Q120 / 150 · MHT CET 2009 · Physics
Ratio of kinetic energy at mean position to potential energy at $A / 2$ of a partide performing SHM
- A$2: 1$
- Correct answerB$4: 1$
- C$8: 1$
- D$1: 1$
Solution
Kinetic energy $K=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)$
At mean position $y=0$
$$
K=\frac{1}{2} m \omega^{2}\left(A^{2}\right)
$$
Potential energy $U=\frac{1}{2} m \omega^{2} y^{2}$
$U$ at $y=\frac{A}{2}$
$$
U=\frac{1}{2} \frac{m A^{2}}{4} \omega^{2}
$$
Dividing Eq (i) by Eq (ii), we get
$$
\frac{\mathrm{KE}}{U}=\frac{4}{1}
$$
Hence, the correct answer is option: 'B'.