# 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 niona

2015-04-02T03:58:24+05:30

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Simple harmonic motion:

A = amplitude = 0.20 m
60 oscillations in 2 minutes = 120 sec.
So frequency = 1/2 oscillation in 1 sec.  => f = 1/2 Hz
time period T = 1/f = 2 seconds.
angular frequency = ω = 2 π f = 2 π / T = π radians
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2.    let the displacement from the mean position during the SHM be:
x (t) = A Cos (ω t + Ф)
= 0.20 Cos ( π t + π/4)  meters
=  020 Cos π (t + 0.25)  meters

velocity v(t) = d x(t) / dt  = - 0.20 π Sin π(t + 0.25)

acceleration = a(t) = d v(t) / dt = - 0.20 π² Cos π(t + 0.25)
or,    a (t) = - π² x(t)
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3.      maximum value of velocity = v(t = 0.25 sec or 1.25 sec)
= 0.20 π  m/sec

maximum value of acceleration = a(t = 0.75 sec or 1.75 sec)
= 0.20 π²  m/sec²