JJEEPrep.app

Physics

Oscillations — JEE Main PYQs

28 previous year questions

Q1JEE Main 2026 Apr 2 Shift 2Medium

The equation of motion of a particle is given by x=asin⁔(50t+Ļ€3)x = a\sin\left(50t + \frac{\pi}{3}\right) cm. The particle will come to rest at time t1t_1 and it will have zero acceleration at time t2t_2. The t1t_1 and t2t_2 respectively are ________.

  1. A

    π300\frac{\pi}{300} s, π75\frac{\pi}{75} s

  2. B

    π75\frac{\pi}{75} s, π300\frac{\pi}{300} s

  3. C

    π300\frac{\pi}{300} s, π25\frac{\pi}{25} s

  4. D

    π50\frac{\pi}{50} s, π100\frac{\pi}{100} s

Show answer

Correct option: A

Q2JEE Main 2026 Apr 4 Shift 1MediumNumerical

The velocity of a particle executing simple harmonic motion along xx-axis is described as v2=50āˆ’x2v^2 = 50 - x^2, where xx represents displacement. If the time period of motion is x7\frac{x}{7} s, the value of xx is _________.

Show answer

Answer: 44

Q3JEE Main 2026 Apr 4 Shift 2Medium

A uniform disc of radius RR and mass MM is free to oscillate about the axis AA as shown in the figure. For small oscillations the time period is __________.

(gg is acceleration due to gravity)

[Figure: A uniform disc of mass M and radius R hanging from a horizontal axis A that is tangent to the disc at its topmost point; the disc oscillates about this axis.]

  1. A

    2Ļ€5R4g2\pi\sqrt{\frac{5R}{4g}}

  2. B

    2Ļ€2R3g2\pi\sqrt{\frac{2R}{3g}}

  3. C

    2Ļ€3R2g2\pi\sqrt{\frac{3R}{2g}}

  4. D

    2Ļ€3Rg2\pi\sqrt{\frac{3R}{g}}

Show answer

Correct option: A

Q4JEE Main 2026 Apr 5 Shift 2Medium

Match List - I with List - II.

List - I | List - II

A. sin⁔2ωt\sin^2 \omega t — I. Periodic with time period T=πωT = \dfrac{\pi}{\omega} but not simple harmonic motion (SHM)

B. sin⁔3(2ωt)\sin^3 (2\omega t) — II. Periodic with time period T=2πωT = \dfrac{2\pi}{\omega} but Not SHM

C. sin⁔(ωt)+cos⁔(πωt)\sin(\omega t) + \cos(\pi \omega t) — III. Periodic with time period T=πωT = \dfrac{\pi}{\omega} and SHM

D. cos⁔ωt+cos⁔2ωt\cos \omega t + \cos 2\omega t — IV. Non-periodic

Choose the correct answer from the options given below :

  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

Show answer

Correct option: A

Q5JEE Main 2026 Apr 6 Shift 1Medium

A particle is executing simple harmonic motion. Its amplitude is AA and time period is 5 sec. The time required by it to move from x=Ax = A to x=A2x = \frac{A}{\sqrt{2}} is ________ sec.

  1. A

    1/41/4

  2. B

    5/45/4

  3. C

    5/85/8

  4. D

    3/83/8

Show answer

Correct option: C

Q6JEE Main 2026 Apr 6 Shift 2Medium

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=10g = 10 m/s2^2)

  1. A

    550Ļ€\dfrac{5\sqrt{50}}{\pi} and 8Ɨ10āˆ’38 \times 10^{-3}

  2. B

    550Ļ€\dfrac{5\sqrt{50}}{\pi} and 88

  3. C

    105010\sqrt{50} and 2Ɨ10āˆ’32 \times 10^{-3}

  4. D

    550Ļ€\dfrac{5\sqrt{50}}{\pi} and 16Ɨ10āˆ’316 \times 10^{-3}

Show answer

Correct option: A

Q7JEE Main 2026 Apr 8 Shift 2Easy

The frequency of oscillation of a mass mm suspended by a spring is ν1\nu_1. If the length of the spring is cut to half, the same mass oscillates with frequency ν2\nu_2. The value of ν2/ν1\nu_2/\nu_1 is ________.

  1. A

    11

  2. B

    22

  3. C

    2\sqrt{2}

  4. D

    3\sqrt{3}

Show answer

Correct option: C

Q8JEE Main 2025 Apr 2 Shift 1Medium

A particle is subjected to two simple harmonic motions as :

x1=7sin⁔5t cmx_1=\sqrt{7}\sin 5t~\mathrm{cm}

and x2=27sin⁔(5t+Ļ€3)cmx_2=2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right)\mathrm{cm}

where xx is displacement and tt is time in seconds. The maximum acceleration of the particle is xƗ10āˆ’2Ā msāˆ’2x\times10^{-2}~\mathrm{ms^{-2}}. The value of xx is :

  1. A

    25725\sqrt{7}

  2. B

    575\sqrt{7}

  3. C

    125125

  4. D

    175175

Show answer

Correct option: D

Q9JEE Main 2025 Apr 3 Shift 1Hard

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

(μ\mu = coefficient of friction between the two blocks)

[Figure: block M on a frictionless table connected to a wall by a spring of constant k; smaller block m rests on top of M with friction coefficient μ\mu between them]

A. The time period of small oscillation of the two blocks is T=2Ļ€(m+M)kT = 2\pi\sqrt{\frac{(m+M)}{k}}

B. The acceleration of the blocks is a=āˆ’kxM+ma = -\frac{kx}{M+m} (xx = displacement of the blocks from the mean position)

C. The magnitude of the frictional force on the upper block is mμ∣x∣M+m\frac{m\mu|x|}{M+m}

D. The maximum amplitude of the upper block, if it does not slip, is μ(M+m)gk\frac{\mu(M+m)g}{k}

E. Maximum frictional force can be μ(M+m)g\mu(M + m)g.

Choose the correct answer from the options given below:

  1. A

    A, B, C Only

  2. B

    B, C, D Only

  3. C

    C, D, E Only

  4. D

    A, B, D Only

Show answer

Correct option: D

Q10JEE Main 2025 Jan 22 Shift 2Medium

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

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.

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 the correct explanation of (A)

  2. B

    Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

  3. C

    (A) is true but (R) is false

  4. D

    (A) is false but (R) is true

Show answer

Correct option: B

Q11JEE Main 2025 Jan 23 Shift 1Medium

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×10āˆ’2y\pi \times 10^{-2} s, where the value of yy is

(Acceleration due to gravity, g=10Ā m/s2g = 10\ \mathrm{m/s^2}, density of water =103Ā kg/m3= 10^3\ \mathrm{kg/m^3})

  1. A

    [Option lost due to PDF corruption]

  2. B

    [Option lost due to PDF corruption]

  3. C

    [Option lost due to PDF corruption]

  4. D

    [Option lost due to PDF corruption]

Show answer

Correct option: D

Q12JEE Main 2025 Jan 24 Shift 1Medium

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 Dd\frac{D}{d} is

  1. A

    154\frac{15}{4}

  2. B

    2525

  3. C

    165\frac{16}{5}

  4. D

    1010

Show answer

Correct option: B

Q13JEE Main 2025 Jan 24 Shift 2Medium

A particle oscillates along the xx-axis according to the law, x(t)=x0sin⁔2(t2)x(t) = x_0 \sin^2\left(\frac{t}{2}\right) where x0=1x_0 = 1 m. The kinetic energy (K) of the particle as a function of xx is correctly represented by the graph

  1. A

    [Graph: K vs x; K rises smoothly from zero at x = 0 to a maximum at x = ½ and falls to zero at x = 1 (inverted-parabola-like arch)]

  2. B

    [Graph: K vs x; K is zero at x = 0 and x = 1 with a sharp cusp-like peak at x = ½]

  3. C

    [Graph: K vs x; K is maximum at x = 0 and x = 1 and dips to zero at x = ½ with a sharp cusp]

  4. D

    [Graph: K vs x; K is maximum at x = 0 and x = 1 and decreases smoothly to zero at x = ½ (smooth U-shaped valley)]

Show answer

Correct option: A

Q14JEE Main 2025 Jan 28 Shift 2Medium

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_0 and initial momentum p0p_0 is enough to determine the position and momentum at any time t for a simple harmonic motion with a given angular frequency ω\omega.

Reason (R) : The amplitude and phase can be expressed in terms of x0x_0 and p0p_0.

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 the correct explanation of (A)

  2. B

    Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

  3. C

    (A) is true but (R) is false

  4. D

    (A) is false but (R) is true

Show answer

Correct option: A

Q15JEE Main 2024 Apr 4 Shift 2Medium

In simple harmonic motion, the total mechanical energy of given system is EE. If mass of oscillating particle PP is doubled then the new energy of the system for same amplitude is:

[Figure: a spring of stiffness kk hanging from a ceiling with a block of mass mm (particle PP) attached at its lower end]

  1. A

    E2E\sqrt{2}

  2. B

    E/2E/\sqrt{2}

  3. C

    2E2E

  4. D

    EE

Show answer

Correct option: D

Q16JEE Main 2024 Apr 4 Shift 2MediumNumerical

The displacement of a particle executing SHM is given by x=10sin⁔(wt+Ļ€3)mx = 10\sin\left(wt + \frac{\pi}{3}\right) m. The time period of motion is 3.14 s. The velocity of the particle at t=0t = 0 is ______ m/sm/s.

Show answer

Answer: 10

Q17JEE Main 2024 Apr 5 Shift 1Medium

A simple pendulum doing small oscillations at a place R height above earth surface has time period of T1=4T_1 = 4 s. T2T_2 would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of earth] :

  1. A

    T1=T2T_1 = T_2

  2. B

    2T1=T22T_1 = T_2

  3. C

    3T1=2T23T_1 = 2T_2

  4. D

    2T1=3T22T_1 = 3T_2

Show answer

Correct option: C

Q18JEE Main 2024 Apr 6 Shift 1EasyNumerical

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.

Show answer

Answer: 12

Q19JEE Main 2024 Apr 8 Shift 2MediumNumerical

An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of (25Ļ€)\left(\frac{25}{\pi}\right) Hz. At the position x=0.04x = 0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is _________ cm.

Show answer

Answer: 6

Q20JEE Main 2024 Apr 9 Shift 1MediumNumerical

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of 4 m, 2Ā msāˆ’12\ \mathrm{ms^{-1}} and 16Ā msāˆ’216\ \mathrm{ms^{-2}} at a certain instant. The amplitude of the motion is x\sqrt{x} m where xx is ________.

Show answer

Answer: 17

Q21JEE Main 2024 Apr 9 Shift 2MediumNumerical

A particle of mass 0.50 kgkg executes simple harmonic motion under force F=āˆ’50Ā (Nmāˆ’1) xF = -50~(\mathrm{Nm^{-1}})\,x. The time period of oscillation is x35\frac{x}{35} s. The value of xx is ________ .

(Given π=227\pi = \frac{22}{7})

Show answer

Answer: 22

Q22JEE Main 2024 Feb 1 Shift 1Medium

If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g=Ļ€2Ā m/s2g = \pi^2\ \mathrm{m/s^2}, then the length of the second's pendulum at a height h = 2R from the surface of earth will be, :

  1. A

    29\dfrac{2}{9} m

  2. B

    49\dfrac{4}{9} m

  3. C

    89\dfrac{8}{9} m

  4. D

    19\dfrac{1}{9} m

Show answer

Correct option: D

Q23JEE Main 2024 Feb 1 Shift 2EasyNumerical

A mass m is suspended from a spring of negligible mass and the system oscillates with a frequency f1f_1. The frequency of oscillations if a mass 9 m is suspended from the same spring is f2f_2. The value of f1f2\frac{f_1}{f_2} is ________.

Show answer

Answer: 3

Q24JEE Main 2024 Jan 27 Shift 1MediumNumerical

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} cm, where α=\alpha = ________.

Show answer

Answer: 12

Q25JEE Main 2024 Jan 29 Shift 1EasyNumerical

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\frac{x}{8}, where x=x = ________.

Show answer

Answer: 9

Q26JEE Main 2024 Jan 30 Shift 2MediumNumerical

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 4m4m, then the time period of small oscillations will be _______ ss. [take g=Ļ€2Ā msāˆ’2g = \pi^2\ ms^{-2}]

Show answer

Answer: 8

Q27JEE Main 2024 Jan 31 Shift 1MediumNumerical

A particle performs simple harmonic motion with amplitude AA. Its speed is increased to three times at an instant when its displacement is 2A3\frac{2A}{3}. The new amplitude of motion is nA3\frac{nA}{3}. The value of nn is ________.

Show answer

Answer: 7

Q28JEE Main 2024 Jan 31 Shift 2MediumNumerical

The time period of simple harmonic motion of mass MM in the given figure is παM5k\pi\sqrt{\frac{\alpha M}{5k}}, where the value of α\alpha is ________.

[Figure: mass M suspended from the ceiling by a combination of springs — two springs of constant k in parallel, in series with another spring of constant k, all in parallel with a spring of constant k]

Show answer

Answer: 12