- JEE Main 2019: A particle is executing simple harmonic motion (SHM). Its amplitude is A and time period is T. The magnitude of the force acting on the particle at the instant when its displacement is half the amplitude is given by:
(A) 3/4 mAω2
(B) 1/4 mAω2
(C) 1/2 mAω2
(D) mAω2
The answer is (C)
The force acting on a particle executing SHM is given by:
F = -mω2x
where,
- m is the mass of the particle
- ω is the angular frequency of the oscillation
- x is the displacement of the particle from its equilibrium position
When the displacement of the particle is half the amplitude, x = A/2. Substituting this value into the equation for the force, we get:
F = -mω2(A/2) = -mω2A/2
This is the magnitude of the force acting on the particle at the instant when its displacement is half the amplitude.
