JEE Main2015PhysicsOscillationsActual
A simple harmonic oscillator of angular frequency 2 rad s - 1 is acted upon by an external force F = sin ⁡ t N . If the oscillator is at rest in its equilibrium position at t = 0 , its position at later times is proportional to:
Options
- Asin t + 1 2 cos 2 t
- Bc o s t - 1 2 sin 2 t
- Csin t - 1 2 sin 2 t
- Dsin t + 1 2 sin 2 t
Correct answer
C. sin t - 1 2 sin 2 t
Step-by-step solution
It is given that oscillator at rest at t = 0 i.e. at t = 0 , v = 0 . So, in option, we can check by putting v = d x d t = 0 . (1) I f   x ∝ sin ⁡ t + 1 2 cos ⁡ 2 t ⟹ v ∝ cos ⁡ t + 1 2 × 2   ( - sin ⁡ 2 t ) ⟹     a t   t = 0 ,   v ∝ 1 - 0 ≠ 0 (2) I f   x   ∝ cos ⁡ t - 1 2 sin ⁡ t ⟹     v ∝ - sin ⁡ t - 1 2 cos ⁡ t ⟹ a t   t = 0 ,   v ∝ -