Question:
A particle is executing simple harmonic motion with time period \(T\) and amplitude \(A\). The distance travelled by the particle in \(\frac{T}{12}\) time starting from rest is:
A particle is executing simple harmonic motion with time period \(T\) and amplitude \(A\). The distance travelled by the particle in \(\frac{T}{12}\) time starting from rest is:
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For SHM starting from extreme position:
\[
x=A\cos\omega t
\]
and distance travelled from the extreme point is:
\[
A-x
\]
Always identify whether the motion starts from mean position or extreme position.
Updated On: Jun 17, 2026
- \( \dfrac{A(\sqrt{3}-\sqrt{2})}{2} \)
- \( \dfrac{A(2-\sqrt{3})}{2} \)
- \( \dfrac{2A}{2-\sqrt{3}} \)
- \( \dfrac{2A}{\sqrt{3}-\sqrt{2}} \)
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The Correct Option is B
Solution and Explanation
Concept:
In Simple Harmonic Motion (SHM), if a particle starts from the extreme position, its displacement at time \(t\) is:
\[
x=A\cos \omega t
\]
where
\[
\omega=\frac{2\pi}{T}
\]
The distance travelled from the extreme position is:
\[
A-x
\]
Step 1: Find angular frequency. \[ \omega=\frac{2\pi}{T} \] Given: \[ t=\frac{T}{12} \] Therefore, \[ \omega t=\frac{2\pi}{T}\times \frac{T}{12} =\frac{\pi}{6} \]
Step 2: Find displacement after \(\frac{T}{12}\) time. Since motion starts from extreme position, \[ x=A\cos\frac{\pi}{6} \] \[ x=A\left(\frac{\sqrt3}{2}\right) \] \[ x=\frac{\sqrt3 A}{2} \]
Step 3: Compute distance travelled. Initially particle is at: \[ x=A \] Final position: \[ x=\frac{\sqrt3A}{2} \] Hence distance travelled: \[ A-\frac{\sqrt3A}{2} \] \[ =\frac{2A-\sqrt3A}{2} \] \[ =\frac{A(2-\sqrt3)}{2} \] Hence the required answer is: \[ \boxed{\frac{A(2-\sqrt3)}{2}} \]
Step 1: Find angular frequency. \[ \omega=\frac{2\pi}{T} \] Given: \[ t=\frac{T}{12} \] Therefore, \[ \omega t=\frac{2\pi}{T}\times \frac{T}{12} =\frac{\pi}{6} \]
Step 2: Find displacement after \(\frac{T}{12}\) time. Since motion starts from extreme position, \[ x=A\cos\frac{\pi}{6} \] \[ x=A\left(\frac{\sqrt3}{2}\right) \] \[ x=\frac{\sqrt3 A}{2} \]
Step 3: Compute distance travelled. Initially particle is at: \[ x=A \] Final position: \[ x=\frac{\sqrt3A}{2} \] Hence distance travelled: \[ A-\frac{\sqrt3A}{2} \] \[ =\frac{2A-\sqrt3A}{2} \] \[ =\frac{A(2-\sqrt3)}{2} \] Hence the required answer is: \[ \boxed{\frac{A(2-\sqrt3)}{2}} \]
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