The force constant of an oscillating simple p | vedclass

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(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

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(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

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