Question

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

(a) 1/ 2π√3

(b) 2π√3

(c) 2π/√3

(d) √3/2π

The question was asked by my college director while I was bunking the class.

The origin of the question is Velocity and Acceleration of a Particle Moving with Simple Harmonic Motion topic in section Simple Harmonic Motion of Machine Kinematics

Answer 1

The correct option is (b) 2π√3

Best explanation: The magnitudes of the velocity and acceleration of the particle when its displacement is ‘y’ are ω√(A2 –y2) and ω2y respectively. Equating them, ω√(A2 –y2) = ω2y, from which ω = [√(A2 –y2)]/y = √(4 –1) = √3. Period T = 2π/ω = 2π/√3.