The amplitude of a simple pendulum is $10 \mathrm{~cm}$. When the pendulum is at a displacement of $4…
The amplitude of a simple pendulum is $10 \mathrm{~cm}$. When the pendulum is at a displacement of $4 \mathrm{~cm}$ from the mean position, the ratio of kinetic and potential energies at that point is
$5.25$
$2.5$
$4.5$
$7.5$
Solution
$\begin{aligned}
& \text { } \mathrm{K} \cdot \mathrm{E}=\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(A^2-x^2\right) \\
& \mathrm{P} . \mathrm{E}=\frac{1}{2} k x^2=\frac{1}{2} m \omega^2 x^2
\end{aligned}$
Therefore, the ratio
$\begin{aligned}
& \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\frac{1}{2} m \omega^2\left(A^2-x^2\right)}{\frac{1}{2} m \omega^2 x^2} \Rightarrow \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{\mathrm{A}^2-\mathrm{x}^2}{\mathrm{x}^2} \\
& \Rightarrow \frac{10^2-4^2}{4^2}=\frac{84}{16} \\
& \Rightarrow \frac{\mathrm{KE}}{\mathrm{PE}}=\frac{21}{4}=5.25
\end{aligned}$