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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

A block of mass m on a spring of constant k executes SHM of amplitude A. As it passes the mean position an equal mass m is gently placed on it and sticks. The new amplitude is:

Answer: A/ 2.

  • A A/2
  • B A√2
  • C A/√2
  • D 2A

Correct answer: C. A/√2

Explanation: At the mean position speed v = Aω. Momentum conservation for the sudden sticking gives new speed v' = v/2 with mass 2m. New ω' = √(k/2m) = ω/√2, so A' = v'/ω' = (Aω/2)·(√2/ω) = A/√2.

Displacement, Velocity, Acceleration in SHMx(t)v(t)a(t)v leads x by 90°; a is exactly out of phase with x (a = −ω²x)

In SHM, displacement and velocity are 90° out of phase (v is maximum when x=0, and zero when x is extreme), while acceleration is always exactly opposite in sign to displacement, consistent with a = −ω²x.

Concept context

Simple harmonic motion, spring-mass, pendulum, energy in SHM, and resonance.

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