Question

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

• A. 44565
• B. 44563
• C. 44569
• D. 1

Answer 1

Answer: (A)

44565

Answer 2

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 $\frac{\text{Potential}\,\text{energy}}{\text{Total}\,\text{energy}}={{\left( \frac{a}{2} \right)}^{2}}\times \frac{1}{{{a}^{2}}}=\frac{1}{4}$