A simple pendulum of length L and mass (bob) | vedclass

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

Question

(D) For a simple pendulum oscillating in a vertical plane,the forces acting on the bob are the tension $T$ in the string (directed towards the pivot)

Vedclass · Accepted Answer

Both $(b)$ and $(c)$

Vedclass · Answer

(D) For a simple pendulum oscillating in a vertical plane,the forces acting on the bob are the tension $T$ in the string (directed towards the pivot) and the gravitational force $Mg$ (directed vertically downwards).$1$. Radial direction: The net force towards the center of the circular path is the centripetal force,which is given by $\frac{Mv^2}{L}$. The radial component of the gravitational force is $Mg \cos 	heta$ (acting away from the pivot). Thus,the net radial force is $T - Mg \cos 	heta$. Equating these,we get $T - Mg \cos 	heta = \frac{Mv^2}{L}$. This matches option $(b)$.$2$. Tangential direction: The component of the gravitational force acting tangentially to the path is $Mg \sin 	heta$. According to Newton's second law,$F_T = Ma_T$,so $Mg \sin 	heta = Ma_T$. Therefore,the magnitude of the tangential acceleration is $|a_T| = g \sin 	heta$. This matches option $(c)$.Since both $(b)$ and $(c)$ are correct,the correct option is $(d)$.

Similar Questions

$A$ $50 \ cm$ long simple pendulum is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration $\sqrt{3} \ g \ m/s^2$. The time period of small oscillations of the pendulum about its equilibrium position is $.... \ s$ $\left(g=\pi^2 \ m/s^2\right):-$

From a fixed support,two small identical spheres are suspended by means of strings of length $1 \,m$ each. They are pulled aside as shown and then released. $B$ is the mean position. Then the two spheres collide

The time period of a simple pendulum will be doubled if we

Vedclass Test Series

Exam Paper Generator

Online Exam Module