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Question

For speed to be zero.

$\frac{1}{2} K(S-\ell)^{2}=m g(S)$

$\mathrm{S}^{2}+\ell^{2}-2 \mathrm{S}_{\ell}=\frac{2 m g}{K}s$

$\Rightarrow \mathrm{

Vedclass · Accepted Answer

time to come to rest for the first time  $= \,\left( {\frac{\pi }{2}\sqrt {\frac{m}{k}} \, + \,\sqrt {\frac{{2l}}{g}} } \right)$

Vedclass · Answer

For speed to be zero.

$\frac{1}{2} K(S-\ell)^{2}=m g(S)$

$\mathrm{S}^{2}+\ell^{2}-2 \mathrm{S}_{\ell}=\frac{2 m g}{K}s$

$\Rightarrow \mathrm{KS}^{2}-2 \ell K S-2 m g S+K \ell^{2}=0$

$S=\frac{(2 \ell K+2 m g) \pm \sqrt{4\left((K+m g)^{2}-4 K \ell^{2} 	imes kight.}}{2 K}$

$S=\frac{(K \ell+m g) \pm \sqrt{K^{2} \ell^{2}+m^{2} g^{2}+2 \ell K m g-K^{2} \ell^{2}}}{K}$

$S=\frac{(K \ell+m g)+\sqrt{2 m g K \ell+m^{2} g^{2}}}{K}$

Now maximum speed will be at equilibrium position

$\frac{1}{2} m v^{2}+\frac{1}{2} K\left[\frac{m g}{K}ight]^{2}=m g\left(\ell+\frac{m g}{K}ight)$

$\frac{1}{2} m v^{2}+\frac{1}{2} \frac{m^{2} g^{2}}{K}=m g \ell+\frac{m^{2} g^{2}}{K}$

$\frac{1}{2} m v^{2}=m g \ell+\frac{m^{2} g^{2}}{2 K}$

$\mathrm{v}=\sqrt{2 g \ell+\frac{m g^{2}}{K}}$

time of free fall is $\sqrt{\frac{2 \ell}{g}}$

and clearly option $4$ is wrong.

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