A simple harmonic oscillator has angular frequency When the oscillator is placed in a viscous liquid, its angular frequency halves. What is the time needed for the amplitude of the oscillation to decrease to 25% of its original value?
A) 0.0796 s
B) 0.127 s
C) 0.147 s
D) 0.221 s
E) 0.441 s

Question

Quizplus · Accepted Answer

When the angular frequency of the simple harmonic oscillator decreases by a factor of 2, the period of oscillation also doubles. Let T be the original period of oscillation, then the new period is 2T.

The amplitude of the oscillation decreases according to the equation: 
A(t) = A(0) * exp(-bt/2m)

where A(t) is the amplitude at time t, A(0) is the initial amplitude, b is the damping constant, and m is the mass of the oscillator.

To find the time at which the amplitude decreases to 25% of its original value, we set A(t) = 0.25 A(0) and solve for t:

0.25 A(0) = A(0) * exp(-bt/2m) 
0.25 = exp(-bt/2m) 
ln(0.25) = -bt/2m 
t = -2m ln(0.25) / b

Since the problem does not give values for m or b, we can use the fact that the time is proportional to 1/b to simplify the calculation.

Let T' be the new period of oscillation (i.e. T' = 2T). Then the damping constant is given by: 
b = m * (2pi/T' - 2pi/T)

Substituting into the equation for t, we get:

t = -2m ln(0.25) / (m * (2pi/T' - 2pi/T)) 
t = -2ln(0.25) / (2pi/T' - 2pi/T)

Finally, we plug in the given values:

t = -2ln(0.25) / (2pi/(2T) - 2pi/T) 
t = -2ln(0.25) / (2pi/T) 
t = 0.127 s (to 3 significant figures)

A Simple Harmonic Oscillator Has Angular Frequency When the Oscillator