When a mass ' $\mathrm{m}$ ' is suspended from a spring of length ' $\ell^{\prime}$, the length of the…
When a mass ' $\mathrm{m}$ ' is suspended from a spring of length ' $\ell^{\prime}$, the length of the spring
becomes ' $\mathrm{L}^{\prime}$. The mass is pulled down by a distance ' $\mathrm{d}$ ' and released. If the equation of motion of the mass is $\frac{d^{2} x}{d t^{2}}+\mathrm{P}^{2} x=0$, then $\mathrm{P}$ is equal to
$(\mathrm{g}=$ acceleration due to gravity $)$
$\frac{\mathrm{L}-\ell}{\mathrm{g}}$
$\frac{\mathrm{g}}{\mathrm{L}-\ell}$
$\sqrt{\frac{\mathrm{g}}{\mathrm{L}-\ell}}$
$\sqrt{\frac{\mathrm{L}-\ell}{\mathrm{g}}}$
Solution
INTIAL LENGTH $=l$
FINAL LENGTH=L.
NET DISTANCE/LENGTH
PULLED $=L-l$
$\therefore$ TIME PERIOD $=2 \pi \sqrt{\frac{L-\ell}{g}}$ allo, $P=\frac{2 \pi}{T .}$
$\begin{aligned} \therefore P=& \frac{g}{L-l} . \\ &(\text { option }-3) . \end{aligned}$