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⚛️ Physics  ·  Oscillations  ·  NEET & JEE

A pendulum of length L on a trolley that accelerates at a (horizontal) has effective gravity g<sub>eff</sub> = √(g²+a²). Its period is:

Answer: 2π√(L/√(g²+a²)).

  • A 2π√(L/√(g²+a²))
  • B 2π√(L/g)
  • C 2π√(L/(g+a))
  • D 2π√(L/(g-a))

Correct answer: A. 2π√(L/√(g²+a²))

Explanation: On an accelerating trolley, effective g is the vector sum of g (down) and -a (pseudo-force). g<sub>eff</sub> = √(g²+a²), directed at angle arctan(a/g) from vertical. T = 2π√(L/g<sub>eff</sub>).

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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