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A particle moves in such a way that its ...

A particle moves in such a way that its acceleration a= - bx, where x is its displacement from the mean position and b is a constant. The period of its oscillation is

A

`2 pi b`

B

`(2pi)/(b)`

C

`(2pi)/(sqrt(b))`

D

`2pisqrt(b)`

Text Solution

Verified by Experts

The correct Answer is:
C
a =- bx or a +bx=0 comparing it with
`(a^(2)x)/(dt^(2))+omega^(2)x=0`
We get `omega^(2)=b " " therefore (4pi^(2))/(T^(2))=b " " therefore T=(2pi)/(sqrt(b))`
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Knowledge Check

  • A particle moves such that its acceleration a is given by a = -bx , where x is the displacement from equilibrium positionand is a constant. The period of oscillation is

    A
    `2 pi sqrtb`
    B
    `(2 pi)/ sqrtb`
    C
    `2pi/b`
    D
    `2 sqrt(pi/ b)`

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    A
    `2pisqrt(b)`
    B
    `(2pi)/b`
    C
    `(2pi)/sqrt(b)`
    D
    `2 sqrt(pi/b)`

  • A particle moves so that its acceleration a is give by a=bn where x is displacement from equilibrium position and b is non negative real constant the time period of oscillation of the particle is

    A
    `2pisqrt(b)`
    B
    `(2pi)/b`
    C
    `(2pi)/sqrt(b)`
    D
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