Question

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

• A. 3:01
• B. 4:01
• C. 2:01
• D. 8:01

Answer 1

Answer: (A)

3:01

Answer 2

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 E (ii) by E (i), we get
$\frac{K}{U}=\frac{3}{8} \times \frac{8}{1}$
$\frac{K}{U}=\frac{3}{1}$