To solve the problem of finding the ratio of kinetic energy (KE) to potential energy (PE) when the body undergoing Simple Harmonic Motion (SHM) is at position \( x = \frac{A}{2} \), we need to understand the energy characteristics of SHM. Let's go through the steps:
In SHM, the total energy \(E\) of the system is constant and is given by:
\( E = \frac{1}{2} k A^2 \),
where \(k\) is the spring constant and \(A\) is the amplitude.
The potential energy at position \(x\) is:
\( PE = \frac{1}{2} k x^2 \).
Substitute \(x = \frac{A}{2}\) into the potential energy formula:
\( PE = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{2} k \frac{A^2}{4} = \frac{1}{8} k A^2 \).
The kinetic energy is then calculated by subtracting the potential energy from the total energy:
\( KE = E - PE = \frac{1}{2} k A^2 - \frac{1}{8} k A^2 = \frac{4}{8} k A^2 - \frac{1}{8} k A^2 \)\( = \frac{3}{8} k A^2 \).
Thus, the ratio of kinetic energy to potential energy at \(x = \frac{A}{2}\) is:
\[
\text{Ratio} = \frac{KE}{PE} = \frac{\frac{3}{8} k A^2}{\frac{1}{8} k A^2} = \frac{3}{1}
\]
Therefore, the ratio of kinetic energy to potential energy when the body is at position \(x = \frac{A}{2}\) is 3:1. Hence, the correct answer is:
