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Consider a particle moving in simple har...

Consider a particle moving in simple harmonic motion according to the equation
`x=2.0cos(50pit+tan^-1 0.75)`
where x is in centimetre and t in second. The motion is started at t=0. a. When does the particle come to rest for the first time? B. When does the acceleration have its maximum magnitude for the first time? c. When does the particle comes to rest for the second time?

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Verified by Experts

The correct Answer is:
`rarr t=1.6xx10^-2sec`
`rarr t=1.6xx10^-2sec`
`rarrt=6.3xx10^-2s`

`a`. `x=20cos(50pit+tan^-10.75)`
`=2.0cos(50pit+0.643)`
`v=(dx)/(dt)`
`=-100pisin(50kpit0.643)=0`
`rarr sin(50pit+0.643)=0`
As the particle comes to rest for the 1st time
`rarr50pit+0.643=pi`
`rarr t=1.6xx10^-2sec`
`b`. Acceleration
`a=(dv)/(dt)`
`=100pixx50picos(50pit+0.643)`
For maximum acceleration
`cos(50it+0.643)=-1cospi(max)(so a is max)`
`rarr t=1.6xx10^-2ssec`
`c`. When teh particle comes to rest for second time
`50pit+0.643=2pi`
`:.t=6.3xx10^-2s`
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