Question

A body of mass 0.2kg is executing shmwith an amplitude of 2pcm. The maximum force which acts upon it is 0.064N. Calculate it's maximum velocity and period of oscillation.

Answer 1

Answer

Force

$F=mAw^2=0.064N$

So

$Aw^2=0.32$

maximum velocity

$V=Aw=\sqrt{Aw^2 A}\\=\sqrt{0.32*0.02}\\=0.08m/s$

Period of oscillation

$T=\frac{2\pi}{w}=\pi /2$