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. Find the frequency of vibration of the system for small values of the amplitude (small$\mathit{\theta }$ ). Assume the vertical suspension rod of length L is rigid, but ignore its mass.

Short answer

$f=\frac{1}{2\pi }\sqrt{\frac{\left(k{h}^2+mgL\right)}{m{L}^2}}$

Step-by-step solution

Given information

Given that the pendulum has length L, mass M, spring constant k, amplitude $\theta$.

Concept

A quantity expressing a body tendency to resist angular acceleration, which is the sum of the product of the mass of each particle in the body with the square of its distance from the axis of rotation, it is given by:

$\mathit{\tau }\mathbf{=}\mathit{I}\mathit{\alpha }$

Formula for angular frequency

As the pendulum slightly shifted to the left then we have restoring torque it gives as:

$\tau =I\alpha$

So here, we have

$\tau =kxh+mg\sin \theta L$

Now we know that:

$x=h\theta$

Also for small angular displacement:

$\tau =k{h}^2\theta +mgL\theta$

Now we have its moment of inertia about suspension point given as:

$I=m{L}^2$

Thus becomes;

$$\begin{gathered} m{L}^2\alpha =\left(k{h}^2+mgL\right)\theta \\ \alpha =\frac{\left(k{h}^2+mgL\right)}{m{L}^2}\theta \end{gathered}$$

So we have

${\omega }^2=\frac{\left(k{h}^2+mgL\right)}{m{L}^2}$

So angular frequency is given by:

$f=\frac{1}{2\pi }\sqrt{\frac{\left(k{h}^2+mgL\right)}{m{L}^2}}$