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A partical excutes SHM with amplitudeA a...

A partical excutes SHM with amplitude`A` and angular frequency `omega`. At an instant the particle is at a distance `A//5` from the mean position and is moving away from it. Find the time after which it will come back to this position again and also find the time after which it will pass through the mean position.

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figure show the circular motion repesentatuion fot the partical `P` give in problem. The initial situation of partical is shown in the figure. As `P` moves , its initial position when `P` will go up and then come back to its initial position when `P` reaches to the corresponding position in the second quandrant as shown. In this process `P` traversed an angular displacement `theta` with angular velocity `omega` , thus time taken in the process is
`t = (theta)/(omega) = (1)/(omega) [2 cos ^(-1) (1//5)] = (2 cos ^(-1) (1//5))/(omega)`
Now when `P` reaches point `C,P` will each its mean position . Thus, time taken by `P` from its initial position to point `C` is
`t = ((pi - alpha))/(omega) = (pi - sin ^(-1) (1//5))/(omega)`
The same time `P` will take from `A//5` position t mean position through its exterme position.
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