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:

Options

A. Time perid of oscillation is 6t
B. amplitude of motion is 3a
C. time period of oscillations is 8t
D. amplitude of motion is 4a

Answer

A

Explanation

As it starts from rest
`x + A cos omegat`
at t = 0, x = A
at `t = tau, x = A -a`
`t = 2tau`
`x + A -3a`
`A - a = A cos omegat`
`A -3 a = A cos omega(2t) = A(2 cos^(2) omegat - 1)`
`(A-3a) = 2A [((A-a)/(A))^(2)] -A` [Using equation (i)]
` = 2A ((A^(2) + a^(2) - 2Aa))/(A^(2)) - A`
`rArr 2((A^(2) + a^(2) -2Aa))/(A) -A`
`rArr A = 2a` Now, `A - a = A cos omegat`
`a = 2 a cos omegat`
`cos omegat = cos (pi)/(3)`
`omegat = (pi)/(3)rArr (2pi)/(T) t = (pi)/(3)`
`t = (T)/(6)`