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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

    A
    `T/3`
    B
    `T/4`
    C
    `(5T)/6`
    D
    `T/6`

  • A particle starts SHM from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is

    A
    `A/2`
    B
    `A/sqrt2`
    C
    `(Asqrt3)/(2)`
    D
    `(2A)/(sqrt3)`

  • Time period (T) and amplitude (A) are same for two particle which undergoes SHM along the same line. At one particular instant one particle is at phase (3pi)/(2) and the other is at zero,while moving in the same direction. Find the time at which they will cross each other

    A
    `4T//2`
    B
    `3T//8`
    C
    `3T//4`
    D
    `3T//7`
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