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If a simple harmonic motion is represent...

If a simple harmonic motion is represented by `(d^(2)x)/(dt^(2)) + alphax = 0`, its time period is :

A

`(2pi)/(alpha)`

B

`(2pi)/(sqrt(alpha))`

C

`2pialpha`

D

`2pisqrt(alpha)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(d^(2)x)/(dt^(2)) = -alphax………(i)`
We know `a = (d^(2)x)/(dt^(2)) = -omega^(2)x …….(ii)`
From Eq. `(i)` and `(ii)`, we have
`omega^(2) = alpha`
`omega = sqrt(alpha)`
or `(2pi)/(T) = sqrt(alpha) :. T = (2pi)/(sqrt(a))`
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Knowledge Check

  • If a simple harmonic motion is erpresented by (d^(2)x)/(dt^(2))+ax=0 , its time period is.

    A
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    B
    `(2pi)/a`
    C
    `2pisqrta`
    D
    `2pia`

  • The equation of a simple harmonic motion of a particle is (d^(2)x)/(dt^(2)) + 0.2 (dx)/(dt) + 36x = 0 . Its time period is approximately

    A
    `(pi)/(2) sec`
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    `(pi)/(4) sec`
    C
    `(pi)/(3) sec`
    D
    `(pi)/(6)sec`

  • A particle executes simple harmonic motion according to equation 4(d^(2)x)/(dt^(2))+320x=0 . Its time period of oscillation is :-

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    B
    `(pi)/(3sqrt(2))s`
    C
    `(pi)/(2sqrt(5))s`
    D
    `(2pi)/(sqrt(3))s`
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