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Q.
The ratio of kinetic energy to the potential energy of a particle executing $SHM$ at a distance equal to half its amplitude, the distance being measured from its equilibrium position is
Key concept As, potential energy of a particle executing simple harmonic niotion also periodic with period $T / 2$. So, potential eneroy' is zero at the mean position and maximum at the extreme displacements.
Let the amplitude of SHM be $A$.
Now, potential energy of $SHM =\frac{1}{2} \,kx ^{2}$
Here. $x=\frac{A}{2}$
$U=\frac{1}{2} k \frac{A^{2}}{4}\,\,\,\,\,\dots(i)$
Kinetic energy, $K=\frac{1}{2} k A^{2}-\frac{1}{2} k \frac{A^{2}}{4}$
$K=\frac{3}{8} k A^{2}\,\,\,\,\,\dots(ii)$
On dividing Eq. (ii) by Eq. (i), we get
$\frac{K}{U}=\frac{3}{8} \times \frac{8}{1} $
$\frac{K}{U}=\frac{3}{1}$