Question

An elastic string of length $l$ supports a heavy particle of mass m and the system is in equilibrium with elongation produced being e as shown in the figure. The particle is now pulled down below the equilibrium position through a distance $d(\le e)$ and released. The angular frequency and maximum amplitude for SHM is

• A. $\sqrt{\frac{g}{e}},e$
• B. $\sqrt{\frac{g}{l}},2e$
• C. $\sqrt{\frac{g}{d+e}},d$
• D. $\sqrt{\frac{g}{e}},2d$

Answer 1

The correct option is A $\sqrt{\frac{g}{e}},e$

Here, weight of the ball must balance spring force at equilibrium position.

$kx=mgke=mgk=\frac{mg}{e}$

After elongating more, the net restoring force will be,

$F=kda=\frac{Kd}{m}a={\omega }^{2}x\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{mg}{em}}=\sqrt{\frac{g}{e}}T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}$

Now, putting the value of K

$T=2\pi \sqrt{\frac{m\times e}{mg}}T=2\pi \sqrt{\frac{e}{g}}g={\omega }^{2}A$ [At extreme position]

$g=\frac{g}{e}\times AA=e$