A uniform rod of length L and mass M is pivot | vedclass

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Question

The correct option is $C \frac{1}{2 \pi}, \frac{\overline{6 k}}{ M }$

ANSWER:

Spring force acting at the end gives restoring torque

$	au=r F

Vedclass · Accepted Answer

$\frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}$

Vedclass · Answer

The correct option is $C \frac{1}{2 \pi}, \frac{\overline{6 k}}{ M }$

ANSWER:

Spring force acting at the end gives restoring torque

$	au=r F \sin 	heta$

$\Rightarrow 	au_0=-\left(k x \frac{L}{2} \cos 	hetaight) 	imes 2$

Value of displacement of spring,

$x=\frac{L}{2} \sin 	heta$

For the value of angular displacement tends to zero,

$	heta ightarrow 0$

$\Rightarrow \cos 	heta \cong 1, \sin 	heta \cong 	heta$

Then torque,

$\Rightarrow 	au=-k \frac{L}{2} 	imes \frac{L}{2} 	heta=-\frac{k L^2}{2} 	heta \ldots$

From equation $(i)$,

$\Rightarrow 	au=1 a=\frac{k L^2}{2} 	heta$

$\Rightarrow \frac{M L^2}{12} a=-\frac{k L^2}{2} 	heta$

$\Rightarrow a=-\frac{6 k}{M} 	heta=-\omega^2 	heta$

$\Rightarrow \omega=\sqrt{\frac{6 k}{M}}$

Time period,

$T =\frac{2 \pi}{\omega}$

$\Rightarrow T=2 \pi \sqrt{\frac{M}{6 k}}$

Frequency,

$f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{\overline{M k}}{M}}$

Final answer: $(c)$

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