A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential energy (PE) equals kinetic energy (KE), the position of the particle will be
Select the correct option:
A
$\frac{A}{\sqrt{2}}$
B
A
C
$\frac{\mathrm{A}}{2 \sqrt{2}}$
D
$\frac{A}{2}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol. KE = potential energy
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} m v^2=\frac{1}{2} K x^2 \\
& \frac{1}{2} m \omega^2\left(A^2-x^2\right)=\frac{1}{2} K x^2 \\
& A^2-x^2=x^2 \\
& x=\frac{A}{\sqrt{2}}
\end{aligned}
$$
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