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

$A$ simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$. The time period is given by $T = 2\pi \sqrt{\frac{l}{g'}}$,where $g'$ is equal to:

Two masses $M_{A}$ and $M_{B}$ are hung from two strings of length $l_{A}$ and $l_{B}$ respectively. They are executing $SHM$ with frequency relation $f_{A}=2 f_{B}$,then the relation is:

The mass and diameter of a planet are twice those of Earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on Earth?

$A$ simple pendulum of length $1 \,m$ is freely suspended from the ceiling of an elevator. The time period of small oscillations as the elevator moves up with an acceleration of $2 \,m/s^2$ is (use $g=10 \,m/s^2$).

$A$ simple pendulum of length $l$ is made to oscillate with an amplitude of $45^{\circ}$. The acceleration due to gravity is $g$. Let $T_0 = 2 \pi \sqrt{l / g}$. The time period of oscillation of this pendulum will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module