Two simple pendulum whose lengths are 1,m and 121, cm are suspended side by side. Their bobs are pul

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13.Oscillations

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Two simple pendulum whose lengths are $1\,m$ and $121\, cm$ are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum will the two be in phase again

A

$11$

B

$10$

C

$21$

D

$20$

Solution

$\mathrm{N} \sqrt{\ell_{\ell}}=(\mathrm{N}+1) \sqrt{\ell_{\mathrm{s}}}$

$\Rightarrow \mathrm{N} \sqrt{121}=(\mathrm{N}+1) \sqrt{100}$

$\Rightarrow \mathrm{N}(11)+(\mathrm{N}+1) 10$

$\Rightarrow 11 \mathrm{N}=10 \mathrm{N}+10$

$\Rightarrow \mathrm{N}=10$

Standard 11

Physics

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