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a piston executes simple harmonic motion with an amplitude of 0.1m. If it passes through the center of it's motion with a speed of 0.5 m/s, what is the period of oscillation?

User Sergey L
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Answer:

Time period will be 1.26 sec

Step-by-step explanation:

We have given amplitude A = 0.1 M

Speed
(dx)/(dt)=0.5m/sec

The displacement equation of simple harmonic motion is given by


x(t)=Asin\omega t

Differentiating both side


(dx)/(dt)=A\omega cos\omega t

In question it is given that at t=0, x=0 and
(dx)/(dt)=0.5m/sec

So
0.5=0.1\omega cos0


\omega =5sec^(-1)

Now period of oscillation
T=(2\pi )/(\omega )=(2* 3.14)/(5)=1.26sec

User Dan Fuller
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