# Homework Help: Damped oscilating spring

1. Jan 18, 2004

### Klion

Okie, doing homework for physics and I'm stuck. The section is on damped oscillations, question is as follows:

A 10.6 kg object oscillates at end of a vertical spring that has a spring constant of 2.05 * 10^4 N/m. The effect of air resistance is represented by the damping coefficient b=3.00 N*m/s.

a) Calculate the frequency of the damped oscillation.

b) By what percentage does the amplitude of the oscillation decrease in each cycle?

c) Find the time interval that elapses while the energy of the system drops to 5.00% of its initial value.

Now the section in the textbook on this is extremely sparse, there are no examples and basically only one equation. Fortunately the answer is in the back of the book so I can at least tell that I am doing it wrong, but I cannot figure out how to do it correctly.

The equation I believe to be relevant to (a) is

the angular frequency of oscillation is w = sqrt((k/m) - (b/2*m)^2)

and it also says it is convenient to express the angular frequency of a damped oscillator in the form w = sqrt(W0^2) - (b/2m)^2)

where W0 is omega not and W0 = sqrt(k/m)

It seems to me these are both the same equation, but the book distinguishes between the two, and between the two frequences (one as omega not, one as simply omega), and I doubt you would need the answer to find the answer, however I get the same answer regardless. Everything I've tried for a has given me 43.97, while the answer in the back of the book is 7.00 hz.

Any help appreciated. Thx.

-Klion

Last edited: Jan 18, 2004
2. Jan 18, 2004

### jamesrc

Has your class gone over the equations of motion for a damped oscillator? That is, do you know how to solve the differential equation:

$$m\ddot{x} + b\dot{x} + kx = 0$$

The natural frequency is given by:

$$\omega_n = \sqrt{\frac{k}{m}}$$

and the damped frequency is given by:

$\omega_d = \omega_n \sqrt{1-\zeta^2}$

(where zeta is defined by:
$$2\zeta\omega_n = \frac{b}{m}$$
)

It looks like your book labels the natural frequency &omega;o and simply &omega; as the damped natural frequency; same difference. The equation you have written is a valid expression for the damped natural frequency. The expressions for &omega; and &omega;o are not equivalent; you only get the same number because there is so little damping in this system.

Anyway, all of the &omega;'s in these equations are in radians per second. If you want to find a frequency in Hertz, you will have to divide by 2&pi;. How are you doing with the other two parts?

3. Jan 18, 2004

### Klion

ahaha right answer wrong units, duh :(

Hmm I'm working on the second part atm, my thoughts are that now that I have the frequency I also have the period so all I have to do is find the Amplitude at t=0 then find the amplitude at t=T then divide them to find the % difference between the two. Not sure if I have enough information to do that though. The formula it gives is the position as a function of time:

x = A*e ^(-bt/2m) * cos (W*t+phi)
so I figure solve for A and punch in the numbers but I need to find x (and phi?) to do that and not sure how to find it. Only equations I see for position have amplitude in them :/

4. Jan 18, 2004

### jamesrc

Leave A as a variable and start t = 0 at a maximum (choose your initial condition to be that at t = 0, x = A, forcing &phi; = 0).

Now the percent change can be found by comparing the next peak to the peak at t = 0 (compute x(Td)).

The percent change would then be:

$$100\frac{A-Ae^{-\frac{bT_d}{2m}}}{A}$$