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A particle moves with simple harmonic mo...

A particle moves with simple harmonic motion in a straight line. In first `taus`, after starting form rest it travels a destance a, and in next `tau s` it travels 2a, in same direction, then:

A

Amplitude of motion is 3a

B

Time period of oscillations is `8 tau`

C

Amplitude of motion is 4a

D

Time period of oscillations is `6 tau`

Text Solution

Verified by Experts

The correct Answer is:
D
Since ` x = cos omega t ` [ at t = 0 , x = A]
when `t = tau , x = A - a `
`rArr A - a = A cos omega tau rArr cos omega tau = ((A - a)/( A))`
when ` t = 2 tau , x = A - a - 2 a = A - 3a `
`rArr A - ea = A cos 2 omega tau `
`rArr A - ea = A ( 2 cos ^(2) omega tau - 1)`
`rArr A - 3 a = A [ 2 ((A - a)/( A))^(2) - 1]`
`rArr A [ ( 2 A^(2) + 2a^(2) - 4 a A - A^(2))/( A^(2)) ]`
`rArr a = 2a rArr (a)/(A) = (1)/(2)`
`:. A - a = A cos omega tau rArr cos omega tau = (1)/(2) rArr (2 pi)/( T) tau = (pi)/(3)`
`rArr T = 6 tau`
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