The maximum potential energy of a block executing simple harmonic motion is 25 J. A is amplitude of oscillation. At A / 2, the kinetic energy of the block is

Solution

Physics sir · 3 steps

Maximum potential energy is the total energy, and at half the amplitude PE is only a quarter of it.

1. Total energy. At the extreme position x = A the block stops for an instant, so all its energy is potential. The total energy is therefore E = 1/2kA² = 25 J, and it stays the same throughout the motion.
2. PE at A/2. Potential energy goes as x², not as x. Halving the displacement makes it a quarter: U = 1/2k(A/2)² = 1/4 × 25 = 6.25 J.

   K = E - U = 1/2k(A² - x²) = 25 - 6.25 = 18.75 J

3. So. At x = A/2 the kinetic energy is 3/4 of the total, that is 18.75 J, option (c).

The idea to keep. In SHM, PE is x²/A² of the total energy and KE is the rest. At x = A/2, that is a 1 : 3 split between PE and KE.

Answer from the NTA final answer key. Solution by PrepWiser.