Question

A particle is executing a simple harmonic motion with time period \(T\) and amplitude \(A\) and having the origin at \(O\). What is the time difference between its travel from \(O\) to \(A/2\) and from \(A/2\) to \(A\) on one side of the origin?

• A. \(T/2\)
• B. \(T/4\)
• C. \(T/6\)
• D. \(T/12\)

Hint:

Use \[ x=A\sin\theta \] and convert phase difference into time using \[ t=\frac{T}{2\pi}\theta. \] In SHM, equal distances are generally not covered in equal times.

Answer 1

The Correct Option is D

Concept: For SHM, \[ x=A\sin\theta \] and \[ t=\frac{T}{2\pi}\theta. \]

Step 1: Find the time taken from \(O\) to \(A/2\). At \[ x=\frac{A}{2}, \] \[ \sin\theta=\frac12. \] Hence, \[ \theta=\frac{\pi}{6}. \] Therefore, \[ t_1 = \frac{T}{2\pi} \cdot \frac{\pi}{6} = \frac{T}{12}. \]

Step 2: Find the time taken from \(A/2\) to \(A\). At \[ x=A, \] \[ \theta=\frac{\pi}{2}. \] Thus, \[ t_2 = \frac{T}{2\pi} \left( \frac{\pi}{2} - \frac{\pi}{6} \right) \] \[ = \frac{T}{2\pi} \cdot \frac{\pi}{3} \] \[ = \frac{T}{6}. \]

Step 3: Calculate the difference. \[ t_2-t_1 = \frac{T}{6} - \frac{T}{12} \] \[ = \frac{T}{12}. \] \[\begin{aligned} \boxed{\frac{T}{12}} \end{aligned}\] Hence, option \(\mathbf{(D)}\) is correct.