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
Force
F=mAw2=0.064NF=mAw^2=0.064NF=mAw2=0.064N
So
Aw2=0.32Aw^2=0.32Aw2=0.32
maximum velocity
V=Aw=Aw2A=0.32∗0.02=0.08m/sV=Aw=\sqrt{Aw^2 A}\\=\sqrt{0.32*0.02}\\=0.08m/sV=Aw=Aw2A=0.32∗0.02=0.08m/s
Period of oscillation
T=2πw=π/2T=\frac{2\pi}{w}=\pi /2T=w2π=π/2
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