Question:
A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec . The time required by it to move from $x=A$ to $x=\frac{A}{\sqrt{2}}$ is ________ sec.
A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec . The time required by it to move from $x=A$ to $x=\frac{A}{\sqrt{2}}$ is ________ sec.
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Write $x = A\cos\omega t$ and find the time at which $x = A/\sqrt{2}$; compare it with the time period $T$.
Updated On: Oct 26, 2026
- 1/4
- 5/4
- 5/8
- 3/8
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The Correct Option is C
Solution and Explanation
Since the particle starts from the extreme position $x=A$ at $t=0$, its displacement is $x = A\cos\omega t$.
The angular frequency is $\omega = \frac{2\pi}{T}$.
At $x = \frac{A}{\sqrt{2}}$ we need $\cos\omega t = \frac{1}{\sqrt{2}}$, so $\omega t = \frac{\pi}{4}$.
Therefore $t = \frac{\pi}{4\omega} = \frac{\pi}{4}\cdot\frac{T}{2\pi} = \frac{T}{8}$.
With $T = 5$ s, $t = \frac{5}{8}$ s.
The correct option is (3) $5/8$.
The angular frequency is $\omega = \frac{2\pi}{T}$.
At $x = \frac{A}{\sqrt{2}}$ we need $\cos\omega t = \frac{1}{\sqrt{2}}$, so $\omega t = \frac{\pi}{4}$.
Therefore $t = \frac{\pi}{4\omega} = \frac{\pi}{4}\cdot\frac{T}{2\pi} = \frac{T}{8}$.
With $T = 5$ s, $t = \frac{5}{8}$ s.
The correct option is (3) $5/8$.
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