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. \( \dfrac{A(\sqrt{3}-\sqrt{2})}{2} \)
• B. \( \dfrac{A(2-\sqrt{3})}{2} \)
• C. \( \dfrac{2A}{2-\sqrt{3}} \)
• D. \( \dfrac{2A}{\sqrt{3}-\sqrt{2}} \)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Pick the right starting equation.
The particle starts from rest. In SHM, rest happens at the extreme end. When motion begins from the extreme, the displacement follows a cosine curve. \[ x = A\cos\omega t \]

Step 2: Find the angular frequency.
\[ \omega = \frac{2\pi}{T} \]
Step 3: Work out the phase at the given time.
The time given is $t = \tfrac{T}{12}$. \[ \omega t = \frac{2\pi}{T} \times \frac{T}{12} = \frac{\pi}{6} \]
Step 4: Find the new position.
\[ x = A\cos\frac{\pi}{6} = A\left(\frac{\sqrt3}{2}\right) = \frac{\sqrt3 A}{2} \]
Step 5: Get the distance travelled.
It started at $x = A$ and moved inward to $x = \tfrac{\sqrt3 A}{2}$. The distance is the difference. \[ A - \frac{\sqrt3 A}{2} = \frac{2A - \sqrt3 A}{2} \]
Step 6: Write the final answer.
\[ \frac{A(2 - \sqrt3)}{2} \] \[ \boxed{\dfrac{A(2-\sqrt3)}{2}} \]