A gibbon, hanging onto a horizontal tree branch with one arm, swings with a small amplitude. The gibbon's CM is 0.40 m from the branch, and its rotational inertia divided by its mass is

$I / m = 0.25 \mathrm { m } ^ { 2 }$

. Estimate the frequency of oscillation.

Solutions

Verified$\textbf{Concepts: }$

Simple harmonic motion is defined as the vibratory motion that occurs when the restoring force is proportional to the displacement from equilibrium. The net force on the object is given by

$\begin{align*} F_{x} &= - k x \end{align*}$

Where, $k$ is the spring's constant. In order to evaluate the elastic potential energy of the spring, We use the following relation:

$\begin{align*} U &= \dfrac{1}{2} k x^{2} \end{align*}$

Where, $x$ is the displacement. In order to evaluate the total energy of the spring, We use the following relation:

$\begin{align*} E &= \dfrac{1}{2} m v_{x}^{2} + \dfrac{1}{2} k x^{2} \end{align*}$

**Given Data:**

Center of mass, $\text{d}=0.40\ \text{m}$.

Inertia by mass, $\dfrac{\text{I}}{\text{m}}=0.25\ \text{m}^2$.

**To Find:**

We need to find the frequency of oscillation.