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Two pendulums have time period T and 5T/...

Two pendulums have time period T and 5T/4. They starts SHM at the same time from the mean position. What will be the phase difference between them after the bigger pendulum completed one oscillation ?

A

`45^(@)`

B

`90^(@)`

C

`60^(@)`

D

`30^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B
When a bigger pendulum of `T_(1)=5T//4` complete one vibration, the smalle pendulum will complete 5/4 vibration. It means the smaller pendulum will be leading, the bigger pendulum by phase T/4 sec = `(pi)/(2)` rad = `90^(@)`
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Knowledge Check

  • Two simple pendulums have time periods T and 5T//4 . They start vibrating at the same instant from the mean position in the same phase. The phase difference between them when the pendulum with higher time period complets one oscillation is

    A
    `30^(@)`
    B
    `45^(@)`
    C
    `60^(@)`
    D
    `90^(@)`

  • Two simple pendulums have time period T and 5T//4 . They start vibrating at the same instant from the mean position in the same phase. The phase difference (in rad) between them when the smaller pendulum completes one oscillation will be

    A
    `(pi)/(6)`
    B
    `(pi)/(5)`
    C
    `(pi)/(2)`
    D
    `(2pi)/(5)`

  • Two pendulums have time periods T and 5T//4 . They start SHM at the same time from the mean position After how many oscillations of the smaller pendulum they will be again in the same phase:

    A
    `5`
    B
    `4`
    C
    `11`
    D
    `9`

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