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Two particles are in SHM in a straight l...

Two particles are in `SHM` in a straight line about same equilibrium position. Amplitude `A` and time period `T` of both the particles are equal. At time `t=0`, one particle is at displacement `y_(1)=+A` and the other at `y_(2)=-A//2`, and they are approaching towards each other. after what time they cross each other?

A

`(T)/(3)`

B

`(T)/(4)`

C

`(5T)/(6)`

D

`(T)/(6)`

Text Solution

Verified by Experts

The correct Answer is:
D

Total time `=2t = (T)/(4) +(T)/(12)`
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Knowledge Check

  • Two particles are executing SHM in a straight line. Amplitude A and the time period T of both the particles are equal. At time t=0, one particle is at displacement x_(1)=+A and the other x_(2)=(-A/2) and they are approaching towards each other. After what time they across each other? T/4

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