# Homework Help: Out of 2 oscillators, which one decays to .1A first?

1. Oct 15, 2012

### cdot

1. The problem statement, all variables and given/known data

Consider 2 oscillators (mass attached to spring on surface). For one oscillator, the surface is frictionless, but there is viscous damping (f=-bv). For the other, the surface has coefficient of kinetic friction uk, but there is no viscous damping. The masses are both pulled away from the equilibrium the same distance then released. When they first reach the equilibrium position, the magnitude of the viscous damping force for the first oscillator is equal to the magnitude of the frictional force for the second oscillator. which oscillator damps down to 1/10 the initial amplitude more quickly?

2. Relevant equations

F=-bv (viscous damping at slow speeds)
F=uk×N=uk×mg (force of friction)
x=Ae^(-σt)cos(ωt+∅)

3. The attempt at a solution

The decaying envelope function e^(-σt) determines the degree of damping. The term Ae^(-σt) acts as a varying amplitude. I defined Ae^(-σt) to be the amplitude of the oscillator subject to viscous damping and Ae^(-βt) to be the amplitude of the oscillator with no viscous damping. By setting .1A=Ae^(-σt) and .1A=Ae^(-βt) and solving for t in both cases I could get the time taken for each oscillator to decay to .1A in terms of σ and β. Time taken for oscillator subject to viscous damping=t1=ln(.1)/-σ and the time taken for the other oscillator=t2=ln(.1)/-β. I have two equations and four unknowns. I know that for the oscillator subject to viscous damping σ=b/2m where m is the mass and b is the damping coefficient.

Last edited: Oct 15, 2012
2. Oct 15, 2012

### jambaugh

You don't need to solve for the exponential rate coefficients. Rather you need to compare them. The bigger one in magnitude (of beta and sigma) the faster the dampening. Or you might compare the forces throughout a cycle and see which is larger (thus loosing energy more quickly.) This you can do given they are equal at equilibrium. As the system moves past the equilibrium position how do the two types of forces change?

3. Oct 15, 2012

### cdot

The drag force is proportional to the velocity. The velocity is at a maximum at equilibrium so the drag force is at a maximum. As the system moves away from equilibrium the velocity decreases and so would the drag force. The force of friction is the same throughout the cycle so although the two forces are equal at equilibrium the drag force would be less than the force of friction everywhere else in the cycle. The system subject to the force of friction would lose energy quicker right? I think the question was made confusing by the fact that it specifically asked which one would reach .1A. I feel very stupid for spending 3 hours on this problem. Thank you for the help

4. Oct 15, 2012

### jambaugh

Remember the question's answer in and of itself wasn't the point. The understanding of the physics is. The time you spent is not wasted, it merely didn't apply to the specific question.

Oh, and your analysis sounds quite correct.