Question:
A particle executes simple harmonic motion with a time period \(0.6\text{ s}\) and amplitude \(10\text{ cm}\). Then the mean velocity of the particle over the time interval during which it travels a distance \(5\text{ cm}\) starting from the equilibrium position is
A particle executes simple harmonic motion with a time period \(0.6\text{ s}\) and amplitude \(10\text{ cm}\). Then the mean velocity of the particle over the time interval during which it travels a distance \(5\text{ cm}\) starting from the equilibrium position is
Show Hint
For SHM starting from equilibrium,
\[
x=A\sin\omega t
\]
is usually the most convenient form.
Updated On: Jun 25, 2026
- \(1\text{ ms}^{-1}\)
- \(50\text{ cm s}^{-1}\)
- \(10\text{ cm s}^{-1}\)
- \(1\text{ cm s}^{-1}\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the SHM equation.
For SHM, \[ x=A\sin\omega t \] Given: \[ A=10\text{ cm}=0.1\text{ m} \] and \[ T=0.6\text{ s} \] Angular frequency: \[ \omega=\frac{2\pi}{T} \] \[ \omega=\frac{2\pi}{0.6} = \frac{10\pi}{3} \]
Step 2: Find the time to travel \(5\text{ cm}\).
Starting from equilibrium position: \[ x=5\text{ cm}=0.05\text{ m} \] Using \[ x=A\sin\omega t, \] we get \[ 0.05=0.1\sin\omega t \] \[ \sin\omega t=\frac12 \] Therefore, \[ \omega t=\frac{\pi}{6} \] Thus, \[ t=\frac{\pi/6}{10\pi/3} \] \[ t=\frac{\pi}{6}\cdot \frac{3}{10\pi} \] \[ t=\frac{1}{20}\text{ s} \] \[ t=0.05\text{ s} \]
Step 3: Calculate mean velocity.
Mean velocity: \[ v_{\text{mean}}=\frac{\text{displacement}}{\text{time}} \] Displacement from equilibrium to \(5\text{ cm}\): \[ =0.05\text{ m} \] Hence, \[ v_{\text{mean}}=\frac{0.05}{0.05} \] \[ v_{\text{mean}}=1\text{ ms}^{-1} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{1\text{ ms}^{-1}} \]
For SHM, \[ x=A\sin\omega t \] Given: \[ A=10\text{ cm}=0.1\text{ m} \] and \[ T=0.6\text{ s} \] Angular frequency: \[ \omega=\frac{2\pi}{T} \] \[ \omega=\frac{2\pi}{0.6} = \frac{10\pi}{3} \]
Step 2: Find the time to travel \(5\text{ cm}\).
Starting from equilibrium position: \[ x=5\text{ cm}=0.05\text{ m} \] Using \[ x=A\sin\omega t, \] we get \[ 0.05=0.1\sin\omega t \] \[ \sin\omega t=\frac12 \] Therefore, \[ \omega t=\frac{\pi}{6} \] Thus, \[ t=\frac{\pi/6}{10\pi/3} \] \[ t=\frac{\pi}{6}\cdot \frac{3}{10\pi} \] \[ t=\frac{1}{20}\text{ s} \] \[ t=0.05\text{ s} \]
Step 3: Calculate mean velocity.
Mean velocity: \[ v_{\text{mean}}=\frac{\text{displacement}}{\text{time}} \] Displacement from equilibrium to \(5\text{ cm}\): \[ =0.05\text{ m} \] Hence, \[ v_{\text{mean}}=\frac{0.05}{0.05} \] \[ v_{\text{mean}}=1\text{ ms}^{-1} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{1\text{ ms}^{-1}} \]
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