A simple harmonic motion is represented by ` x = 5 sin (10t + 0.25)`
Determine is
(i) amplitude
(ii) Angular frequency
(iii) Frequency
(iv) Time period
(v) Initial phase
[Displacement is measured in metres and time in seconds]

Explanation

Given, ` x= 5 sin(10t + 0.25)`
Comparing the above equation with standard equation for S.H.M.
`x = A sin (omegat + phi_(0))`
we get
(i) Amplitude, A = 5 m
(ii) Angular frequency , `omega = 10 "rad/s"`
(iii) Frequency ,`v = (omega)/(2pi) = (10)/(2pi) Hz = 1.59Hz`
(iv) Time period, `T = (2pi)/(omega) = (1)/(v) = (1)/(1.59) = 0.629s`
(v) Initial phase, `delta = 0.25` rad