# Physics EE help

## Recommended Posts

Hi,

So for my EE, I'm finding the effect of string length on the speed of wave (on the string). I'm using the concept of standing waves on a string and in order to find the speed, I need to know which harmonic the string is playing at. How can I find that?

The formula I'm using is f = nv/2L - where I need to find n - the nth harmonic.

##### Share on other sites

Hi,

So for my EE, I'm finding the effect of string length on the speed of wave (on the string). I'm using the concept of standing waves on a string and in order to find the speed, I need to know which harmonic the string is playing at. How can I find that?

The formula I'm using is f = nv/2L - where I need to find n - the nth harmonic.

If I remember correctly, n (which is the harmonics number) is there to distinguish between different standing wave patterns. So $n = 1$ represents the simplest form of a standing wave. Then the next simplest pattern for a standing wave will be when $n = 2$, and so on. The general rule is as follow:

- If the string is fixed at both ends, then 'n' is equal to the total number of anti-nodes.

- If the string is unfixed at both ends, then 'n' is equal to the total number of nodes.

- If the string is fixed at only one end, then 'n' is equal to the number of nodes, which is also equal to the number of anti-nodes.

So the answer is, do the experiment! Try to create a standing wave with the string, and count how many anti-nodes or nodes there are. From there, use the rules above to calculate the harmonic number.

Btw I don't think the string length has anything to do with the speed of that the wave travels. The speed only depends on the property of the medium (which has to do with the material of the string, rather than how long it is). Also, notice that the harmonic number ( n ) is directly proportional to the frequency. This is a fact that can easily be tested with a simple experiment. Now let's check the maths. Consider a string with both ends fixed, then the formula is:

$\lambda = \frac{2L}{n}$

Now, multiply both sides with the frequency to get the speed of the wave, you'll get:

$v= f\lambda = f \frac{2L}{n} = Constant$

What this essentially means is that, in order to keep the speed of sound constant as you increase the harmonics, the frequency must be increased proportionally as well. And that's exactly the result that you should be getting in your experiment. The speed has nothing to do with the length of the string.

Of course, since I did this like 2 years ago, so my memory is a bit fuzzy and I could be wrong. If anybody spots a mistake, please let me know.

Edited by Vioh

## Create an account

Register a new account