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Two particles are executing simple harmo...

Two particles are executing simple harmonic motion of the same amplitude A and frequency `omega` along the x-axis. Their mean position is separated by distance `X_(0)(X_(0)gt A)`. If the maximum separation between them is `(X_(0)+A)`, the phase difference between their motion is

A

`(pi)/(3)`

B

`(pi)/(4)`

C

`(pi)/(6)`

D

`(pi)/(2)`.

Text Solution

Verified by Experts

The correct Answer is:
A
Let `x_(1)=A sin omega t` and `x_(2)=A sin (omega t+phi)`
`x_(2)-x_(1)=2A cos (omegat+(phi)/(2))sin""(phi)/(2)`
But 2A `sin""(phi)/(2)=Aimpliesphi=(pi)/(3)`
Correct chice : (a).
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Knowledge Check

  • A particle executing a simple harmonic motion has a period of 6 sec.The time taken by the particle to move from the mean position to half the amplitude is

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