Question

A particle is executing SHM on a straight line with zero initial phase. It crosses the point from where the time is considered for the motion at successive intervals t and 2t with a speed v. Find the amplitude of the motion:

• A. $2vt$
• B. $vt$
• C. $\frac{vt}{\pi }$
• D. $\frac{vt}{2\pi }$

Answer 1

The correct option is C

$\frac{vt}{\pi }$

Initial phase $\varphi =0$

Thus the point from where time is considered, is origin.

$T=(t_{OA}+t_{AO})+(t_{OB}+t_{BO})$

$=t+t=2t=\frac{2\pi }{\omega }$

$\omega =\frac{\pi }{t}$

$x=asin\omega t=asin\pi =0$

$v=a\omega cos\omega t=-a\omega =-\frac{a\pi }{t}$

Amplitude = $|a|=\frac{vt}{\pi }$