Correct answer: 50

Step-by-step solution →
Q62·PhysicsNumericalJEE Main 2023
In the figure given below, a block of mass M=490M = 490M=490 g placed on a frictionless table is connected with two springs having same spring constant (K=2 N m−1K = 2\,\text{N m}^{-1}K=2N m−1). If the block is horizontally displaced through 'XXX' m then the number of complete oscillations it will make in 14π14\pi14π seconds will be __________.

Correct answer: 20

Step-by-step solution →
Q63·PhysicsSingle correctJEE Main 2023
The maximum potential energy of a block executing simple harmonic motion is 252525 J. AAA is amplitude of oscillation. At A/2A/2A/2, the kinetic energy of the block is :
  1. (A)18.7518.7518.75 J
  2. (B)9.759.759.75 J
  3. (C)37.537.537.5 J
  4. (D)12.512.512.5 J

Correct answer: (A)

Step-by-step solution →
Q64·PhysicsNumericalJEE Main 2023
The velocity of a particle executing SHM varies with displacement (xxx) as 4v2=50−x24v^{2}=50-x^{2}4v2=50−x2. The time period of oscillation is x7\dfrac{x}{7}7x​ s. The value of xxx is _______. (Take π=227\pi=\dfrac{22}{7}π=722​)

Correct answer: 88

Step-by-step solution →
Q65·PhysicsSingle correctJEE Main 2023
For a simple harmonic motion in a mass-spring system on a frictionless surface, the angular frequency is ω1\omega_1ω1​ when the mass of the block is 1 kg and ω2\omega_2ω2​ when the mass is 2 kg. The ratio ω2ω1\dfrac{\omega_2}{\omega_1}ω1​ω2​​ is:
  1. (A)12\dfrac{1}{\sqrt{2}}2​1​
  2. (B)2\sqrt{2}2​
  3. (C)222
  4. (D)12\dfrac{1}{2}21​

Correct answer: (A)

Step-by-step solution →
Q66·PhysicsNumericalJEE Main 2023
The general displacement of a simple harmonic oscillator is x=Asin⁡ωtx=A\sin\omega tx=Asinωt. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when t=Tβt=\dfrac{T}{\beta}t=βT​. The value of β\betaβ is

Correct answer: 8

Step-by-step solution →
Q67·PhysicsNumericalJEE Main 2023
A particle of mass 250 g250\,g250g executes a simple harmonic motion under a periodic force F=(−25x)NF=(-25x)NF=(−25x)N. The particle attains a maximum speed of 4 m/s4\,m/s4m/s during its oscillation. The amplitude of the motion is _____ cm.

Correct answer: 40

Step-by-step solution →
Q68·PhysicsSingle correctJEE Main 2023
T is the time period of a simple pendulum on the earth's surface. Its time period becomes xTxTxT when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of xxx will be:
  1. (A)444
  2. (B)222
  3. (C)14\dfrac{1}{4}41​
  4. (D)12\dfrac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q69·PhysicsSingle correctJEE Main 2023
A particle executes simple harmonic motion between x=−Ax = -Ax=−A and x=+Ax = +Ax=+A. If time taken by particle to go from x=0x = 0x=0 to A2\frac{A}{2}2A​ is 222 s; then time taken by particle in going from x=A2x = \frac{A}{2}x=2A​ to A is
  1. (A)444 S
  2. (B)1.51.51.5 S
  3. (C)222 S
  4. (D)333 S

Correct answer: (A)

Step-by-step solution →
Q70·PhysicsNumericalJEE Main 2023
A mass mmm attached to free end of a spring executes SHM with a period of 111 s. If the mass is increased by 333 kg the period of oscillation increases by one second, the value of mass mmm is _________ kg.

Correct answer: 1

Step-by-step solution →
Q71·PhysicsNumericalJEE Main 2023
A block of mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is πx\tfrac{\pi}{\sqrt{x}}x​π​ in SI unit. The value of xxx is _______ .

Correct answer: 5

Step-by-step solution →
Q72·PhysicsIntegerJEE Advanced 2022
On a frictionless horizontal plane, a bob of mass m = 0.1 kg is attached to a spring with natural length l0_00​=0.1 m. The spring constant is k1_11​ = 0.009Nm−1^{-1}−1 when the length of the spring l>l0_00​ and is k2_22​ = 0.016Nm−1^{-1}−1 when l < l0_00​. Initially the bob is released from l=0.15m . Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is T=(n π) s, then the integer closest to n is _______

Correct answer: 6

Step-by-step solution →
Q73·PhysicsSingle correctJEE Main 2022
The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination α\alphaα, is given by :
  1. (A)2πL/(gcos⁡α)2\pi\sqrt{L/(g\cos\alpha)}2πL/(gcosα)​
  2. (B)2πL/(gsin⁡α)2\pi\sqrt{L/(g\sin\alpha)}2πL/(gsinα)​
  3. (C)2πL/g2\pi\sqrt{L/g}2πL/g​
  4. (D)2πL/(gtan⁡α)2\pi\sqrt{L/(g\tan\alpha)}2πL/(gtanα)​

Correct answer: (A)

Step-by-step solution →
Q74·PhysicsNumericalJEE Main 2022
The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is 10 s. If the metallic bob is immersed in water, then the new time period becomes 5x5\sqrt{x}5x​ s. The value of x will be ________.

Correct answer: 5

Step-by-step solution →
Q75·PhysicsNumericalJEE Main 2022
The potential energy of a particle of mass 4 kg in motion along the x-axis is given by U=4(1−cos⁡4x)U = 4(1 - \cos 4x)U=4(1−cos4x) J. The time period of the particle for small oscillation (sin⁡θ≃θ)(\sin\theta \simeq \theta)(sinθ≃θ) is (πK)\left(\dfrac{\pi}{K}\right)(Kπ​) s. The value of K is ........

Correct answer: 2

Step-by-step solution →
Q76·PhysicsSingle correctJEE Main 2022
Two identical positive charges Q each are fixed at a distance of ‘2a’ apart from each other. Another point charge q0_{0}0​ with mass ‘m’ is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge q0_{0}0​ executes SHM. The time period of oscillation of charge q0_{0}0​ will be :
  1. (A)4π3ε0ma3q0Q\sqrt{\dfrac{4\pi^{3}\varepsilon_{0}ma^{3}}{q_{0}Q}}q0​Q4π3ε0​ma3​​
  2. (B)q0Q4π3ε0ma3\sqrt{\dfrac{q_{0}Q}{4\pi^{3}\varepsilon_{0}ma^{3}}}4π3ε0​ma3q0​Q​​
  3. (C)2π2ε0ma3q0Q\sqrt{\dfrac{2\pi^{2}\varepsilon_{0}ma^{3}}{q_{0}Q}}q0​Q2π2ε0​ma3​​
  4. (D)8π3ε0ma3q0Q\sqrt{\dfrac{8\pi^{3}\varepsilon_{0}ma^{3}}{q_{0}Q}}q0​Q8π3ε0​ma3​​

Correct answer: (A)

Step-by-step solution →
Q77·PhysicsNumericalJEE Main 2022
A mass 0.9 kg, attached to a horizontal spring, executes SHM with an amplitude A1A_1A1​. When this mass passes through its mean position, then a smaller mass of 124 g is placed over it and both masses move together with amplitude A2A_2A2​. If the ratio A1A2\frac{A_1}{A_2}A2​A1​​ is αα−1\frac{\alpha}{\alpha-1}α−1α​, then the value of α\alphaα will be ____ .

Correct answer: 16

Step-by-step solution →
Q78·PhysicsNumericalJEE Main 2022
As per given figures, two springs of spring constants K and 2K are connected to mass m. If the period of oscillation in figure (a) is 3s, then the period of oscillation in figure (b) will be x\sqrt{x}x​ s. The value of x is__________ .

Correct answer: 2

Step-by-step solution →
Q79·PhysicsSingle correctJEE Main 2022
When a particle executes simple Harmonic motion, the nature of graph of velocity as function of displacement will be :
  1. (A)Circular
  2. (B)Ellipitical
  3. (C)Sinusoidal
  4. (D)Straight line

Correct answer: (B)

Step-by-step solution →
Q80·PhysicsSingle correctJEE Main 2022
In figure (A), mass '2 m' is fixed on mass 'm' which is attached to two springs of spring constant k. In figure (B), mass 'm' is attached to two spring of spring constant 'k' and '2k'. If mass 'm' in (A) and (B) are displaced by distance 'x' horizontally and then released, then time period T1T_1T1​ and T2T_2T2​ corresponding to (A) and (B) respectively follow the relation.
  1. (A)T1T2=32\frac{T_1}{T_2} = \frac{3}{\sqrt{2}}T2​T1​​=2​3​
  2. (B)T1T2=32\frac{T_1}{T_2} = \sqrt{\frac{3}{2}}T2​T1​​=23​​
  3. (C)T1T2=23\frac{T_1}{T_2} = \sqrt{\frac{2}{3}}T2​T1​​=32​​
  4. (D)T1T2=23\frac{T_1}{T_2} = \frac{\sqrt{2}}{3}T2​T1​​=32​​

Correct answer: (A)

Step-by-step solution →
Q81·PhysicsNumericalJEE Main 2022
A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is x\sqrt{x}x​ cm. The value of x is___________.

Correct answer: 700

Step-by-step solution →
Q82·PhysicsSingle correctJEE Main 2022
The motion of a simple pendulum excuting S.H.M. is represented by following equation. Y=Asin⁡(πt+ϕ)Y = A \sin(\pi t + \phi)Y=Asin(πt+ϕ), where time is measured in second. The length of pendulum is :
  1. (A)97.23 cm
  2. (B)25.3 cm
  3. (C)99.4 cm
  4. (D)406.1 cm

Correct answer: (C)

Step-by-step solution →
Q83·PhysicsNumericalJEE Main 2022
A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is _________ s.

Correct answer: 1

Step-by-step solution →
Q84·PhysicsSingle correctJEE Main 2022
The equation of a particle executing simple harmonic motion is given by x=sin⁡π(t+13)x=\sin\pi\left(t+\dfrac{1}{3}\right)x=sinπ(t+31​) m . At t = 1s, the speed of particle will be (Given : π=3.14\pi=3.14π=3.14)
  1. (A)000 cm s−1^{-1}−1
  2. (B)157157157 cm s−1^{-1}−1
  3. (C)272272272 cm s−1^{-1}−1
  4. (D)314314314 cm s−1^{-1}−1

Correct answer: (B)

Step-by-step solution →
Q85·PhysicsSingle correctJEE Main 2022
The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is :
  1. (A)6 s
  2. (B)8 s
  3. (C)12s
  4. (D)36 s

Correct answer: (D)

Step-by-step solution →
Q86·PhysicsSingle correctJEE Main 2022
Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with g6\frac{g}{6}6g​ vertically upwards then the time period will be : (where g = acceleration due to gravity)
  1. (A)65T\sqrt{\frac{6}{5}}T56​​T
  2. (B)56T\sqrt{\frac{5}{6}}T65​​T
  3. (C)67T\sqrt{\frac{6}{7}}T76​​T
  4. (D)76T\sqrt{\frac{7}{6}}T67​​T

Correct answer: (C)

Step-by-step solution →
Q87·PhysicsSingle correctJEE Main 2021
A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments are performed?
  1. (A)10 m/s2^{2}2
  2. (B)5 m/s2^{2}2
  3. (C)4 m/s2^{2}2
  4. (D)9.8 m/s2^{2}2

Correct answer: (C)

Step-by-step solution →
Q88·PhysicsSingle correctJEE Main 2021
For a body executing S.H.M. : (a) Potential energy is always equal to its K.E. (b) Average potential and kinetic energy over any given time interval are always equal. (c) Sum of the kinetic and potential energy at any point of time is constant. (d) Average K.E. in one time period is equal to average potential energy in one time period. Choose the most appropriate option from the options given below :
  1. (A)(c) and (d)
  2. (B)only (c)
  3. (C)(b) and (c)
  4. (D)only (b)

Correct answer: (A)

Step-by-step solution →
Q89·PhysicsSingle correctJEE Main 2021
A bob of mass 'm' suspended by a thread of length lll undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density 14\frac{1}{4}41​ times that of the bob and the length of the thread is increased by 1/3rd1/3^{rd}1/3rd of the original length, then the time period of the simple harmonic oscillations will be :-
  1. (A)T
  2. (B)32\frac{3}{2}23​ T
  3. (C)34\frac{3}{4}43​ T
  4. (D)43\frac{4}{3}34​ T

Correct answer: (D)

Step-by-step solution →
Q90·PhysicsNumericalJEE Main 2021
A particle of mass 1 kg is hanging from a spring of force constant 100 Nm−1^{-1}−1. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is Tx\frac{T}{x}xT​ .The value of x is ____.

Correct answer: 8

Step-by-step solution →
Q91·PhysicsSingle correctJEE Main 2021
The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure :
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q92·PhysicsNumericalJEE Main 2021
Two simple harmonic motion, are represented by the equations y1=10sin⁡(3πt+π3)y_1 = 10 \sin\left(3\pi t + \frac{\pi}{3}\right)y1​=10sin(3πt+3π​) y2=5(sin⁡3πt+3cos⁡3πt)y_2 = 5(\sin 3\pi t + \sqrt{3}\cos 3\pi t)y2​=5(sin3πt+3​cos3πt) Ratio of amplitude of y1y_1y1​ to y2y_2y2​ = x : 1. The value of x is _________ .

Correct answer: 1

Step-by-step solution →
Q93·PhysicsSingle correctJEE Main 2021
If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is:
  1. (A)86.4 s
  2. (B)4.32 s
  3. (C)43.2 s
  4. (D)8.64 s

Correct answer: (C)

Step-by-step solution →
Q94·PhysicsNumericalJEE Main 2021
Two simple harmonic motions are represented by the equations x1=5sin⁡(2πt+π4)x_1 = 5 \sin\left(2\pi t + \frac{\pi}{4}\right)x1​=5sin(2πt+4π​) and x2=52 (sin⁡2πt+cos⁡2πt)x_2 = 5\sqrt{2}\,(\sin 2\pi t + \cos 2\pi t)x2​=52​(sin2πt+cos2πt). The amplitude of second motion is ............. times the amplitude in first motion.

Correct answer: 2

Step-by-step solution →
Q95·PhysicsSingle correctJEE Main 2021
A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is 3E4\frac{3E}{4}43E​ then its displacement 'y' is given by :
  1. (A)y = a2\frac{a}{2}2a​
  2. (B)y = a2\frac{a}{\sqrt{2}}2​a​
  3. (C)y = a32\frac{a\sqrt{3}}{2}2a3​​
  4. (D)y = a

Correct answer: (A)

Step-by-step solution →
Q96·PhysicsSingle correctJEE Main 2021
An object of mass 0.5 kg executing simple harmonic motion. Its amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant t = T4\frac{T}{4}4T​ s starting from mean position. Assume that the initial phase of the oscillation is zero.
  1. (A)6.2 × 10−3^{-3}−3 J
  2. (B)0.62 J
  3. (C)6.2 × 103^{3}3 J
  4. (D)1.2 × 103^{3}3 J

Correct answer: (B)

Step-by-step solution →
Q97·PhysicsNumericalJEE Main 2021
A particle executes simple harmonic motion represented by displacement function as x(t) = A sin(ωt + φ) If the position and velocity of the particle at t = 0s are 2cm and 2ω cm s−1^{-1}−1 respectively, then its amplitude is x√2 cm where the value of x is __________.

Correct answer: 2

Step-by-step solution →
Q98·PhysicsSingle correctJEE Main 2021
In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.
  1. (A)12\frac{1}{2}21​
  2. (B)34\frac{3}{4}43​
  3. (C)13\frac{1}{3}31​
  4. (D)14\frac{1}{4}41​

Correct answer: (B)

Step-by-step solution →
Q99·PhysicsNumericalJEE Main 2021
In the reported figure, two bodies A and B of masses 200g and 800g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be _______ rad /s when k = 20 N / m.

Correct answer: 10

Step-by-step solution →
Q100·PhysicsSingle correctJEE Main 2021
T0T_0T0​ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to 116\frac{1}{16}161​ times of its initial value, the modified time period is :
  1. (A)T0T_0T0​
  2. (B)8π T08\pi\,T_08πT0​
  3. (C)4 T04\,T_04T0​
  4. (D)14T0\frac{1}{4}T_041​T0​

Correct answer: (D)

Step-by-step solution →
Q101·PhysicsSingle correctJEE Main 2021
The motion of a mass on a spring, with spring constant K is as shown in figure. The equation of motion is given by x(t)=Asin⁡ωt+Bcos⁡ωtx(t)=A\sin\omega t+B\cos\omega tx(t)=Asinωt+Bcosωt with ω=Km\omega=\sqrt{\frac{K}{m}}ω=mK​​. Suppose that at time t = 0, the position of mass is x(0)x(0)x(0) and velocity υ(0)\upsilon(0)υ(0), then its displacement can also be represented as x(t)=Ccos⁡(ωt−ϕ)x(t)=C\cos(\omega t-\phi)x(t)=Ccos(ωt−ϕ), where C and ϕ\phiϕ are :
  1. (A)C=υ(0)2ω2+x(0)2, ϕ=tan⁡−1(x(0)ωυ(0))C=\sqrt{\frac{\upsilon(0)^2}{\omega^2}+x(0)^2},\ \phi=\tan^{-1}\left(\frac{x(0)\omega}{\upsilon(0)}\right)C=ω2υ(0)2​+x(0)2​, ϕ=tan−1(υ(0)x(0)ω​)
  2. (B)C=2υ(0)2ω2+x(0)2, ϕ=tan⁡−1(x(0)ω2υ(0))C=\sqrt{\frac{2\upsilon(0)^2}{\omega^2}+x(0)^2},\ \phi=\tan^{-1}\left(\frac{x(0)\omega}{2\upsilon(0)}\right)C=ω22υ(0)2​+x(0)2​, ϕ=tan−1(2υ(0)x(0)ω​)
  3. (C)C=υ(0)2ω2+x(0)2, ϕ=tan⁡−1(υ(0)x(0)ω)C=\sqrt{\frac{\upsilon(0)^2}{\omega^2}+x(0)^2},\ \phi=\tan^{-1}\left(\frac{\upsilon(0)}{x(0)\omega}\right)C=ω2υ(0)2​+x(0)2​, ϕ=tan−1(x(0)ωυ(0)​)
  4. (D)C=2υ(0)2ω2+x(0)2, ϕ=tan⁡−1(υ(0)x(0)ω)C=\sqrt{\frac{2\upsilon(0)^2}{\omega^2}+x(0)^2},\ \phi=\tan^{-1}\left(\frac{\upsilon(0)}{x(0)\omega}\right)C=ω22υ(0)2​+x(0)2​, ϕ=tan−1(x(0)ωυ(0)​)

Correct answer: (C)

Step-by-step solution →
Q102·PhysicsSingle correctJEE Main 2021
A particle is making simple harmonic motion along the X-axis. If at a distances x1x_1x1​ and x2x_2x2​ from the mean position the velocities of the particle are υ1\upsilon_1υ1​ and υ2\upsilon_2υ2​ respectively. The time period of its oscillation is given as :
  1. (A)T=2πx22−x12υ12−υ22T = 2\pi \sqrt{ \dfrac{x_2^{2} - x_1^{2}}{\upsilon_1^{2} - \upsilon_2^{2}} }T=2πυ12​−υ22​x22​−x12​​​
  2. (B)T=2πx22+x12υ12−υ22T = 2\pi \sqrt{ \dfrac{x_2^{2} + x_1^{2}}{\upsilon_1^{2} - \upsilon_2^{2}} }T=2πυ12​−υ22​x22​+x12​​​
  3. (C)T=2πx22−x12υ12+υ22T = 2\pi \sqrt{ \dfrac{x_2^{2} - x_1^{2}}{\upsilon_1^{2} + \upsilon_2^{2}} }T=2πυ12​+υ22​x22​−x12​​​
  4. (D)T=2πx22+x12υ12+υ22T = 2\pi \sqrt{ \dfrac{x_2^{2} + x_1^{2}}{\upsilon_1^{2} + \upsilon_2^{2}} }T=2πυ12​+υ22​x22​+x12​​​

Correct answer: (A)

Step-by-step solution →
Q103·PhysicsNumericalJEE Main 2021
A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is 1a\dfrac{1}{a}a1​ s. The value of 'a' to the nearest integer is _______ .

Correct answer: 6

Step-by-step solution →
Q104·PhysicsSingle correctJEE Main 2021
The function of time representing a simple harmonic motion with a period of πω\frac{\pi}{\omega}ωπ​ is :
  1. (A)sin(ωt) + cos (ωt)
  2. (B)cos(ωt) + cos (2ωt) + cos (3ωt)
  3. (C)sin⁡2(ωt)\sin^{2}(\omega t)sin2(ωt)
  4. (D)3cos⁡(π4−2ωt)3\cos\left(\frac{\pi}{4}-2\omega t\right)3cos(4π​−2ωt)

Correct answer: (D)

Step-by-step solution →
Q105·PhysicsNumericalJEE Main 2021
Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig.1 shows one of them and Fig.2 shows their series combination. The ratios of time period of oscillation of the two SHM is TbTa=x\frac{\mathrm{T}_b}{\mathrm{T}_a} = \sqrt{\mathrm{x}}Ta​Tb​​=x​, where value of x is _______. (Round off to the Nearest Integer)

Correct answer: 2

Step-by-step solution →
Q106·PhysicsSingle correctJEE Main 2021
For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal ?
  1. (A)x=0x = 0x=0
  2. (B)x=±Ax = \pm Ax=±A
  3. (C)x=±A2x = \pm \frac{A}{\sqrt{2}}x=±2​A​
  4. (D)x=A2x = \frac{A}{2}x=2A​

Correct answer: (C)

Step-by-step solution →
Q107·PhysicsSingle correctJEE Main 2021
Two particles A and B of equal masses are suspended from two massless springs of spring constants K1_{1}1​ and K2_{2}2​ respectively.If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is
  1. (A)K2K1\frac{K_{2}}{K_{1}}K1​K2​​
  2. (B)K1K2\frac{K_{1}}{K_{2}}K2​K1​​
  3. (C)K1K2\sqrt{\frac{K_{1}}{K_{2}}}K2​K1​​​
  4. (D)K2K1\sqrt{\frac{K_{2}}{K_{1}}}K1​K2​​​

Correct answer: (D)

Step-by-step solution →
Q108·PhysicsSingle correctJEE Main 2021
Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be :
  1. (A)3 T\sqrt{3}\,T3​T
  2. (B)T3\frac{T}{\sqrt{3}}3​T​
  3. (C)32 T\sqrt{\frac{3}{2}}\,T23​​T
  4. (D)23 T\sqrt{\frac{2}{3}}\,T32​​T

Correct answer: (D)

Step-by-step solution →
Q109·PhysicsSingle correctJEE Main 2021
Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to half of its initial value ? (ln 2 = 0.693)
  1. (A)34.65 s
  2. (B)17.32 s
  3. (C)0.034 s
  4. (D)15.01 s

Correct answer: (A)

Step-by-step solution →
Q110·PhysicsNumericalJEE Main 2021
Time period of a simple pendulum is T. The time taken to complete 58\frac{5}{8}85​ oscillations starting from mean position is αβ\frac{\alpha}{\beta}βα​T. The value of α\alphaα is _____________.

Correct answer: 7

Step-by-step solution →
Q111·PhysicsNumericalJEE Main 2021
A particle excutes S.H.M with amplitude 'a' and time period T. The displacement of the particle when its speed is half of maximum speed is x a2\frac{\sqrt{x}\,a}{2}2x​a​. The value of x is ______________

Correct answer: 3

Step-by-step solution →
Q112·PhysicsSingle correctJEE Main 2021
A particle executes S.H.M., the graph of velocity as a function of displacement is:
  1. (A)a circle
  2. (B)a parabola
  3. (C)an ellipse
  4. (D)a helix

Correct answer: (C)

Step-by-step solution →
Q113·PhysicsSingle correctJEE Main 2021
If two similar springs each of spring constant K1_{1}1​ are joined in series, the new spring constant and time period would be changed by a factor :
  1. (A)12\frac{1}{2}21​, 2\sqrt{2}2​
  2. (B)14\frac{1}{4}41​, 222\sqrt{2}22​
  3. (C)12\frac{1}{2}21​, 222\sqrt{2}22​
  4. (D)14\frac{1}{4}41​, 2\sqrt{2}2​

Correct answer: (A)

Step-by-step solution →
Q114·PhysicsSingle correctJEE Main 2021
Given below are two statements : Statement (1) :- A second's pendulum has a time period of 1 second. Statement (2) :- It takes precisely one second to move between the two extreme positions. In the light of the above statements, choose the correct answer from the options give below.
  1. (A)Both Statement I and Statement II are false
  2. (B)Statement I is true but Statement II is false
  3. (C)Statement I is false but Statement II is true
  4. (D)Both Statement I and Statement II is true

Correct answer: (C)

Step-by-step solution →
Q115·PhysicsSingle correctJEE Main 2021
The point A moves with a uniform speed along the circumference of a circle of radius 0.36m and cover 30o^{o}o in 0.1s. The perpendicular projection 'P' form 'A' on the diameter MN represents the simple harmonic motion of 'P'. The restoration force per unit mass when P touches M will be: A 0.1s 0.36m30o^{o}o MN OP
  1. (A)100 N
  2. (B)50 N
  3. (C)9.87 N
  4. (D)0.49 N

Correct answer: (C)

Step-by-step solution →
Q116·PhysicsSingle correctJEE Main 2021
If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :
  1. (A)2π2ms−22\pi^2 \mathrm{ms}^{-2}2π2ms−2
  2. (B)16m/s216 \mathrm{m/s}^216m/s2
  3. (C)9.8ms−29.8 \mathrm{ms}^{-2}9.8ms−2
  4. (D)π2ms−2\pi^2 \mathrm{ms}^{-2}π2ms−2

Correct answer: (A)

Step-by-step solution →
Q117·PhysicsSingle correctJEE Main 2021
Y = A sin(ωt + φ0) is the time – displacement equation of a SHM, At t = 0 the displacement of A the particle is Y = and it is moving along negative x-direction. Then the initial phase angle φ0 2 will be.
  1. (A)π6\dfrac{\pi}{6}6π​
  2. (B)π3\dfrac{\pi}{3}3π​
  3. (C)2π3\dfrac{2\pi}{3}32π​
  4. (D)5π6\dfrac{5\pi}{6}65π​

Correct answer: (D)

Step-by-step solution →
Q118·PhysicsSingle correctJEE Main 2021
Two identical spring of spring constant '2K' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. Then time period of oscillations of this system is: 2km2k mmmm
  1. (A)πmk\pi\sqrt{\dfrac{m}{k}}πkm​​
  2. (B)πm2k\pi\sqrt{\dfrac{m}{2k}}π2km​​
  3. (C)2πmk2\pi\sqrt{\dfrac{m}{k}}2πkm​​
  4. (D)2πm2k2\pi\sqrt{\dfrac{m}{2k}}2π2km​​

Correct answer: (A)

Step-by-step solution →
Q119·PhysicsSingle correctJEE Main 2021
In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be -
  1. (A)AMM+mA\sqrt{\frac{M}{M+m}}AM+mM​​
  2. (B)AMM−mA\sqrt{\frac{M}{M-m}}AM−mM​​
  3. (C)AM−mMA\sqrt{\frac{M-m}{M}}AMM−m​​
  4. (D)AM+mMA\sqrt{\frac{M+m}{M}}AMM+m​​

Correct answer: (A)

Step-by-step solution →
Q120·PhysicsSingle correctJEE Main 2021
When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is :
  1. (A)elliptical
  2. (B)parabolic
  3. (C)straight line
  4. (D)circular

Correct answer: (A)

Step-by-step solution →
Q121·PhysicsNumericalJEE Advanced 2020
One end of a spring of negligible unstretched length and spring constant k is fixed at the origin (0,0). A point particle of mass m carrying a positive charge q is attached at its other end. The entire system is kept on a smooth horizontal surface. When a point dipole p⃗\vec{p}p​ pointing towards the charge q is fixed at the origin, the spring gets stretched to a length ℓ\ellℓ and attains a new equilibrium position (see figure below). If the point mass is now displaced slightly by Δℓ≪ℓ\Delta\ell \ll \ellΔℓ≪ℓ from its equilibrium position and released, it is found to oscillate at frequency 1δkm\frac{1}{\delta}\sqrt{\frac{k}{m}}δ1​mk​​. The value of δ is ______.

Correct answer: 0.50 OR 3.13 TO 3.15

Step-by-step solution →
Q122·PhysicsSingle correctJEE Main 2020
An object of mass m is suspended at the end of a massless wire of length L and area of cross – selection, A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is:
  1. (A)f=12πmLYAf=\dfrac{1}{2\pi}\sqrt{\dfrac{mL}{YA}}f=2π1​YAmL​​
  2. (B)f=12πYAmLf=\dfrac{1}{2\pi}\sqrt{\dfrac{YA}{mL}}f=2π1​mLYA​​
  3. (C)f=12πmAYLf=\dfrac{1}{2\pi}\sqrt{\dfrac{mA}{YL}}f=2π1​YLmA​​
  4. (D)f=12πYLmAf=\dfrac{1}{2\pi}\sqrt{\dfrac{YL}{mA}}f=2π1​mAYL​​

Correct answer: (B)

Step-by-step solution →
Q123·PhysicsSingle correctJEE Main 2020
When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by y(t)=y0sin⁡2ωty(t) = y_{0}\sin^{2}\omega ty(t)=y0​sin2ωt, where 'y' is measured from the lower end of unstretched spring. Then ω is:
  1. (A)12gy0\frac{1}{2}\sqrt{\frac{g}{y_{0}}}21​y0​g​​
  2. (B)gy0\sqrt{\frac{g}{y_{0}}}y0​g​​
  3. (C)g2y0\sqrt{\frac{g}{2y_{0}}}2y0​g​​
  4. (D)2gy0\sqrt{\frac{2g}{y_{0}}}y0​2g​​

Correct answer: (C)

Step-by-step solution →
Q124·PhysicsSingle correctJEE Main 2020
A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period T1_11​ and, T(ii) back and forth in a direction perpendicular to its plane, with a period T2_22​. The ratio T1T2\dfrac{T_1}{T_2}T2​T1​​ will be:
  1. (A)23\dfrac{2}{\sqrt{3}}3​2​
  2. (B)33\dfrac{3}{\sqrt{3}}3​3​
  3. (C)23\dfrac{\sqrt{2}}{3}32​​
  4. (D)23\dfrac{2}{3}32​

Correct answer: (A)

Step-by-step solution →
Q125·PhysicsSingle correctJEE Main 2020
A block of mass m attached to a massless spring is performing oscillatory motion of amplitude 'A' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fAfAfA. The value of f is:
  1. (A)12\dfrac{1}{2}21​
  2. (B)12\dfrac{1}{\sqrt{2}}2​1​
  3. (C)2\sqrt{2}2​
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q126·PhysicsSingle correctJEE Main 2019
A spring whose unstretched length is ℓ\ellℓ has a force constant k. The spring is cut into two pieces of unstretched lengths ℓ1\ell_{1}ℓ1​ and ℓ2\ell_{2}ℓ2​ where, ℓ1=nℓ2\ell_{1} = n\ell_{2}ℓ1​=nℓ2​ and n is an integer. The ratio k1_{1}1​/k2_{2}2​ of the corresponding force constants, k1_{1}1​ and k2_{2}2​ will be:
  1. (A)n
  2. (B)1/n21/n^{2}1/n2
  3. (C)n2n^{2}n2
  4. (D)1/n1/n1/n

Correct answer: (D)

Step-by-step solution →
Q127·PhysicsSingle correctJEE Main 2019
A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. the time period of the pendulum is given by
  1. (A)2πLg2+(qEm)22π\sqrt{\dfrac{L}{\sqrt{g^{2} + \left(\dfrac{qE}{m}\right)^{2}}}}2πg2+(mqE​)2​L​​
  2. (B)2πL(g+qEm)2π\sqrt{\dfrac{L}{\left(g + \dfrac{qE}{m}\right)}}2π(g+mqE​)L​​
  3. (C)2πL(g−qEm)2π\sqrt{\dfrac{L}{\left(g - \dfrac{qE}{m}\right)}}2π(g−mqE​)L​​
  4. (D)2πLg2−q2E2m22π\sqrt{\dfrac{L}{\sqrt{g^{2} - \dfrac{q^{2}E^{2}}{m^{2}}}}}2πg2−m2q2E2​​L​​

Correct answer: (A)

Step-by-step solution →
Q128·PhysicsSingle correctJEE Main 2019
The displacement of a damped harmonic oscillator is given by x(t)=e−0.1tcos⁡(10πt+φ)x(t) = e^{-0.1t}\cos(10\pi t + \varphi)x(t)=e−0.1tcos(10πt+φ). Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:
  1. (A)13 s
  2. (B)27 S
  3. (C)4 s
  4. (D)7 s

Correct answer: (D)

Step-by-step solution →
Q129·PhysicsSingle correctJEE Main 2019
A simple pendulum oscillating in air has period T. The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is 116\dfrac{1}{16}161​th of the material of the bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is:
  1. (A)2T1102T\sqrt{\dfrac{1}{10}}2T101​​
  2. (B)2T1142T\sqrt{\dfrac{1}{14}}2T141​​
  3. (C)4T1154T\sqrt{\dfrac{1}{15}}4T151​​
  4. (D)4T1144T\sqrt{\dfrac{1}{14}}4T141​​

Correct answer: (C)

Step-by-step solution →
Q130·PhysicsSingle correctJEE Main 2019
A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 11000\frac{1}{1000}10001​ of the original amplitude is close to:
  1. (A)10s
  2. (B)100s
  3. (C)50s
  4. (D)20s

Correct answer: (D)

Step-by-step solution →
Q131·PhysicsSingle correctJEE Main 2019
A simple harmonic motion is represented by : y=5(sin⁡3πt+3cos⁡3πt)y = 5(\sin 3\pi t + \sqrt{3} \cos 3\pi t)y=5(sin3πt+3​cos3πt) cm The amplitude and time period of the motion are :
  1. (A)10 cm, 23\frac{2}{3}32​ s
  2. (B)10 cm, 32\frac{3}{2}23​ s
  3. (C)5 cm, 32\frac{3}{2}23​ s
  4. (D)5 cm, 23\frac{2}{3}32​ s

Correct answer: (A)

Step-by-step solution →
Q132·PhysicsSingle correctJEE Main 2019
Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length l and mass m. the rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:
  1. (A)12π3km\frac{1}{2\pi}\sqrt{\frac{3k}{m}}2π1​m3k​​
  2. (B)12π2km\frac{1}{2\pi}\sqrt{\frac{2k}{m}}2π1​m2k​​
  3. (C)12π6km\frac{1}{2\pi}\sqrt{\frac{6k}{m}}2π1​m6k​​
  4. (D)12πkm\frac{1}{2\pi}\sqrt{\frac{k}{m}}2π1​mk​​

Correct answer: (C)

Step-by-step solution →
Q133·PhysicsSingle correctJEE Main 2019
A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1_11​. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2_22​ then:
  1. (A)K2_22​ = 2K1_11​
  2. (B)K2_22​ = K12\frac{K_1}{2}2K1​​
  3. (C)K2_22​ = K14\frac{K_1}{4}4K1​​
  4. (D)K2_22​ = K1_11​

Correct answer: (A)

Step-by-step solution →
Q134·PhysicsSingle correctJEE Main 2019
A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential energy (PE) equals kinetic energy (KE), the position of the particle will be
  1. (A)A2\dfrac{A}{2}2A​
  2. (B)A22\dfrac{A}{2\sqrt{2}}22​A​
  3. (C)A2\dfrac{A}{\sqrt{2}}2​A​
  4. (D)A

Correct answer: (C)

Step-by-step solution →
Q135·PhysicsSingle correctJEE Main 2019
A rod of mass ‘M’ and length ‘2L’ is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of ‘m’ are attached at distance ‘L/2’ from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to:
  1. (A)0.77
  2. (B)0.57
  3. (C)0.37
  4. (D)0.17

Correct answer: (C)

Step-by-step solution →
Q136·PhysicsSingle correctJEE Main 2019
A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in a equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is
  1. (A)2Fmk\dfrac{2F}{\sqrt{mk}}mk​2F​
  2. (B)Fπmk\dfrac{F}{\pi\sqrt{mk}}πmk​F​
  3. (C)πFmk\dfrac{\pi F}{\sqrt{mk}}mk​πF​
  4. (D)Fmk\dfrac{F}{\sqrt{mk}}mk​F​

Correct answer: (D)

Step-by-step solution →
Q137·PhysicsSingle correctJEE Main 2019
Two masses mmm and m2\dfrac{m}{2}2m​ are connected at the two ends of a massless rigid rod of length ℓ\ellℓ. The rod is suspended by a thin wire of torsional constant kkk at the centre of mass of the rod-mass system (see figure). Because of torsional constant kkk, the restoring torque is τ=kθ\tau = k\thetaτ=kθ for angular displacement θ\thetaθ. If the rod is rotated by θ0\theta_0θ0​ and released, the tension in it when it passes through its mean position will be
  1. (A)3kθ02ℓ\dfrac{3k\theta_0^2}{\ell}ℓ3kθ02​​
  2. (B)2kθ02ℓ\dfrac{2k\theta_0^2}{\ell}ℓ2kθ02​​
  3. (C)kθ02ℓ\dfrac{k\theta_0^2}{\ell}ℓkθ02​​
  4. (D)kθ022ℓ\dfrac{k\theta_0^2}{2\ell}2ℓkθ02​​

Correct answer: (C)

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Q138·PhysicsNumericalJEE Advanced 2018
A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0 N m−12.0\ N\,m^{-1}2.0 Nm−1 and the mass of the block is 2.0 kg2.0\ kg2.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass 1.0 kg1.0\ kg1.0 kg moving with a speed of 2.0 m s−12.0\ m\ s^{-1}2.0 m s−1 collides elastically with the first block. The collision is such that the 2.0 kg2.0\ kg2.0 kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time is __________.

Correct answer: 2.09

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Q139·PhysicsMultiple correctJEE Advanced 2016
A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x0x_{0}x0​. Consider two cases: (i) when the block is at x0x_{0}x0​ ; and (ii) when the block is at x=x0+Ax = x_{0} + Ax=x0​+A. In both the cases, a particle with mass m (< M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass m is placed on the mass M?
  1. (A)The amplitude of oscillation in the first case changes by a factor of Mm+M\sqrt{\frac{M}{m+M}}m+MM​​, whereas in the second case it remains unchanged
  2. (B)The final time period of oscillation in both the cases is same
  3. (C)The total energy decreases in both the cases
  4. (D)The instantaneous speed at x0x_{0}x0​ of the combined masses decreases in both the cases

Correct answer: (A), (B), (D)

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Q140·PhysicsMultiple correctJEE Advanced 2015
Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω1\omega_{1}ω1​ and ω2\omega_{2}ω2​ and have total energies E1E_{1}E1​ and E2E_{2}E2​, respectively. The variations of their momenta ppp with positions xxx are shown in the figures. If ab=n2\dfrac{a}{b} = n^{2}ba​=n2 and aR=n\dfrac{a}{R} = nRa​=n, then the correct equation(s) is(are)
  1. (A)E1ω1=E2ω2E_{1}\omega_{1} = E_{2}\omega_{2}E1​ω1​=E2​ω2​
  2. (B)ω2ω1=n2\dfrac{\omega_{2}}{\omega_{1}} = n^{2}ω1​ω2​​=n2
  3. (C)ω1ω2=n2\omega_{1}\omega_{2} = n^{2}ω1​ω2​=n2
  4. (D)E1ω1=E2ω2\dfrac{E_{1}}{\omega_{1}} = \dfrac{E_{2}}{\omega_{2}}ω1​E1​​=ω2​E2​​

Correct answer: (B), (D)

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Q141·PhysicsMultiple correctJEE Advanced 2013
A particle of mass mmm is attached to one end of a mass-less spring of force constant kkk, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t=0t = 0t=0 with an initial velocity u0u_{0}u0​. When the speed of the particle is 0.5u00.5u_{0}0.5u0​. It collides elastically with a rigid wall. After this collision,
  1. (A)the speed of the particle when it returns to its equilibrium position is u0u_{0}u0​.
  2. (B)the time at which the particle passes through the equilibrium position for the first time is t=πmkt = \pi\sqrt{\frac{m}{k}}t=πkm​​.
  3. (C)the time at which the maximum compression of the spring occurs is t=4π3mkt = \frac{4\pi}{3}\sqrt{\frac{m}{k}}t=34π​km​​.
  4. (D)the time at which the particle passes through the equilibrium position for the second time is t=5π3mkt = \frac{5\pi}{3}\sqrt{\frac{m}{k}}t=35π​km​​.

Correct answer: (A), (D)

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Oscillations — frequently asked

How many questions from Oscillations appear in JEE?

Oscillations has appeared in 117 of the last 186 JEE Main and JEE Advanced papers — about 63% of them — contributing 141 questions in total across those papers.

Is Oscillations an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 63% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Oscillations questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Physics chapters

  • Properties of Solids and Liquids 344
  • Current Electricity 309
  • Rotational Motion 266
  • Geometrical Optics 259
  • Kinematics 255
  • Magnetic Field of Current 211
  • Electromagnetic Waves 208
  • Thermodynamics 201

All 28 Physics chapters →

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