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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

The time period of a spring-mass system with spring constant k and mass m is:

Answer: 2π√(m/k).

  • A 2π√(m/k)
  • B 2π√(k/m)
  • C π√(m/k)
  • D 2πm/k

Correct answer: A. 2π√(m/k)

Explanation: T = 2π√(m/k). A stiffer spring (larger k) gives a shorter period; a heavier mass gives a longer period.

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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