MCQ
The $ P.E.$ of a particle executing $SHM$ at a distance $x$ from its equilibrium position is
- ✓$\frac{1}{2}m{\omega ^2}{x^2}$
- B$\frac{1}{2}m{\omega ^2}{a^2}$
- C$\frac{1}{2}m{\omega ^2}({a^2} - {x^2})$
- DZero
✓
Answer
Correct option: A.
$\frac{1}{2}m{\omega ^2}{x^2}$
a
We know that potential energy is given by,
We know that potential energy is given by,
$PE =\frac{1}{2} kx ^2$
and we know, $\omega^2=\frac{k}{m}$
or, $k =\omega^2 m$
Using value of $k$ in equation $(1)$,
$PE =\frac{1}{2} m \omega^2 x ^2$ is our required answer.
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