Question

The ratio of kinetic energy and total energy of a particle in simple harmonic motion, at a point where the displacement is 30% of its amplitude is:

• A. 91 : 100
• B. 49 : 100
• C. 81 : 100
• D. 51 : 100

Hint:

In SHM, the ratio of kinetic to total energy depends on the displacement. Use $KE = \frac{1}{2} k (A^2 - x^2)$ and $E = \frac{1}{2} k A^2$ to find the ratio at any point.

Answer 1

The Correct Option is C

In simple harmonic motion, total energy $E = \frac{1}{2} k A^2$, where $A$ is the amplitude and $k$ is the spring constant.
At displacement $x$, potential energy $PE = \frac{1}{2} k x^2$, and kinetic energy $KE = E - PE = \frac{1}{2} k (A^2 - x^2)$.
Given $x = 0.3A$, we have $x^2 = (0.3A)^2 = 0.09A^2$.
Thus, $KE = \frac{1}{2} k (A^2 - 0.09A^2) = \frac{1}{2} k (0.91A^2) = 0.91 \times \frac{1}{2} k A^2 = 0.91E$.
The ratio of kinetic energy to total energy is $KE : E = 0.91E : E = 0.91 : 1 = 91 : 100$.