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 $\sqrt{\frac{\pi m R}{n Y A}} .$ If the circumference of the ring be same as it was initially, then find $n$. (Neglect loss of energy)

Answer 1

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