Question

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?

Answer 1

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`