Question

A particle executes SHM with a time period of $16$s. At time $t=2s$, the particle crosses the mean position while at $t=4s$, its velocity is $4m{s}^{-1}$. The amplitude of motion in metre is?

• A. $\sqrt{2}\pi$
• B. $16\sqrt{2}\pi$
• C. $32\sqrt{2}/\pi$
• D. $4/\pi$

Answer 1

Answer: (C)

$32\sqrt{2}/\pi$

$x=\alpha sin(\frac{2\pi }{T}t+\varphi )$

$t=2s,x=0,T=16s$

$0=\alpha sin(\frac{\pi }{4}+\varphi )$

$\varphi =-\frac{\pi }{4}$

equ of shm is

$x=asin(\frac{2\pi }{T}t-\frac{\pi }{4})$

$t=4s,v=4m/s$

$v=\frac{dx}{dt}=a\times \frac{2\pi }{t}cos(\frac{2\pi }{T}t-\frac{\pi }{4})$

$4=\alpha \times \frac{2\pi }{16}cos(\frac{\pi }{2}-\frac{\pi }{4})$

$4=\alpha \times \frac{2\pi }{16}\times \frac{\pi }{8}\times \frac{1}{\sqrt{2}}$

$\alpha =\frac{32\sqrt{2}}{\pi }$