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.
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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
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
==================
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
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