Question

A simple harmonic oscillator executes motion whose amplitude is 0.20m and it completes 60 oscillations in 2 mints.
1. Calculate its time period and angular frequency.
2. If the initial phase is 45°, write expressions for instantaneous displacement, velocity and acceleration.
3. Also calculate the maximum values of velocity and acceleration of the oscillator.

Answer 1

Amplitude = A = 0.20 m
 
frequency f of oscillations = 60 / 120 seconds = 0.50 Hz

angular frequency ω = 2 π f = π rad/sec

time period = T = 1/f  or  = 120 sec / 60 oscillations = 2 sec per oscillation

Initial phase Ф for the displacement is given as π/4 radians.

         x = A Sin (ω t + Ф)              for SHM
           x = 0.20 Sin (π t + π/4)  meters.
     velocity v = dx/dt = A ω Cos (ω t + Ф)
                   =  0.20 π Cos (π t + π/4)    m/sec
  acceleration a = dv/dt = - A ω² Cos (ω t + Ф)
                      =  - 0.20 π² Sin (πt + π/4)

  maximum velocity = A ω = 0.20 π  m/s
  maximum acceleration = Aω² = 0.20 π²  m/sec²