Two pendulum have time periods $T$ and $5T/4$. They start $SHM$ at the same time from the mean position. After how many oscillations of the smaller pendulum they will be again in the same phase
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$\mathrm{n}_{1} \mathrm{T}_{1}=(\mathrm{n}+1) \mathrm{T}_{2}$
$\mathrm{nT}=\frac{\Sigma \mathrm{T}}{4}(\mathrm{n}+1)$
$\mathrm{nT}=\frac{5 \mathrm{T}}{4} \mathrm{n}-\frac{5 \mathrm{T}}{4}$
$\frac{\mathrm{nT}}{4}=\frac{5 \mathrm{T}}{4}$
$\mathrm{n}=5$
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