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Correct Answer – D
Let `x_(1) = A_(1)sinomega_(1)t` and `x_(2) = A_(2)sinomega_(2)t`
Two pendulums will vibrate in same phase again when there phase difference `(omega_(2) – omega_(1))t = 2pi`
`rArr ((2pi)/(T_(2)) – (2pi)/(T_(1)))t = 2pi`
`rArr (sqrt((g)/(1)) – sqrt((g)/(1.44)))n xx T_(1) = 2pi` (where `n` is number of vibrations completed by ionger pendulum)
`rArr (sqrt((g)/(1)) – sqrt((g)/(1.44)))n xx 2pisqrt((1.44)/(g)) = 2pi rArr n = 5`
Thus after `5` vibrations of longer pendulum they will again start swinging in same phase.
Correct Answer – D
Let `x_(1) = A_(1)sinomega_(1)t` and `x_(2) = A_(2)sinomega_(2)t`
Two pendulums will vibrate in same phase again when there phase difference `(omega_(2) – omega_(1))t = 2pi`
`rArr ((2pi)/(T_(2)) – (2pi)/(T_(1)))t = 2pi`
`rArr (sqrt((g)/(1)) – sqrt((g)/(1.44)))n xx T_(1) = 2pi` (where `n` is number of vibrations completed by ionger pendulum)
`rArr (sqrt((g)/(1)) – sqrt((g)/(1.44)))n xx 2pisqrt((1.44)/(g)) = 2pi rArr n = 5`
Thus after `5` vibrations of longer pendulum they will again start swinging in same phase.