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A body makes angular SHM of amplitude (p...

A body makes angular SHM of amplitude `(pi)/(10)` rad. And time period 0.05 s. IF the body is at displacement `theta=(pi)/(10)` rad. At t=0, then write the equation giving the angular displacement as a function of time

Text Solution

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Let the required equation be
`theta=theta_0sin(omegat+delta)`
here `theta_0=amplitude =pi/10rad`
omega=(2pi)/T=(2pi)/(0.05s)=40pis^-1`
So that `theta=(pi/10rad)sin[(40pis^-1)t+delta]`…………….i
`At t0, theta=pi/10rad. Putting in i
`=pi/10(pi/10)sindelta`
or `sindelta=1`
or `delta=pi/2`
thus by i
`theta=(pi/10rad)sin[(40pis^-1)t+pi/2]`
`=(pi/10rad)cos[(40pis^-1)t]`
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