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If the second pendulum bob is thrown at ...

If the second pendulum bob is thrown at velocity `v_(2)` at an angle `beta` from mean position, but on other side of mean position , find the phase diffrence in the two SHMs now as show in the figure.

Text Solution

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In this case , amplitudes of the two SHMs will still remain the same and are given by Eqs (i) and (ii) but when we represent the two SHM s on their corresponding circular motion , the position of the partical of the principal in circular motion , the position of the partical in circular motion in second pendulum is now difference as shown in figure .

As shown in figure (a) and (b) the initial phase of the two pendulum bobs are
`phi_(1) = pi - sin ^(-1) ((l alpha)/(A_(1))) and phi_(2) = pi + sin ^(-1) ((l beta)/(A_(2)))`
As `omega` for both SHm are same, their phase difference remain constant so it is given as
`Delta phi_(2) - pi_(1) = sin ^(-1) ((l beta)/(A_(2))) + sin ^(-1) ((l alpha)/(A_(1)))`
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