Question:

A particle, doing simple harmonic motion, at a distance \(3\,cm\) from mean position has acceleration \(12\,cm/s^2\). What is its time period?

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Use \( a = \omega^2 x \) to directly find angular frequency in SHM.
Updated On: Apr 16, 2026
  • \(0.5\,s\)
  • \(1\,s\)
  • \(2\,s\)
  • \(3.14\,s\)
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The Correct Option is D

Solution and Explanation

Concept: In SHM: \[ a = \omega^2 x \]

Step 1:
Substitute values.
\[ 12 = \omega^2 \cdot 3 \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2 \]

Step 2:
Find time period.
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi = 3.14\,s \]
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