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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. Which of the correct statement?
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The correct answer is:
$\frac{K}{U}=3$
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 $]$
$\therefore \quad U=\frac{1}{2} k\left(\frac{a}{2}\right)^2 \Rightarrow U=\frac{1}{2} k \frac{a^2}{4}$ $\ldots(\mathrm{i})$
Now, kinetic energy of this body is given by
$K=\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(a^2-x^2\right)$
$\Rightarrow \quad K=\frac{1}{2} K\left(a^2-x^2\right)$
Given, $\quad x=\frac{a}{2}$
$\therefore \quad K=\frac{1}{2} K\left(a^2-\frac{a^2}{4}\right) \Rightarrow K=\frac{1}{2} k \times \frac{3 a^2}{4}$ $\ldots(\mathrm{ii})$
From Eqs. (i) and (ii), we get
$\frac{K}{U}=\frac{\frac{1}{2} k \times \frac{3 a^2}{4}}{\frac{1}{2} k \times \frac{a^2}{4}}=\frac{3}{1}$
$U=\frac{1}{2} k x^2$
Here, $x$ is the displacement of the body from mean position.
Given,
$x=\frac{a}{2}$ [ $a=$ amplitude $]$
$\therefore \quad U=\frac{1}{2} k\left(\frac{a}{2}\right)^2 \Rightarrow U=\frac{1}{2} k \frac{a^2}{4}$ $\ldots(\mathrm{i})$
Now, kinetic energy of this body is given by
$K=\frac{1}{2} m v^2=\frac{1}{2} m \omega^2\left(a^2-x^2\right)$
$\Rightarrow \quad K=\frac{1}{2} K\left(a^2-x^2\right)$
Given, $\quad x=\frac{a}{2}$
$\therefore \quad K=\frac{1}{2} K\left(a^2-\frac{a^2}{4}\right) \Rightarrow K=\frac{1}{2} k \times \frac{3 a^2}{4}$ $\ldots(\mathrm{ii})$
From Eqs. (i) and (ii), we get
$\frac{K}{U}=\frac{\frac{1}{2} k \times \frac{3 a^2}{4}}{\frac{1}{2} k \times \frac{a^2}{4}}=\frac{3}{1}$
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