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Question

The time periods of oscillation of two simple pendulums are 1 sec, 1.2 sec. Initially both are in same phase of oscillation. The minimum number of oscillations made by longer pendulum when they are again in same phase is:

A
5
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B
6
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C
10
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D
12
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Solution

The correct option is A 5
t(ω1ω2)=2π
t=T1T2T2T1
=1×1.21.21
t=6
no. of oscillations made lag longer pendulum =61.2
=5

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