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. 2 : 1
• B. 3 : 1
• C. 8 : 1
• D. 4 : 1

Answer 1

The correct option is B

3 : 1

$KE=\frac{1}{2}m{\omega }^2(A^2-y^2),y=\frac{A}{2}U=\frac{1}{2}m{\omega }^2{y}^2\frac{KE}{U}=\frac{A^2-y^2}{y^2}=\frac{A^2-\frac{A^2}{4}}{\frac{A^2}{4}}=\frac{3A^2}{4}\times \frac{4}{A^2}=\frac{3}{1}$