Ratio of the periods of a damped and undamped oscillator

[Dustgil]

Homework Statement

Given: The amplitude of a damped harmonic oscillator drops to 1/e of its initial value after n complete cycles. Show that the ratio of period of the oscillation to the period of the same oscillator with no damping is given by

$$\frac {T_d} {T_0} = \sqrt {1+ \frac {1} {4\pi^2n^2}} = 1+\frac {1} {8\pi^2n^2}$$

where the approximation in the last expression is valid if n is large.

Homework Equations

I think

$$\frac {\Delta E} {E} = \frac {T_d} {\tau}$$

The Attempt at a Solution

So the energy of the damped oscillator is equal to

$$E=\frac {kA^2} {2}$$

And the change in energy over n cycles is equal to

$$\Delta E= \frac {kA^2} {2} - \frac {k(\frac {A} {e})^2} {2} = \frac {kA^2} {2} (1-\frac {1} {e^2})$$

So

$$\frac {\frac {kA^2} {2} (1-\frac {1} {e^2})} {\frac {kA^2} {2}} = \frac {nT_d} {\tau}$$

This is as far as I get, meaningfully. I'm struggling to find a way to relate the period of the damped and undamped oscillator that makes e drop out. Any thoughts?

 

[Orodruin]

I think

Why would this hold?

I suggest that you instead look at the actual solution to the damped oscillator.

 

[Dustgil]

Why would this hold?

I suggest that you instead look at the actual solution to the damped oscillator.

Sorry, I guess I did give that route a shot.

$$x = Ae^{-\gamma t}$$

$$\frac {A} {e} = Ae^{-\gamma t}$$ so the exponent must equal 1 after n cycles. From there on I'm stuck. I'm not sure how to put n into the exponent to make it give the desired behavior. I must be missing an equation or two..

 

[Dustgil]

I guess I'm just confused on how to pull the period of oscillation from that equation.

 

[Orodruin]

Sorry, I guess I did give that route a shot.

x=Ae−γt​

This is not the solution, it is just the amplitude. You need the full solution to the damped oscillator differential equation.

It is correct that $\gamma t$ must be equal to one after $n$ periods, but to know the period you must know the oscillatory part of the solution. You cannot do this looking at the amplitude only.

 

[ehild]

See http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html.

 

[Dustgil]

I appreciate the help. I've got it now I think. Here's what I found.

The equation is

$$\frac {A} {e}=Ae^{-\gamma T_d}cos(\omega T_d + \phi)$$

In order for this equation to equal A/e,

$$\gamma T_d=1$$

$$\omega_d T_d=2\pi n$$

$$T_d= \frac {2 \pi n} {\omega_d}= \frac {1} {\gamma}$$

Which makes the damped frequency equal to

$$\omega_d = 2\pi n \gamma$$

We can then relate the frequency of the damped oscillator to the frequency of the undamped through the equation

$$\omega_d = \sqrt{\omega_0^2 - \gamma^2}$$

$$2\pi n \gamma = \sqrt{\omega_0^2 - \gamma^2}$$

$$\omega_0 = \gamma \sqrt{1+4pi^2n^2}$$

Then the ratio is:

$$\frac {T_d} {T_0} = \frac {\frac {2\pi} {\omega_d}} {\frac {2\pi} {\omega_0}}=\frac {w_0} {w_d}= \frac {\sqrt{1+4\pi^2n^2}} {2\pi n}=\sqrt{1+ \frac {1} {4\pi^2n^2}}$$

 

[Orodruin]

Right. I assume you also understand the working behind the approximation
$$
\sqrt{1+\frac{1}{4\pi^2 n^2}} \simeq 1 + \frac{1}{8\pi^2 n^2},
$$
which was part of the question (and is an approximation, not an equality, valid for large $n$).