Physics 2022 JAMB Past Questions - Question 18: A boy of mass m, suspended from a spring, is put into simple harmonic motion. If the motion has amplitude A and the spring constant k, the maximum potential energy of the mass is

Correct Answer

B

Explanation

The maximum potential energy of the mass in simple harmonic motion can be found using the formula for potential energy in a spring, which is given by \( PE = \frac{1}{2} kx^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

In simple harmonic motion, the maximum displacement from the equilibrium position is equal to the amplitude, so \( x = A \). Substituting this into the formula for potential energy, we get:

\( PE = \frac{1}{2} kA^2 \)

Therefore, the maximum potential energy of the mass in simple harmonic motion is \( \frac{1}{2} kA^2 \).