JEE Main 2023 · 13 April · evening · Single correct · +4 / −1
A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is :
- (a) 1/√2 A
- (b) 1/2 A
- (c) 2 A
- (d) √(2 A)
Answer and solution
Answer (a) 1/√2 A
Solution
Physics sir · 3 steps
Total energy stays fixed, so when KE equals PE, each one is exactly half of it.
- Energy at any point. At distance x from the mean position, U = 1/2kx². The total energy is E = 1/2kA² and does not change, so K = E - U.
- Set them equal. If K = U, then K + U = 2U = E. So the potential energy is half the total energy.
1/2kx² = 1/2(1/2kA²) ⇒ x² = A²/2
- So. x = A/√2 ≈ 0.71A, which is option (a).
The idea to keep. Potential energy grows as x², not as x. Half the energy is reached at A/√2, not at A/2. At A/2 the potential energy is only one quarter of the total.
Answer from the NTA final answer key. Solution by PrepWiser.