PYQ NEET- Introduction To Periodic Motion

• NEET 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.