Question

A particle moves such that its acceleration a is given by a = -bx, where x is the displacement from equilibrium position and b is a constant. The period of oscillation is: 

1. \(\sqrt{2\pi/b}\)
2. \(2\pi/\sqrt{b}\)
3. 2π/b 
4. \(2\sqrt{(\pi /b)}\)

Answer 1

Correct Answer - Option 2 : \(2\pi/\sqrt{b}\)

Concept:

• Simple harmonic motion occurs when the restoring force is directly proportional to the displacement from equilibrium.

F α -x

Where F = force and x = the displacement from equilibrium.

The equation of SHM is given by:

x = A sinωt

where x is the distance from the mean position at any time t, A is amplitude, t is time, and ω is the angular frequency.

Explanation:

We have a = -bx

For simple harmonic motion

x = A sinωt

comparing ω² = b, ω = √b

\(\omega=\frac{2\pi}{t}= \sqrt{b}\)

Or we can say \(t = \frac{2\pi}{\sqrt{b}}\)