Skip to content
Question 62 of 66

Q.A pendulum is hung in a very high building, oscillates to and fro motion freely like a simple harmonic oscillator. If the acceleration of the bob is 16 ms^-2 at a distance of 4 m from the mean position, then the time period is:

(a) 2 pi s
(b) 2 s
(c) pi s
(d) 1 s
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2025MCQ· 1mImportance★★★★★
94% · 62/66 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

From a = omega^2 x with a=16 ms^-2 and x=4 m, omega=2 rad/s, giving a time period T = 2*pi/omega = pi seconds.

For simple harmonic motion, the acceleration is always directed toward the mean (equilibrium) position and its magnitude is proportional to the displacement from that position:

a = omega^2 x

where omega is the angular frequency of oscillation.

Given a = 16 ms^-2 when x = 4 m: …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.