Zaymiey

⚛️ Physics  ·  Oscillations  ·  NEET & JEE

A physical pendulum (rigid body) has MI I about pivot and its CM is at distance d. Its time period is:

Answer: 2π√(I/Mgd).

  • A 2π√(I/Mgd)
  • B 2π√(Mgd/I)
  • C 2π√(I/Md)
  • D 2π√(d/g)

Correct answer: A. 2π√(I/Mgd)

Explanation: T = 2π√(I/Mgd). For a simple pendulum, I = mL² and d = L, giving T = 2π√(L/g).

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.

Read the full Oscillations notes →