Question

The Young's modulus of material of a thin ring shaped elastic body is $Y$. The mass of the ring is $m$, area of cross-section is $A$ and its initial radius is $R$. The ring is little elongated, then left alone at time $\sqrt{\frac{\pi mR}{nYA}}.$ If the circumference of the ring be same as it was initially, then find $n$. (Neglect loss of energy)

Answer 1

Answer: 8

$Y=\frac{T/A}{x/R}$
$T=\left(\frac{YA}{R}\right)x$
Now, $2T\theta =\left(\frac{m}{2\pi }\times 2\theta \right)\frac{d^{2}x}{d{t}^2}$
$T=\frac{m}{2\pi }\frac{d^{2}x}{d{t}^2}$
Since $x$ and $T$ are in opposite direction,
$\therefore \frac{m}{2\pi }\frac{d^{2}x}{d{t}^2}=-\frac{YA}{R}x$
$\frac{d^{2}x}{d{t}^2}=-\frac{2\pi YA}{mR}x$
$\omega =\sqrt{\frac{2\pi YA}{mR}}$
$T=2\pi \sqrt{\frac{mR}{2\pi YA}}$
Time from extreme to mean position
$=\frac{T}{4}=\frac{\pi }{2}\sqrt{\frac{mR}{2\pi YA}}=\sqrt{\frac{\pi mR}{8YA}}$