Question

A particle is performing a linear simple harmonic motion of amplitude 'A'. When it is midway between its mean and extreme position, the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

Options

• `(2pi)/sqrt3`s

• `sqrt3/(2pi)`s

• `2pisqrt3`s

• `1/(2pisqrt3)`s

Answer 1

`(2pi)/sqrt3`s

Explanation:

In linear simple harmonic motion, the velocity of particle is given by

v = `omega sqrt("A"^2 - x^2)`     ....(i)

where, ω = angular frequency

A = maximum displacement of amplitude

and X = displacement from mean position.

The acceleration of a particle in simple harmonic motion, (SHM) is given by

a = ω²x     ...(ii)

Here, x = `"A"/2`

Also, v = a    ....(given)

`omega sqrt("A"^2 - x^2) = omega^2 x`   ....[from Eqs. (i) and (ii), we get]

`=> sqrt("A"^2 - "A"^2/4) = omega xx "A"/2`

`=> (sqrt3 "A")/2 = omega xx "A"/2`

`=> (2pi)/"T" = sqrt3    ....[because omega = (2pi)/"T"]`

`=> "T" = (2pi)/sqrt3`s