Question

When the displacement is half the amplitude the ratio of potential energy to the total energy is:

• A. 1/4
• B. 1/2
• C. 1/8
• D. 1

Answer 1

Answer: (A)

Potential energy is given by $=\frac{1}{2}m{a}^2{\omega }^2{\sin }^2\omega t$ ?(i) Total energy $=\frac{1}{2}m{a}^2{\omega }^2$ ?(ii) Now from (i), (ii) and (iii), we get $\frac{\text{Potential}\text{energy}}{\text{Total}\text{energy}}={\sin }^2\omega t=\frac{x^2}{a^2}$ Now for $x=\frac{a}{2},$ we get $\frac{\text{Potential}\text{energy}}{\text{Total}\text{energy}}={\left(\frac{a}{2}\right)}^2\times \frac{1}{a^2}=\frac{1}{4}$