Question

Write the equation of an S.H.M. of amplitude 5cm if 150 oscillations are performed in

one minute and the initial phase is 45°
​

Answer 1

Time period of the given S.H.M. = 60/150

= 2/5

= 0.4 sec.

Φ = 45° or π/4

Now, Angular frequency (ω) = 2π/T = 5π

Now, Using the general equation of S.H.M. ,

x = ASin(ωt + Φ)

x = 5Sin(5πt + π/4)

This is the required equation of the S.H.M.

Hope it helps.