Question

Find out the amplitude and time period from equation of SHM y=a sin omega t​

Answer 1

Answer: Amplitude a , time period  2 π / ω

Explanation:

We know that in SHM particles retrace the same path for a period of time, so that time at which the particle returns back to the same state is called time period.

The amplitude for a particle undergoing the SHM

y = a sin ω t​ is obtained by the displacement of that particle at time t = 0 and

t = T ( where T is the time period ).

We know the time period T = 2 π / ω ;

where ω is the angular frequency of the particle.

We can find the amplitude as,  y (t = 0) = y(t = T= 2 π / ω) = a

[since sin 0 = sin 2π  = 0 ]