Question:
A point mass of \(400\ \text{g}\) executes S.H.M. under a force
\[
F=-(10\ \text{Nm}^{-1})x
\]
If it crosses the centre of oscillation with a speed of \(10\ \text{ms}^{-1}\), the amplitude of motion is:
A point mass of \(400\ \text{g}\) executes S.H.M. under a force
\[
F=-(10\ \text{Nm}^{-1})x
\]
If it crosses the centre of oscillation with a speed of \(10\ \text{ms}^{-1}\), the amplitude of motion is:
Show Hint
In S.H.M.,
\[
F=-kx,\quad \omega=\sqrt{\frac{k}{m}}
\]
and the maximum speed at mean position is
\[
v_{\max}=\omega A
\]
Use these formulas together to find amplitude.
Updated On: Jun 25, 2026
- \(2\ \text{m}\)
- \(4\ \text{m}\)
- \(0.4\ \text{m}\)
- \(0.5\ \text{m}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Identify the given quantities.
Mass of the particle: \[ m=400\ \text{g}=0.4\ \text{kg} \] Restoring force: \[ F=-kx \] Given: \[ k=10\ \text{Nm}^{-1} \] Speed at the centre of oscillation: \[ v_{\max}=10\ \text{ms}^{-1} \] In S.H.M., speed is maximum at the mean position.
Step 2: Find angular frequency.
For simple harmonic motion, \[ \omega=\sqrt{\frac{k}{m}} \] Substituting values, \[ \omega=\sqrt{\frac{10}{0.4}} \] \[ =\sqrt{25} \] \[ =5\ \text{rad/s} \]
Step 3: Use the formula for maximum speed in S.H.M.
In S.H.M., \[ v_{\max}=\omega A \] where \[ A=\text{amplitude} \] So, \[ A=\frac{v_{\max}}{\omega} \] Substituting values, \[ A=\frac{10}{5} \] \[ A=2\ \text{m} \]
Step 4: Final conclusion.
Hence, the amplitude of motion is \[ \boxed{2\ \text{m}} \]
Mass of the particle: \[ m=400\ \text{g}=0.4\ \text{kg} \] Restoring force: \[ F=-kx \] Given: \[ k=10\ \text{Nm}^{-1} \] Speed at the centre of oscillation: \[ v_{\max}=10\ \text{ms}^{-1} \] In S.H.M., speed is maximum at the mean position.
Step 2: Find angular frequency.
For simple harmonic motion, \[ \omega=\sqrt{\frac{k}{m}} \] Substituting values, \[ \omega=\sqrt{\frac{10}{0.4}} \] \[ =\sqrt{25} \] \[ =5\ \text{rad/s} \]
Step 3: Use the formula for maximum speed in S.H.M.
In S.H.M., \[ v_{\max}=\omega A \] where \[ A=\text{amplitude} \] So, \[ A=\frac{v_{\max}}{\omega} \] Substituting values, \[ A=\frac{10}{5} \] \[ A=2\ \text{m} \]
Step 4: Final conclusion.
Hence, the amplitude of motion is \[ \boxed{2\ \text{m}} \]
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