Question:
For a particle executing simple harmonic motion with amplitude $A$ and time period $T$ along x-axis, the time taken by the particle to move from $x = 0$ to $x = A$ is
For a particle executing simple harmonic motion with amplitude $A$ and time period $T$ along x-axis, the time taken by the particle to move from $x = 0$ to $x = A$ is
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In SHM:
- Motion from mean position to extreme takes $\frac{T}{4}$
- Full cycle takes time $T$
Updated On: Apr 30, 2026
- $\frac{T}{2}$
- $\frac{T}{3}$
- $\frac{T}{4}$
- $\frac{T}{8}$
- $\frac{T}{6}$
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The Correct Option is C
Solution and Explanation
Concept:
In simple harmonic motion:
\[
x = A\sin(\omega t)
\]
Time period:
\[
T = \frac{2\pi}{\omega}
\]
Step 1: Use SHM equation.
\[ x = A\sin(\omega t) \] At $x = A$: \[ A = A\sin(\omega t) \Rightarrow \sin(\omega t) = 1 \]
Step 2: Find time.
\[ \omega t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{2\omega} \]
Step 3: Substitute $\omega = \frac{2\pi}{T}$.
\[ t = \frac{\pi}{2} \cdot \frac{T}{2\pi} = \frac{T}{4} \]
Step 1: Use SHM equation.
\[ x = A\sin(\omega t) \] At $x = A$: \[ A = A\sin(\omega t) \Rightarrow \sin(\omega t) = 1 \]
Step 2: Find time.
\[ \omega t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{2\omega} \]
Step 3: Substitute $\omega = \frac{2\pi}{T}$.
\[ t = \frac{\pi}{2} \cdot \frac{T}{2\pi} = \frac{T}{4} \]
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