The ratio of the period of oscillation of a c | vedclass

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(B) For a simple pendulum,the time period is given by $T_1 = 2\pi \sqrt{\frac{l}{g}}$.For a conical pendulum,the radius of the circular path is $r = l

Vedclass · Accepted Answer

$\sqrt{\cos 	heta}$

Vedclass · Answer

(B) For a simple pendulum,the time period is given by $T_1 = 2\pi \sqrt{\frac{l}{g}}$.For a conical pendulum,the radius of the circular path is $r = l \sin 	heta$.The forces acting are $T \sin 	heta = \frac{mv^2}{r}$ and $T \cos 	heta = mg$.Dividing these equations,we get $	an 	heta = \frac{v^2}{rg}$.Substituting $r = l \sin 	heta$,we have $	an 	heta = \frac{v^2}{lg \sin 	heta}$,which gives $v = \sqrt{lg \sin 	heta 	an 	heta}$.The time period for the conical pendulum is $T_2 = \frac{2\pi r}{v} = \frac{2\pi l \sin 	heta}{\sqrt{lg \sin 	heta 	an 	heta}}$.Simplifying this,$T_2 = 2\pi \sqrt{\frac{l \sin 	heta}{g 	an 	heta}} = 2\pi \sqrt{\frac{l \cos 	heta}{g}}$.Therefore,the ratio $\frac{T_2}{T_1} = \frac{2\pi \sqrt{\frac{l}{g} \cos 	heta}}{2\pi \sqrt{\frac{l}{g}}} = \sqrt{\cos 	heta}$.

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

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