Wave Speed on a Spring: Is It True?

[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!

 

[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

 

[Hootenanny]

I don't understand why TeX doesn't works!

Your $$brackets need to be in lowercase <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /> Also to write 1/L you need to write \frac{1}{L} <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" />$$

 

[Doc Al]

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?

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).

 

[lpfr]

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

 

[lpfr]

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

 

[Hootenanny]

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

Well, you learn something new everyday ... thanks

 

[Doc Al]

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.)

 

[lpfr]

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.)

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

 

[Doc Al]

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.)

 

[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.

 

[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!