Question

The angular frequency of a simple pendulum is $\omega \mathrm{rad}/\sec$. Now, if the length is made one-fourth of the original length, the angular frequency becomes

• A. $\frac{\omega }{2}$
• B. $2\omega$
• C. $4\omega$
• D. $\omega$

Answer 1

The correct option is B

$2\omega$

Step 1: Given

Given, the angular frequency of a simple pendulum is $\omega \mathrm{rad}/\sec$.

The reduced length of the pendulum is $\frac{l}{4}$ where $l$ is the initial length of the pendulum.

Step 2: Formula used

We know that, $\omega =\sqrt{\frac{g}{l}}$, where $\omega$ is the angular frequency, $g$ is the acceleration due to gravity and $l$ is the length of the pendulum.

Step 3: Calculate the new frequency

Thus the new angular frequency of the pendulum will be,

$\begin{array}{rcl}{\omega }_0& =& \sqrt{\frac{g}{{\displaystyle \frac{l}{4}}}}\\ & =& \sqrt{\frac{4g}{l}}\\ & =& 2\sqrt{\frac{g}{l}}\\ \therefore {\omega }_0& =& 2\omega \end{array}$

Hence, option B is correct.