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Two particles are executing SHM in a str...

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`

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 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`

  • A particle executing SHM with time period T and amplitude A. The mean velocity of the particle averaged over quarter oscillation, is

    A
    `(A)/(4T)`
    B
    `(2A)/(T)`
    C
    `(3A)/(T)`
    D
    `(4A)/(T)`

  • A particle executing SHM has amplitude of 1m and time period pi sec . Velocity of particle when displacement is 0.8m is

    A
    `1.2 m//s`
    B
    `0.6 m//s`
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    `2.4 m//s`
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    `0.3m//s`
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