Question

Consider a simple harmonic motion (SHM). Let $K$ and $U$ be kinetic energy and potential energy when the displacement in SHM is one -half $\left(\frac{1}{2}\right)$ the amplitude. The correct statement is

• A. $\frac{K}{U}=1$
• B. $\frac{K}{U}=\frac{1}{2}$
• C. $\frac{K}{U}=\frac{4}{3}$
• D. $\frac{K}{U}=3$

Answer 1

Answer: (D)

Potential energy of a body performing SHM is given by
$U=\frac{1}{2}k{x}^2$
Here, $x$ is the displacement of the body from mean position.
Given, $x=\frac{a}{2}[a=$ amplitude $]$
$\therefore U=\frac{1}{2}k{\left(\frac{a}{2}\right)}^2$
$\Rightarrow U=\frac{1}{2}k\frac{a^2}{4}..(i)$
Now, kinetic energy of this body is given by
$K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2\left(a^2-x^2\right)$
$\Rightarrow K=\frac{1}{2}K\left(a^2-x^2\right)$
Given, $x=\frac{a}{2}$
$\therefore K=\frac{1}{2}K\left(a^2-\frac{a^2}{4}\right)$
$\Rightarrow K=\frac{1}{2}k\times \frac{3a^2}{4}$ с....(ii)
From Eqs. (i) and (ii), we get
$\frac{K}{U}=\frac{\frac{1}{2}k\times \frac{3a^2}{4}}{\frac{1}{2}k\times \frac{a^2}{4}}=\frac{3}{1}$