# Homework Help: Resonance Speed

1. Dec 2, 2013

### woaini

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

If a 0.505m long wire is excited into its lowest electrical resonance by a 2.2E7 Hz electrical oscillator, what is the ratio of the speed of the electrical current to that of light? Assume that the wire is like a tube with both ends closed.

f=2.2E7
L=0.505
v=?

2. Relevant equations

f=nv/2L
f=v/λ

3. The attempt at a solution

How do I solve this problem if I do not have λ or n? How would the equations work?

2. Dec 2, 2013

### CWatters

Hint: How would you find λ if this was a problem involving sound waves in an organ pipe? You would do a drawing showing a standing wave with nodes and antinodes then calculate the wavelength.

3. Dec 2, 2013

### woaini

How do I know how any nodes and antinodes to draw?

4. Dec 2, 2013

### CWatters

The problem states "lowest electrical resonance"

5. Dec 2, 2013

### woaini

Does that mean n=1?

6. Dec 2, 2013

7. Dec 2, 2013

### woaini

It says the fundamental frequency for a closed cylinder is f=v/4L

Therefore the speed I determine using the formula is...

v= f*4L = 2.2E7 * 4 *0.505 = 444400000 m/s

And the ratio of it to the speed of light is:

444400000 m/s / 3E8 m/s = 0.148

Does this look correct?

8. Dec 3, 2013

### CWatters

I think that's the formulae for a cylinder with one end open and one closed.

9. Dec 3, 2013

### woaini

So how do I determine the formula for something closed on both ends?

10. Dec 3, 2013

### CWatters

"Closed ends on a tube form a node..."

Draw it?

If you looked at the link I gave earlier it shows a drawing for one end open and one end closed. You can see how the formula is derived from that drawing. Do a new drawing with both ends closed and apply same ideas.

11. Dec 3, 2013

### woaini

Oh I see, the formula should be f=v/2L

12. Dec 3, 2013

### woaini

22220000m/s is the new speed I get?

13. Dec 3, 2013

### CWatters

Remember they ask for a ratio.

14. Dec 3, 2013

### woaini

Alright makes sense now.