IIT / JEE ADVANCED – ULTRA-EXTREME PROBLEMS

Oscillations & Simple Harmonic Motion (SET–5)


Problem 1: SHM with Time-Dependent Equilibrium (Topper Trap)

A mass m is attached to a spring of constant k and oscillates horizontally. The support of the spring moves as y = vt with constant velocity. Find the time period of oscillation observed from ground frame.

Solution:

Let x be displacement from instantaneous equilibrium. Restoring force depends only on extension relative to moving support.

Equation of motion:

m d²x/dt² = −kx

Hence time period remains:

T = 2π √(m/k)

Answer: T = 2π √(m/k)

Elite Insight:
Constant velocity changes position, not acceleration → no change in frequency.


Problem 2: SHM with Gradually Increasing Mass

A spring–mass system oscillates with mass increasing slowly due to rain falling uniformly. How does frequency vary with time?

Solution:

f = (1/2π) √(k/m)

As mass increases, frequency decreases gradually.

Answer: Frequency decreases with time

Exam Trap:
Students assume symmetry like sand leakage → opposite effect here.


Problem 3: Maximum Rate of Change of Energy

During SHM, at what displacement is the rate of change of kinetic energy maximum?

Solution:

Rate of change of KE = Power = Fv

F = kx, v = ω√(A² − x²)

P ∝ x√(A² − x²)

Maximum occurs at:

x = A/√2

Answer: x = A/√2


Problem 4: SHM Validity under Arbitrary Force

Force on a particle is given by:

F(x) = −k(x − x³)

Is the motion SHM for small oscillations?

Solution:

For small x, x³ ≪ x

So F ≈ −kx

Hence restoring force ∝ displacement.

Answer: Yes, SHM for small oscillations

Topper Rule:
Always expand force near equilibrium point.


Problem 5: Energy Ratio with Phase Velocity

At an instant, phase angle θ satisfies tanθ = 2. Find the ratio of kinetic to potential energy.

Solution:

In SHM:

KE/E = sin²θ
PE/E = cos²θ

tanθ = 2 ⇒ sin²θ / cos²θ = 4

Therefore:

KE : PE = 4 : 1

Answer: 4 : 1


Problem 6: Frequency Matching (Resonance Concept)

A system executes SHM with natural frequency ω₀. A periodic force of frequency ω is applied. For which condition does amplitude become maximum?

Answer: ω = ω₀

This is the condition of resonance.


SET–5 FINAL TAKEAWAYS

  • Acceleration (not velocity) controls oscillation nature
  • Mass variation directly alters frequency
  • Maximum rates occur at balance points
  • Small-x approximation is a powerful tool
  • Phase relations unlock energy ratios

IIT / JEE Tough Problems – SET–5 Completed ✅

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🔹 Simple Harmonic Motion (SHM) — Extended Series (30–56)

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