A simple pendulum of length 'l' and a bob of | vedclass

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

Question

(B) The tension $T$ in the string of a simple pendulum at any angle $	heta$ is given by the radial force balance equation:$T = mg \cos 	heta + \frac

Vedclass · Accepted Answer

$mg\left[1+\left(\frac{A}{l}\right)^2\right]$

Vedclass · Answer

(B) The tension $T$ in the string of a simple pendulum at any angle $	heta$ is given by the radial force balance equation:$T = mg \cos 	heta + \frac{mv^2}{l}$For small oscillations,$\cos 	heta \approx 1$ and the velocity $v$ is maximum at the mean position $(	heta = 0)$.Thus,the maximum tension $T_{\max}$ occurs at the mean position:$T_{\max} = mg + \frac{mv_{\max}^2}{l}$For a simple pendulum,the angular frequency is $\omega = \sqrt{\frac{g}{l}}$.The maximum velocity in $S$.$H$.$M$. is $v_{\max} = A\omega$.Substituting $v_{\max}$ into the tension equation:$T_{\max} = mg + \frac{m(A\omega)^2}{l}$$T_{\max} = mg + \frac{m A^2 \omega^2}{l}$Since $\omega^2 = \frac{g}{l}$,we get:$T_{\max} = mg + \frac{m A^2 (g/l)}{l}$$T_{\max} = mg + \frac{m A^2 g}{l^2}$$T_{\max} = mg \left[1 + \left(\frac{A}{l}ight)^2ight]$

$A$ simple pendulum of length '$l$' and a bob of mass '$m$' is executing $S$.$H$.$M$. of small amplitude '$A$'. The maximum tension in the string will be ($g=$ acceleration due to gravity).

Similar Questions

Two simple pendulums have first $(A)$ bob of mass $M_1$ and length $L_1$,second $(B)$ of mass $M_2$ and length $L_2$. Given $M_1 = M_2$ and $L_1 = 2 L_2$. If their total energies are the same,then which of the following statements is correct?

The time period of a simple pendulum of length $L$ as measured in an elevator descending with acceleration $\frac{g}{3}$ is

In a stationary lift,the time period of a simple pendulum is $T$. If the lift starts accelerating downwards with an acceleration of $\frac{g}{4}$,then the new time period of the pendulum will be:

Two pendulums have time periods $T$ and $\frac{5T}{4}$. They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has completed one oscillation (in $^o$)?

$A$ simple pendulum is suspended from the ceiling of a lift. When the lift is at rest,its period is $T$. With what acceleration $a$ should the lift be accelerated upward in order to reduce the period to $\frac{T}{2}$? (Take $g$ as the acceleration due to gravity.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module