Question

A pendulum of length L and mass M has a spring of force constant k connected to it at a distance h below its point of suspension (as shown in the following figure). Find the frequency of vibration of the system for small values of the amplitude (small ?). Assume that the vertical suspension of length L is rigid, but ignore its mass. (Use any variable or symbol stated above along with the following as necessary: g and ?.) f =_________.

Answer 1

Answer:

$f = \frac{1}{2\pi}\sqrt{\frac{(kh^2 + mgL)}{mL^2}}$

Explanation:

As the pendulum is slightly shifted to the left then we have restoring torque on it given as

$\tau = I \alpha$

so here we have

$\tau = kx(h) + mg sin\theta L$

here we know that

$x = h\theta$

also for small angular displacement

$\tau = kh^2\theta + mgL\theta$

now we have its moment of inertia about suspension point given as

$I = mL^2$

now it is given as

$mL^2 \alpha = (kh^2 + mgL)\theta$

$\alpha = \frac{(kh^2 + mgL)}{mL^2}\theta$

so we have

$\omega^2 = \frac{(kh^2 + mgL)}{mL^2}$

so angular frequency is given as

$f = \frac{1}{2\pi}\sqrt{\frac{(kh^2 + mgL)}{mL^2}}$