Question:
Time period of a simple pendulum is $\mathrm{T}$. The time taken to complete $5 / 8$ oscillations
starting from mean position is $\frac{\alpha}{\beta} \mathrm{T}$. The value of $\alpha$ is .........
Solution:
$\frac{5}{8}$ th of oscillation $=\left(\frac{1}{2}+\frac{1}{8}\right)^{\text {th }}$ of oscillation
$\pi+\theta=\omega t$
$\pi+\frac{\pi}{6}=\left(\frac{2 \pi}{T}\right) \mathrm{t}$
$\frac{7 \pi}{6}=\left(\frac{2 \pi}{T}\right) t$
$t=\frac{7 \mathrm{~T}}{12}$
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