JEE Main 2021 · 18 March · evening · Single correct · +4 / −1
The function of time representing a simple harmonic motion with a period of π/ω is :
- (a) cos(ωt) + cos(2ωt) + cos(3ωt)
- (b) sin²(ωt)
- (c) sin(ωt) + cos(ωt)
- (d) 3cos( π/4 - 2ω t )
Answer and solution
Answer (d) 3cos( π/4 - 2ω t )
Solution
Physics sir · 3 steps
Write each option as one sinusoid, then check it is centred on zero with period π/ω.
- What SHM needs. An SHM displacement is a single sinusoid, x = A cos(ω' t + φ), about the mean position x = 0. Then d²x/dt² = -ω'² x and the period is T = 2π/ω'. A period of π/ω needs ω' = 2ω.
Option Rewritten SHM about x = 0? Period (a) three cosines with frequencies ω, 2ω, 3ω no, it is not a single sinusoid 2π/ω (b) 1/2 - 1/2 cos(2ω t) no, it oscillates about 1/2 π/ω (c) √2 sin(ω t + π/4) yes 2π/ω (d) 3 cos(2ω t - π/4) yes π/ω - Check option (d). Cosine is even, so 3 cos(π/4 - 2ω t) = 3 cos(2ω t - π/4). Differentiating twice gives d²x/dt² = -4ω² x. This is SHM with ω' = 2ω, so T = 2π/2ω = π/ω.
- So. Only (d) is a simple harmonic motion about x = 0 with period π/ω. The answer is (d).
The idea to keep. Reduce the function to one A cos(ω' t + φ) and read off T = 2π/ω'. A mix of frequencies is not SHM, and a constant term moves the mean position off zero.
Answer from the NTA final answer key. Solution by PrepWiser.