JEE Main 2021 · 17 March · morning · Single correct · +4 / −1
For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal ?
- (a) x = A/2
- (b) x = ± A
- (c) x = ± A/√2
- (d) x = 0
Answer and solution
Answer (c) x = ± A/√2
Solution
Physics sir · 3 steps
Equal KE and PE means each is half the total energy, so set PE to half of 1/2kA².
- Total energy. In SHM the total energy stays fixed at E = 1/2kA². At displacement x the potential energy is U = 1/2kx², and the rest, K = 1/2k(A² - x²), is kinetic.
- Set them equal. If K = U, together they make E, so each one is exactly E/2.
1/2kx² = 1/2(1/2kA²) ⇒ x² = A²/2
- So. x = ± A/√2. The oscillator passes through this energy split on both sides of the mean position, so both signs count. The answer is (c).
The idea to keep. Potential energy grows as x², not as x. Half the energy as PE needs x² = A²/2, which puts you at A/√2, not at A/2.
Answer from the NTA final answer key. Solution by PrepWiser.