The ratio of period of oscillation of the con | vedclass

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Question

For Simple Pendulum,

Time period $\left(T_1ight)=2 \pi \sqrt{\frac{l}{g}}-(j)$ $r=l \sin 	heta$ (from figure)

$T \sin 	heta=\frac{m v^2}{r}$

Vedclass · Accepted Answer

$\sqrt {\cos 	heta } $

Vedclass · Answer

For Simple Pendulum,

Time period $\left(T_1ight)=2 \pi \sqrt{\frac{l}{g}}-(j)$ $r=l \sin 	heta$ (from figure)

$T \sin 	heta=\frac{m v^2}{r}$ and $T \cos 	heta=m g$

Dividing them, we get, $	an 	heta=v^2 / r g$

Putting $r=l \sin 	heta, 	an 	heta=\frac{v^2}{(l \sin 	heta) g}$

$\Rightarrow \quad v=\sqrt{	an 	heta \sin 	heta l g}$

$T_2=\frac{2 \pi r}{v}=\frac{2 \pi l \sin 	heta}{\sqrt{	an 	heta \sin 	heta l g}} 	ext { [for conical] }$

$\Rightarrow T_2=2 \pi \sqrt{\frac{l}{g} \frac{\sin 	heta}{	an 	heta}}$

$\Rightarrow T_2=2 \pi \sqrt{\frac{l}{g} \cos 	heta}$

$\frac{T_2}{T_1}=\frac{2 \pi \sqrt{\frac{l}{g} \cos 	heta}}{2 \pi \sqrt{\frac{l}{g}}}=\sqrt{\cos 	heta}$

The ratio of period of oscillation of the conical pendulum to that of the simple pendulum is : (Assume the strings are of the same length in the two cases and $\theta$ is the angle made by the string with the verticla in case of conical pendulum)

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