A stone of mass $2 \mathrm{~kg}$ is whirled in a horizontal circle attached at the end of a $1.5 \mathrm{~m}$ long string. If the string makes an angle of $30^{\circ}$ with the vertical, compute its period.
Q 52.1
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Data : $L=1.5 \mathrm{~m}, \theta=30^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
The period of the conical pendulum,
$
\begin{aligned}
T & =2 \pi \sqrt{\frac{L \cos \theta}{g}}=2 \times 3.142 \times \sqrt{\frac{1.5 \cos 30^{\circ}}{10}} \\
& =6.284 \times \sqrt{\frac{1.5 \times 0.866}{10}}=6.284 \sqrt{\frac{1.299}{10}} \\
& =2.265 \mathrm{~s} \quad
\end{aligned}
$
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