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 : 1
• B. 4 : 1
• C. 2 : 1
• D. 8 : 1

Answer 1

Answer: (A)

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 $SHM=\frac{1}{2}k{x}^2$
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
$\frac{K}{U}=\frac{3}{8}\times \frac{8}{1}$
$\frac{K}{U}=\frac{3}{1}$