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Two simple pendulums of length 1m and 16...

Two simple pendulums of length `1m` and `16m` respectively are both given small displacements in the same direction at the same instant. After how many oscillations of the shorter pendulum will, the two pendulums vibrate in the same phase ?

A

`(4)/(3)`

B

`(3)/(4)`

C

`4`

D

`(1)/(16)`

Text Solution

Verified by Experts

The correct Answer is:
A
We know that time period is directly proportional to the square root of its length. Two different pedulums can be in phase at the earliest with a difference in the number of oscillations being a maximum of one. `NT_("shortest") = (N-1)T_("Longer")`
`Nsqrt(1) =(N-1) sqrt(16), N = (N-1) 4 = 4 N-4`
`3N = 4, N = (4)/(3)`
i.e., when `4` oscillations of the shortest and three of the longer pendulum are over they will be in phase. So, choice (a) is correct and rest of the choices are wrong.
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Knowledge Check

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