Question

A simple harmonic oscillator has a time period of 2 s. What will be the change in the phase 0.25 s after leaving the mean position?

Answer 1

it is given that,

time period, T = 2 sec

time taken to leave the mean position of particle , t = 0.25

use formula, $\Theta=\left(\frac{2\pi}{T}\right)t$

here $\Theta$ denotes phase difference.

so, $\Theta$ = (2π/2) × 0.25 = π/4

hence, phase difference = π/4

hence, A simple harmonic oscillator has a time period of 2 s. π/4 be the change in the phase 0.25 s after leaving the mean position.

Answer 2

Answer:

time period= 2seconds

y= Asin (¤)

¤= 2pi/T*t

= 2pi/2*0.25

=2pi/8

=pi/4