JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Elasticity & SHM

One full oscillation covers a path of four amplitudes, so completing $\frac{5}{8}$ of an oscillation means the bob travels a path of $\frac{5}{8}\times 4A=\frac{5A}{2}$, two and a half amplitudes. Starting from the mean position, the first amplitude of travel carries the bob out to an extreme in a quarter period, $\frac{T}{4}$, and the next amplitude brings it back to the mean in a further $\frac{T}{4}$, so two amplitudes are covered in $\frac{T}{2}$. For the remaining half amplitude the displacement grows as $x=A\sin \frac{2\pi t}{T}$, and reaching $x=\frac{A}{2}$ needs $\frac{2\pi t}{T}=\frac{\pi }{6}$, giving $t=\frac{T}{12}$. The total time is $\frac{T}{2}+\frac{T}{12}=\frac{7T}{12}$, which has the form $\frac{\alpha }{\beta }T$ with $\alpha =7$.