A simple pendulum oscillates in a vertical pl | vedclass

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

Question

(D) At the mean position,the tension $T$ is given by $T = mg + \frac{mv^2}{l}$,where $l$ is the length of the string.Given that $T = 3mg$,we have $mg

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(D) At the mean position,the tension $T$ is given by $T = mg + \frac{mv^2}{l}$,where $l$ is the length of the string.Given that $T = 3mg$,we have $mg + \frac{mv^2}{l} = 3mg$,which simplifies to $\frac{mv^2}{l} = 2mg$,so $v^2 = 2gl$.Let $	heta$ be the maximum angular displacement from the vertical. By the principle of conservation of energy,the kinetic energy at the mean position must equal the potential energy at the maximum displacement: $\frac{1}{2}mv^2 = mgl(1 - \cos 	heta)$.Substituting $v^2 = 2gl$,we get $\frac{1}{2}m(2gl) = mgl(1 - \cos 	heta)$.This simplifies to $mgl = mgl(1 - \cos 	heta)$,which implies $1 = 1 - \cos 	heta$,so $\cos 	heta = 0$.Therefore,$	heta = 90^\circ$.

$A$ simple pendulum oscillates in a vertical plane. When it passes through the mean position,the tension in the string is $3$ times the weight of the pendulum bob. What is the maximum displacement of the pendulum string with respect to the vertical in degrees?

Similar Questions

$A$ simple pendulum executing $S.H.M.$ is falling freely along with its support. Then:

The average speed of the bob of a simple pendulum oscillating with a small amplitude $A$ and time period $T$ is

$A$ simple pendulum of period $T$ has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate,its period will

The period of a simple pendulum is measured as $T$ in a stationary lift. If the lift moves upwards with an acceleration of $5g$,the new period will be:

$A$ simple pendulum of length $2 \ m$ is given a horizontal push through an angular displacement of $60^{\circ}$. If the mass of the bob is $200 \ g$,the angular velocity of the bob will be (Take acceleration due to gravity $g = 10 \ m/s^2$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module