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A particle in SHM has a period of 4s .It...

A particle in SHM has a period of 4s .It takes time `t_(1)` to start from mean position and reach half the amplitude. In another case it takes a time `t_(2)` to start from extreme position and reach half the amplitude. Find the ratio `t_(1)//t_(2)`

A

`(t_(1))/(t_(2))`

B

`(t_(1))/(t_(2))`

C

`(t_(1))/(t_(2)) =2`

D

`(t_(1))/(t_(2)) = 3`

Text Solution

Verified by Experts

The correct Answer is:
B
`(A)/(2) = A sin (omegat_(1)), (A)/(2) = A cos (omegat_(2))`
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Knowledge Check

  • A particle executing SHM of amplitude 4 cm and T=4 s .The time taken by it to move from position to half the amplitude is

    A
    `1 sec `
    B
    `1// sec`
    C
    `2//3sec `
    D
    `sqrt(3//2) sec`

  • A particle executes SHM with a time period of 4 s . Find the time taken by the particle to go from its mean position to half of its amplitude . Assume motion of particle to start from mean position.

    A
    1/3 s
    B
    2/3 s
    C
    4/3 s
    D
    5/3 s

  • A particle executing SHM of amplitude 4 cm and T = 4s. The time taken by it to move from positive extreme position to half the amplitude is

    A
    1 s
    B
    `1/3s`
    C
    `2/3s`
    D
    `sqrt(3/2)s`