# 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.

1
by sweta1234

2015-09-10T21:22:03+05:30

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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

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