If the string of a conical pendulum makes an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a conical pendulum,the time period $T$ is given by $T = 2\pi \sqrt{\frac{h}{g}}$,where $h$ is the vertical height of the pendulum bob from the

Vedclass · Accepted Answer

$\sin 	heta$

Vedclass · Answer

(A) For a conical pendulum,the time period $T$ is given by $T = 2\pi \sqrt{\frac{h}{g}}$,where $h$ is the vertical height of the pendulum bob from the point of suspension.If the string of length $l$ makes an angle $\alpha$ with the vertical,then $h = l \cos \alpha$.In this question,the angle $	heta$ is given with the horizontal. Therefore,the angle with the vertical is $\alpha = 90^\circ - 	heta$.Substituting this into the expression for height: $h = l \cos(90^\circ - 	heta) = l \sin 	heta$.Thus,the time period is $T = 2\pi \sqrt{\frac{l \sin 	heta}{g}}$.Squaring both sides,we get $T^2 = 4\pi^2 \frac{l \sin 	heta}{g}$.Since $4\pi^2$,$l$,and $g$ are constants,we have $T^2 \propto \sin 	heta$.

If the string of a conical pendulum makes an angle $\theta$ with the horizontal,then the square of its time period is proportional to

Similar Questions

An object of mass $m$ moves with constant speed in a circular path of radius $R$ under the action of a force of constant magnitude $F$. The kinetic energy of the object is ............

$A$ string of length '$L$' fixed at one end carries a mass '$m$' at the other end. The string makes $\frac{3}{\pi}$ r.p.s. around the vertical axis through the fixed end. The tension in the string is (in $mL$)

In a conical pendulum,the bob of mass $m$ moves in a horizontal circle of radius $r$ with uniform speed $V$. The string of length $L$ describes a cone of semi-vertical angle $\theta$. The centripetal force acting on the bob is $(g = \text{acceleration due to gravity})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module