Question

A particle executes SHM with amplitude \( A \) and time period \( T \). The time taken to move from equilibrium to \( \frac{A}{2} \) is:

• A. \(\frac{T}{12}\)
• B. \(\frac{T}{8}\)
• C. \(\frac{T}{6}\)
• D. \(\frac{T}{4}\)

Hint:

In SHM, use \( x = A \sin \omega t \) and solve for \( t \) when given displacement. Recall \( \omega = \frac{2\pi}{T} \).

Answer 1

The Correct Option is C

In simple harmonic motion (SHM), the displacement \( x \) at time \( t \) is given by: \[ x = A \sin \omega t \] where \( \omega = \frac{2\pi}{T} \) is the angular frequency. Given that the particle moves from equilibrium (i.e., \( x=0 \) at \( t=0 \)) to \( x = \frac{A}{2} \) in time \( t \): \[ \frac{A}{2} = A \sin \omega t \implies \sin \omega t = \frac{1}{2} \] Therefore, \[ \omega t = \sin^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{6} \] Recall \(\omega = \frac{2\pi}{T}\), so: \[ \frac{2\pi}{T} t = \frac{\pi}{6} \implies t = \frac{\pi}{6} \times \frac{T}{2\pi} = \frac{T}{12} \] However, this is the time taken to reach \( +\frac{A}{2} \) for the first time after starting at 0 (equilibrium). Note: The particle moves from 0 to \( \frac{A}{2} \) within this time. Hence, the time taken is: \[ t = \frac{T}{12} \] This matches option (A). Wait, the options given in the question are different from the final result. The correct time from equilibrium to \( \frac{A}{2} \) is \( \frac{T}{12} \). If your options differ, the answer is option (A) \(\frac{T}{12}\).