Question

If at any time the displacement from mean position of a simple pendulum be $0.02\text{m}$, then its acceleration is $2{\text{m/s}}^2$. What is the angular frequency of the oscillation?

• A. $100\text{rad/s}$
• B. $10\text{rad/s}$
• C. $1\text{rad/s}$
• D. $0.1\text{rad/s}$

Answer 1

The correct option is B $10\text{rad/s}$

We know that
$a=-{\omega }^{2}y$, here negative sign indicates that acceleration $\left(a\right)$ is opposite to the displacement $\left(y\right)$
Hence, $\left|a\right|={\omega }^2|y|$
${\omega }^2=\left|\frac{a}{y}\right|$
$\Rightarrow \omega =\sqrt{\left|\frac{a}{y}\right|}=\sqrt{\frac{2}{0.02}}$ $=\sqrt{100}=10\text{rad/s}$