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The amplitude of a particle in SHM is 5 ...

The amplitude of a particle in SHM is 5 cms and its time period is `pi`. At a displacement of 3 cms from its mean position the velocity in cms/sec will be :

A

8

B

12

C

2

D

16

Text Solution

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The correct Answer is:
A
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The amplitude and periodic time of SHM are 5 cm and 6 s, respectively. What is the phase at a distance of 2.5 cm from the mean position ?

The acceleration of a particle performing SHM is 12 cm //s^(2) at a distance of 3 cm from the mean position . Calculate its time - period . Hint : a = omega^(2) x a = 12 cm//s^(2) , x=3cm Find omega T= (pi)/(omega)

Knowledge Check

  • A particle is in SHM with time period 3 sec. Calculate the displacement of particle from mean position after t = 0.5 sec.

    A
    `(1)/(2) xx` amplitude
    B
    `(1)/(4) xx` amplitude
    C
    `(sqrt(3))/(2) xx` amplitude
    D
    `(2)/(sqrt(3)) xx` amplitude

  • The amplitude and the periodic time of a S.H.M. are 5 cm and 6 sec respectively. At a distance of 2.5 cm away from the mean position, the phase will be

    A
    `5 pi//12`
    B
    `pi//4`
    C
    `pi//3`
    D
    `pi//6`

  • The acceleration of a particle performing S.H.M. is 12 cm // sec^(2) cm at a distance of 3 cm form the mean position. Its period is

    A
    0.5 sec
    B
    1.0 sec
    C
    2.0 sec
    D
    3.14 sec

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