Question:
A particle vibrating simple harmonically has an acceleration of \(16 \mathrm{~cm} / \mathrm{s}^2\), when it is at a distance of 4 cm from the mean position. Its periodic time is
A particle vibrating simple harmonically has an acceleration of \(16 \mathrm{~cm} / \mathrm{s}^2\), when it is at a distance of 4 cm from the mean position. Its periodic time is
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Use a = omega squared times x to find omega and then the period.
Updated On: Nov 13, 2026
1 s
2.572 s
3.142 s
6.028 s
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
In S.H.M. the acceleration is \(a=\omega^{2}x\).
Step 2: Find omega
\(\omega^{2}=\dfrac{16}{4}=4\Rightarrow\omega=2\ \text{rad/s}\).
Step 3: Time period
\(T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{2}=\pi\ \text{s}\approx3.142\ \text{s}\).
Final Answer:
The time period is about 3.142 s. \[ \boxed{\text{(C) }\text{3.142 s}} \]
In S.H.M. the acceleration is \(a=\omega^{2}x\).
Step 2: Find omega
\(\omega^{2}=\dfrac{16}{4}=4\Rightarrow\omega=2\ \text{rad/s}\).
Step 3: Time period
\(T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{2}=\pi\ \text{s}\approx3.142\ \text{s}\).
Final Answer:
The time period is about 3.142 s. \[ \boxed{\text{(C) }\text{3.142 s}} \]
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