Physics - SHM Question with Solution | TestHub

Let the amplitude of S.H.M. be $A$ and displacement be $x$.

Given $x=\frac{A}{2}$.

Kinetic energy of the particle is $K=\frac{1}{2}m{\omega }^2(A^2-x^2)$.

Potential energy of the particle is $U=\frac{1}{2}m{\omega }^2{x}^2$.

The ratio of kinetic energy to potential energy is:

$\frac{K}{U}=\frac{A^2-x^2}{x^2}$

Substituting $x=\frac{A}{2}$:

$\frac{K}{U}=\frac{A^2-{\left(\frac{A}{2}\right)}^2}{{\left(\frac{A}{2}\right)}^2}$

$\frac{K}{U}=\frac{A^2-\frac{A^2}{4}}{\frac{A^2}{4}}$

$\frac{K}{U}=\frac{\frac{3A^2}{4}}{\frac{A^2}{4}}=3$

The ratio is $3:1$.

Answer: $3:1$