Question

A particle is performing SHM of amplitude \(A\). Find the time required for particle to go from mean position to \( \dfrac{A}{\sqrt{2}} \). Time period of SHM is \(5\) sec :-

• A. \( \dfrac{5}{4} \) sec.
• B. \( \dfrac{5}{12} \) sec.
• C. \( \dfrac{5}{8} \) sec.
• D. \( \dfrac{5}{6} \) sec.

Answer 1

The Correct Option is C

Concept: In Simple Harmonic Motion (SHM), \[ x = A\sin\theta \] where \( \theta \) is the phase angle. Time corresponding to phase change is \[ t = \frac{\theta}{\omega} \] and \[ \omega = \frac{2\pi}{T} \] Step 1: Find phase corresponding to given positions.} At mean position, \[ x = 0 \Rightarrow \theta = 0 \] At \( x = \frac{A}{\sqrt{2}} \), \[ \frac{A}{\sqrt{2}} = A\sin\theta \] \[ \sin\theta = \frac{1}{\sqrt{2}} \] \[ \theta = \frac{\pi}{4} \] Thus phase covered \[ \Delta\theta = \frac{\pi}{4} \]
Step 2: Find time taken.} \[ t=\frac{\Delta\theta}{\omega} \] \[ t=\frac{\pi/4}{2\pi/T} \] \[ t=\frac{\pi}{4}\times\frac{T}{2\pi} \] \[ t=\frac{T}{8} \]
Step 3: Substitute \(T=5\) s.} \[ t=\frac{5}{8}\text{ sec} \] Final Result \[ t=\frac{5}{8}\text{ sec} \]