Physics - SHM Question with Solution | TestHub

Given amplitude $A=3$ cm.

The kinetic energy of a particle in SHM is given by $K=\frac{1}{2}m{\omega }^2(A^2-x^2)$ and the potential energy is $U=\frac{1}{2}m{\omega }^2{x}^2$.

Given that kinetic energy is $25\%$ more than potential energy:

$K=U+\frac{25}{100}U=\frac{5}{4}U$

$\frac{1}{2}m{\omega }^2(A^2-x^2)=\frac{5}{4}\left(\frac{1}{2}m{\omega }^2{x}^2\right)$

$A^2-x^2=\frac{5}{4}{x}^2$

$A^2=\frac{9}{4}{x}^2$

$x=\frac{2}{3}A$

Substituting $A=3$ cm:

$x=\frac{2}{3}\times 3=2$ cm.

Answer: $2$