JEE Main 2021 · 27 July · morning · Single correct · +4 / −1
A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is 3E/4 then its displacement 'y' is given by :
- (a) y = a
- (b) y = a/√2
- (c) y = a√3/2
- (d) y = a/2
Answer and solution
Answer (d) y = a/2
Solution
Physics sir · 3 steps
If KE is 3E/4, then PE must be E/4. Potential energy goes as y², so y is half the amplitude.
- Find the PE. Total energy stays fixed in SHM, so U = E - K = E - 3E/4 = E/4.
- Compare with E. Potential energy is U = 1/2ky² and total energy is E = 1/2ka². Their ratio depends only on y/a.
U/E = y²/a² = 1/4
- So. y² = a²/4, which gives y = a/2. That is option (d).
The idea to keep. In SHM the fraction of energy stored as PE is (y/a)². Find the PE share first, then take the square root to get the displacement.
Answer from the NTA final answer key. Solution by PrepWiser.