NEET Oscillations PYQs with Detailed Solutions | 2013–2025

NEET Oscillations PYQs — Solutions (2013–2025)

2025

Q1: A box containing sand oscillates on a spring. As the sand leaks slowly, how do the amplitude and frequency of oscillation change over time?

Solution:

• For a spring–mass system:
  • Angular frequency: $\omega = \sqrt{\frac{k}{m}}$ω=mk​​
  • Amplitude depends on initial displacement; if no external force, amplitude stays constant (ideal spring).
• As mass decreases (sand leaks):
  • Frequency increases because $\omega = \sqrt{k/m}$ω=k/m​ → smaller m ⇒ larger ω
  • Amplitude remains approximately constant.
    ✅ Answer: Frequency increases, amplitude roughly constant.

Q2: Two masses attached to separate springs move with equal maximum speed. Determine the ratio of their amplitudes.

Solution:

• Maximum speed in SHM: $v_{max} = \omega A$vmax​=ωA
• Given $v_{max1} = v_{max2}$vmax1​=vmax2​, so $\omega_1 A_1 = \omega_2 A_2$ω1​A1​=ω2​A2​
• Angular frequency: $\omega = \sqrt{k/m}$ω=k/m​
• Therefore, $A_1 / A_2 = \omega_2 / \omega_1 = \sqrt{m_2/k_2} / \sqrt{m_1/k_1}$A1​/A2​=ω2​/ω1​=m2​/k2​​/m1​/k1​​
  ✅ Answer: $A_1 : A_2 = \sqrt{m_2/k_2} : \sqrt{m_1/k_1}$A1​:A2​=m2​/k2​​:m1​/k1​​

2024

Q3: A particle in SHM has equal kinetic and potential energy. At what displacement from the mean position does this happen?

Solution:

• Total energy: $E = \frac{1}{2} k A^2$E=21​kA2
• KE = PE → $\frac{1}{2} k (A^2 – x^2) = \frac{1}{2} k x^2$21​k(A2−x2)=21​kx2
• Solve: $A^2 – x^2 = x^2$A2−x2=x2 → $2x^2 = A^2$2×2=A2 → $x = \frac{A}{\sqrt{2}}$x=2​A​
  ✅ Answer: $x = A/\sqrt{2}$x=A/2​

Q4: Displacement: $x = 5 \sin(\pi t + \pi/3)$x=5sin(πt+π/3). Find amplitude and period.

Solution:

• Amplitude $A = 5$A=5
• Angular frequency: $\omega = \pi$ω=π rad/s
• Period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2\,s$T=ω2π​=π2π​=2s
  ✅ Answer: Amplitude = 5 units, Period = 2 s

Q5: Pendulum length halved, mass tripled. Find new period.

Solution:

• Pendulum formula: $T = 2\pi \sqrt{\frac{L}{g}}$T=2πgL​​
• Mass does not affect T
• New length $L’ = L/2$L′=L/2 → $T’ = 2\pi \sqrt{\frac{L/2}{g}} = T/\sqrt{2}$T′=2πgL/2​​=T/2​
  ✅ Answer: New period = $T/\sqrt{2}$T/2​

2023

Q6: Pendulum in liquid (density ¼ of bob). Period changes?

Solution:

• Effective mass decreases → period decreases slightly
• Formula (qualitative): $T = 2\pi \sqrt{\frac{L}{g_{\text{effective}}}}$T=2πgeffective​L​​
  ✅ Answer: Period slightly less than in air

Q7: Given x–t graph, find acceleration at a time.

Solution:

• SHM acceleration: $a = -\omega^2 x$a=−ω2x
• Find x at given t from graph → plug in formula
  ✅ Answer: $a = -\omega^2 x(t)$a=−ω2x(t)

2022

Q8: Match graphs to SHM situations → Use:

• PE max at extremes, KE max at mean
• Friction reduces amplitude (damped)
• Spring-mass vs pendulum distinction

Q9: Non-periodic function example → e.g., $y = t^2$y=t2 is not periodic

Q10: Two pendulums, lengths L₁ & L₂: meet again at mean after n vibrations →

• Use LCM method: Number of oscillations = LCM(T₁,T₂)/T₂

2021

Q11: Frequency of potential energy variation:

• In SHM: $E_{PE} = \frac{1}{2} k A^2 \sin^2(\omega t)$EPE​=21​kA2sin2(ωt)
• PE oscillates twice as fast → $f_{PE} = 2 f_{SHM}$fPE​=2fSHM​

Q12: Spring stretched 5 cm by 10 N, mass 2 kg. Find T:

• k = F/x = 10 / 0.05 = 200 N/m
• $T = 2\pi \sqrt{m/k} = 2\pi \sqrt{2/200} = 0.628 s$T=2πm/k​=2π2/200​=0.628s

2020

Q13: Phase difference between displacement and acceleration:

• $a = -\omega^2 x$a=−ω2x → Acceleration leads displacement by π (180°)

2017

Q14: Velocity = acceleration → Solve: $v = ω√(A^2-x^2), a = ω^2 x$v=ω√(A2−x2),a=ω2x → find x → then T

2016

Q15: Mass–spring period changes with mass → Solve $T_1 = 2π√(m/k), T_2 = 2π√((m+Δm)/k)$T1​=2π√(m/k),T2​=2π√((m+Δm)/k) → solve for m

2015

Q16: T in terms of α and β: $v_{max} = β, a_{max} = α → ω = α/β, T = 2π/ω$vmax​=β,amax​=α→ω=α/β,T=2π/ω

Q17: Superposition: $y = a \sin ω t + b \cos ω t = R \sin(ω t + φ)$y=asinωt+bcosωt=Rsin(ωt+φ) → Still SHM

Q18: Velocities at x₁, x₂ → use $v = ω√(A^2 – x^2)$v=ω√(A2−x2) → solve for ω → then T = 2π/ω

2014–2013 & Older

Phase differences, momentum graphs, acceleration graphs can be drawn using these equations

Use same SHM formulas:

$x = A \sin(ω t + φ)$x=Asin(ωt+φ)

$v = dx/dt = ω√(A^2 – x^2)$v=dx/dt=ω√(A2−x2)

$a = d^2x/dt^2 = -ω^2 x$a=d2x/dt2=−ω2x

KE = ½ m v², PE = ½ k x², E = constant