# Wave speed on a spring

1. Apr 18, 2007

### zergju

Hi i have a problem.
My physics teacher tells us that a longitudinal wave moving on a spring has speed v=(kL/u)^2 where k is spring constant n L e length of spring u=m/L of that spring..
which i think its unbelievable but the teacher told me its true..
I think the speed of e wave got nothing to do with spring's length.. Which this equation indicates that if u just use a spring 2 times longer, the speed of the wave will be 2 times faster.. which is unbelievable..
Am I wrong or what?

Thank you for all ur help!

2. Apr 18, 2007

### lpfr

I agree with you, it is incredible that the speed depend on total length.
But I made a short derivation adapting the general formula for fluids and solids to a spring and I found:
$$v=\sqrt{{KL\over \mu}}$$
(I think that there was a little typo error in your formula.)
The difference between a spring and a fluid or a solid is that the force that you need to do to compress a column of some distance $$\Delta L$$ varies as $${1\over L}$$. That is diminishes with total length. In the case of a spring, the force is always $$k\Delta L$$, independent of total length.

I will derive the formula from the beginning, to see if I obtain the same result... or, at least, to understand this surprising result. I will post the result tomorrow or after tomorrow

Last edited: Apr 18, 2007
3. Apr 18, 2007

### Hootenanny

Staff Emeritus
Your [TEX] brackets need to be in lowercase Also to write 1/L you need to write \frac{1}{L}

4. Apr 18, 2007

### Staff: Mentor

You are misunderstanding the equation. The wave speed depends not on the length of the spring, but on the mass per length. Using a spring twice as long does not change the mass/length, so the wave speed is unaffected (as long as the tension remains the same).

5. Apr 18, 2007

### lpfr

Thanks Hootenanny!
\frac{1}{L} is in LaTex {1\over L} is in TeX The two forms work.

6. Apr 18, 2007

### lpfr

OK Doc_Al. Would you please shows us the correct formula?

7. Apr 18, 2007

### Hootenanny

Staff Emeritus
Well, you learn something new everyday ... thanks

8. Apr 18, 2007

### Staff: Mentor

I would write the formula for wave speed as:
$$v=\sqrt{{T / \mu}}$$

And for a spring, that becomes:
$$v=\sqrt{{k \Delta L / \mu}}$$

As long as you keep the tension constant, doubling the length by adding a second spring (of same mass/length) should give the same speed.

Of course stretching the spring will change both tension and mass/length.

Am I missing something? (I may have to rethink this, as I am thinking of transverse waves.)

9. Apr 18, 2007

### lpfr

Yes, we are not talking about transverse waves but about longitudinal ones.

10. Apr 18, 2007

### Staff: Mentor

longitudinal waves

My bad! As lpfr stated, the speed of a longitudinal wave on a spring is given by:
$$v=\sqrt{{kL / \mu}}$$

Nonetheless, my earlier point remains that the speed is independent of the length of the spring as long as the tension remains fixed. Note that k is the spring constant for the spring of length L. Add a second spring and the new spring constant becomes k/2 while the new length becomes 2L--thus the speed remains the same. (Of course, if you stretch that same spring to twice its length you change both L and $\mu$, thus changing the speed.)

11. Apr 19, 2007

### lpfr

What I said is that, if you adapt the formula for solids and fluids, you obtain the formula I gave.
But this formula is certainly wrong. As zergju stated, it is physically unacceptable that the speed depend on the length of the spring. It is also in contradiction with relativity: you could know le length of the spring in less time than needed by light to go and come back to the extremity. And last, if L is big enough, the speed (group speed) could be bigger than c!

Please let my one day to find the time to derive the formula from the beginning.

12. Apr 19, 2007

### lpfr

The speed of longitudinal waves in a spring is:
$$v=\sqrt{{\kappa\over\mu}}$$
$$\mu$$ is the mass per unit length and
$$\kappa={k L}$$ is the spring constant per unit length (measured in N).
Yes, this was the catch: if you cut a length $$\ell$$ of a reel of spring, the constant (N/m) of the length you cut is $$k={\kappa \over \ell}$$.
This is the misleading $${L\over k}$$ that appeared on the formula. L is not the total length of the spring and k is not the constant of all the length of the spring. k is the constant of a length L of spring.
Happily, the speed doesn't depend on the length!