Answer to Question #302567 in Mechanics | Relativity for Zainy

Question #302567

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.

1
Expert's answer
2022-03-03T12:05:38-0500

Answer

Force

F=mAw2=0.064NF=mAw^2=0.064N

So

Aw2=0.32Aw^2=0.32

maximum velocity

V=Aw=Aw2A=0.320.02=0.08m/sV=Aw=\sqrt{Aw^2 A}\\=\sqrt{0.32*0.02}\\=0.08m/s

Period of oscillation


T=2πw=π/2T=\frac{2\pi}{w}=\pi /2





Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment