JEE Main 2022 · 27 June · evening · Single correct · +4 / −1
The equation of a particle executing simple harmonic motion is given by x = sin π ( t + 1/3 )m. At t = 1s, the speed of particle will be
(Given : π = 3.14)
- (a) 0 cm s⁻¹
- (b) 157 cm s⁻¹
- (c) 272 cm s⁻¹
- (d) 314 cm s⁻¹
Answer and solution
Answer (b) 157 cm s⁻¹
Solution
Physics sir · 4 steps
Differentiate x once to get velocity, then put t = 1 s into the cosine and take its size.
- Read the equation. Compare x = sin π(t + 1/3) with x = A sin(ω t + φ). The amplitude is A = 1 m and ω = π rad s⁻¹.
- Velocity. Velocity is the time derivative of displacement. The chain rule brings out a factor of π.
v = dx/dt = π cos π(t + 1/3)
- Put in time. At t = 1 s the phase is π(1 + 1/3) = 4π/3, and cos 4π/3 = -1/2. So v = -π/2 = -3.14/2 = -1.57 m s⁻¹.
- So. Speed is the size of the velocity: 1.57 m s⁻¹ = 157 cm s⁻¹, which is option (b).
The idea to keep. In SHM, speed at an instant comes from v = Aω cos(ω t + φ). Find the phase first, then the cosine. The full Aω is only the maximum speed, reached at the mean position.
Answer from the NTA final answer key. Solution by PrepWiser.