The force constant of an oscillating simple p | vedclass

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

Question

(C) For a simple pendulum,the restoring torque is given by $	au = -mgL \sin 	heta$. For small oscillations,$\sin 	heta \approx 	heta$,so $	au \ap

Vedclass · Accepted Answer

Directly proportional to the mass of the bob

Vedclass · Answer

(C) For a simple pendulum,the restoring torque is given by $	au = -mgL \sin 	heta$. For small oscillations,$\sin 	heta \approx 	heta$,so $	au \approx -mgL 	heta$. The angular acceleration is $\alpha = \frac{	au}{I} = \frac{-mgL 	heta}{mL^2} = -\frac{g}{L} 	heta$. Comparing this with the $SHM$ equation $\alpha = -\omega^2 	heta$,we get $\omega^2 = \frac{g}{L}$. The force constant $k$ for an oscillating system is defined as $k = m \omega^2$. Substituting $\omega^2$,we get $k = m \left(\frac{g}{L}ight)$. Thus,the force constant $k$ is directly proportional to the mass of the bob $m$ and inversely proportional to the length $L$ of the pendulum. Among the given options,the most appropriate relationship is that it is directly proportional to the mass of the bob.

The force constant of an oscillating simple pendulum is

Similar Questions

The time period of a simple pendulum in a stationary lift is $T$. If the lift accelerates with $\frac{g}{6}$ vertically upwards,then the new time period will be:
(where $g =$ acceleration due to gravity)

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

$A$ simple pendulum hangs from the ceiling of a car. If the car accelerates with a uniform acceleration,the frequency of the simple pendulum will

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module