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A simple pendulum have time periods ‘T’ and $\frac{5 T}{4}$ . They start vibrating at the same instant from the mean position in the same phase. The phase difference between them when the pendulum with higher time period completes one oscillation is

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60⁰

30⁰

45⁰

90⁰

answer is D.

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Detailed Solution

\large {\omega _1}\, = \,\frac{{2\pi }}{T},\,\,{\omega _2}\, = \,\frac{{2\pi }}{{\frac{{5T}}{4}}}

$ϕ_{1}$ = Phase of pendulum - 1

\large {\omega _1}t\, = \,\frac{{2\pi }}{T} \times \frac{{5T}}{4}\, = \,\frac{{5\pi }}{2}

$ϕ_{2}$ = Phase of pendulum - 2

\large {\omega _2}t\, = \,\frac{{8\pi }}{{5T}} \times \frac{{5T}}{4}\, = \,2\pi

\large \therefore \,\Delta \phi \, = \,{\phi _1} - {\phi _2}\, = \,\frac{\pi }{2}\, = \,{90^0}

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