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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

`(5T)/(6)`

B

`(T)/(3)`

C

`(T)/(4)`

D

`(T)/(6)`

Text Solution

Verified by Experts

The correct Answer is:
D

Angle covered to meet `theta=60^(@)=(pi)/(3)` rad.
If they cross each other at time t then
`t=(theta)/(2pi)=(pi)/(3xx2pi)T=(T)/(6)`
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DISHA PUBLICATION-OSCILLATIONS -Exercise-2 : Concept Applicator
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