Resonance (Differential Equations Class)

[dashkin111]

[SOLVED] Resonance (Differential Equations Class)

Homework Statement

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?

Homework Equations

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?

 

[jhicks]

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

 

[Kreizhn]

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

 

[dynamicsolo]

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?

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

 

[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.

 

[dashkin111]

BTW it was 485 rpm's