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3-2.Motion in Plane
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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)
A$cos \theta$
B$\sqrt {\cos \theta } $
C$1$
Dnone of these
Solution
For Simple Pendulum,
Time period $\left(T_1\right)=2 \pi \sqrt{\frac{l}{g}}-(j)$ $r=l \sin \theta$ (from figure)
$T \sin \theta=\frac{m v^2}{r}$ and $T \cos \theta=m g$
Dividing them, we get, $\tan \theta=v^2 / r g$
Putting $r=l \sin \theta, \tan \theta=\frac{v^2}{(l \sin \theta) g}$
$\Rightarrow \quad v=\sqrt{\tan \theta \sin \theta l g}$
$T_2=\frac{2 \pi r}{v}=\frac{2 \pi l \sin \theta}{\sqrt{\tan \theta \sin \theta l g}} \text { [for conical] }$
$\Rightarrow T_2=2 \pi \sqrt{\frac{l}{g} \frac{\sin \theta}{\tan \theta}}$
$\Rightarrow T_2=2 \pi \sqrt{\frac{l}{g} \cos \theta}$
$\frac{T_2}{T_1}=\frac{2 \pi \sqrt{\frac{l}{g} \cos \theta}}{2 \pi \sqrt{\frac{l}{g}}}=\sqrt{\cos \theta}$
Time period $\left(T_1\right)=2 \pi \sqrt{\frac{l}{g}}-(j)$ $r=l \sin \theta$ (from figure)
$T \sin \theta=\frac{m v^2}{r}$ and $T \cos \theta=m g$
Dividing them, we get, $\tan \theta=v^2 / r g$
Putting $r=l \sin \theta, \tan \theta=\frac{v^2}{(l \sin \theta) g}$
$\Rightarrow \quad v=\sqrt{\tan \theta \sin \theta l g}$
$T_2=\frac{2 \pi r}{v}=\frac{2 \pi l \sin \theta}{\sqrt{\tan \theta \sin \theta l g}} \text { [for conical] }$
$\Rightarrow T_2=2 \pi \sqrt{\frac{l}{g} \frac{\sin \theta}{\tan \theta}}$
$\Rightarrow T_2=2 \pi \sqrt{\frac{l}{g} \cos \theta}$
$\frac{T_2}{T_1}=\frac{2 \pi \sqrt{\frac{l}{g} \cos \theta}}{2 \pi \sqrt{\frac{l}{g}}}=\sqrt{\cos \theta}$
Standard 11
Physics
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