Resonance (Differential Equations Class)

1. Oct 24, 2007

dashkin111

[SOLVED] Resonance (Differential Equations Class)

1. The problem statement, all variables and given/known data
A front-loading washing machine is mounted on a thick rubber pad that acts like a spring; the weight W = mg (with g = 9.8 m/s^2) of the machine depresses the pad exactly 0.38 cm. When its rotor spins at \omega radians per second, the rotor exerts a vertical force
F_0 cos(omega t)
Newtons on the machine. Neglecting friction, determine at what speed (in revolutions per minute) resonance vibrations will occur?

2. Relevant equations

3. The attempt at a solution

I decided to just set it up like a force equation in physics.

$$F=ma$$
$$kx=mg$$

Now solve for $$\omega$$ which is $$\sqrt{\frac{k}{m}}$$

$$\frac{k}{m}=\frac{g}{x}=\omega^{2}$$

So omega is:

$$\frac{35\sqrt{10}}{2}$$

Transform to rpms

$$RPMS = \omega \frac{(60)}{2\pi}$$

Which to the nearest RPM is 528. But this is wrong. Any clues?

Last edited: Oct 25, 2007
2. Oct 24, 2007

jhicks

Try to recalculate omega, or better give the values you used to find omega.

3. Oct 24, 2007

Kreizhn

Yes, I'm confused as to where you got these numbers. Doing the same calculations I don't get the same result...

4. Oct 24, 2007

dynamicsolo

My suspicion is insufficient precision, for one thing. Also, where did 35/2 come from?

5. Oct 24, 2007

dashkin111

Okay, found my mistake and I'll also show more steps now too.

So from the part:

$$\frac{k}{m}=\frac{g}{x}$$

k/m is omega squared. G is given as 9.8 m/s^2 and x is given as .38 cm, or .0038 m.

Solving for Omega you get:

$$\omega=\sqrt{\frac{k}{m}} = \sqrt{\frac{g}{x}}= \sqrt{\frac{9.8}{.0038}}$$

This is where my mistake is, I mistakenly entered .0032 instead of .0038 in my calculator and got the previous result.

6. Oct 24, 2007

dashkin111

BTW it was 485 rpm's