JEE Main 2024 · 4 April · evening · Single correct · +4 / −1
In simple harmonic motion, the total mechanical energy of given system is E. If mass of oscillating particle P is doubled then the new energy of the system for same amplitude is:
- (a) E / √2
- (b) 2 E
- (c) E √2
- (d) E
Answer and solution
Answer (d) E
Solution
Physics sir · 3 steps
SHM energy is set by the spring and the amplitude, 1/2kA². The mass never enters it.
- Energy at the extreme. At the extreme position the block is momentarily at rest, so all the energy is stored as spring energy. For a vertical spring, measure A from the equilibrium position. Gravity only shifts where that equilibrium is.
E = 1/2mω²A² = 1/2m(k/m)A² = 1/2kA²
- Double the mass. With mass 2m, ω² becomes k/2m, half its old value. The factor m doubles, ω² halves, and their product stays k. The spring constant k and the amplitude A are both unchanged.
- So. The new energy is 1/2kA² = E, unchanged. The answer is (d). Only the motion slows down: the period grows by a factor of √2.
The idea to keep. For a spring-block oscillator, total energy is 1/2kA². Change the mass and only the period changes. Change k or A and the energy changes.
Answer from the NTA final answer key. Solution by PrepWiser.