Standing waves in a string

1. Aug 18, 2008

Pranas Juozas

1. The problem statement, all variables and given/known data
I just want to check the solution of the following problem:

We have got a string whose length is l and fundamental frequency is f. This string is clamped at a point 0.25l. What are possible frequencies of oscillations of this clamped string?

3. The attempt at a solution

So here is the solution:

The fundamental vibrational mode of a stretched string is such that the wavelength is twice the length of the string.
From the original length we can get the speed

v =f(2l) =2fl

From the new length
Wavelength =L= 2(0.75l) = (1.5)l
freq= Wavelength/speed = (1.5)l / 2fl= 0.75/f

So the new fundamental frequency is 0.75(1/f)
The new harmonics are given by the rule:
fn = n(fo)

So they are

fn = n (0.75)(1/f)

2. Aug 19, 2008

dynamicsolo

Ah, but the string isn't 0.75 L long -- its length hasn't changed at all! It is just no longer free to vibrate at the point x = 0.25 L; it can still vibrate on the sections on either side of the clamp.

So the fundamental frequency isn't changed; instead, the only permitted standing waves on the string of length L are those which has a node at 0.25 L

3. Aug 19, 2008

Pranas Juozas

So the speed is still v = 2fl.

The possible wavelengths now are:

a = (2n - 1)* wavelenght/4

here n = 1,2,3,4...; a = 0.25l or 0.75l

So the possible frequencies now are:

f(possible) = 2f(2n-1)/3 or f(possible) = 2f(2n-1);

4. Aug 19, 2008

dynamicsolo

The picture is this: the string with length L is fastened at both ends, so its permitted standing waves would be $$\lambda_n = \frac{2L}{n} , n = 1, 2, 3, ...$$. This would cause the string to have nodes at $$x_k = k \cdot \frac{L}{n} , k = 0, 1, 2, ..., n$$.

The clamp at x = L/4 only fixes the string at that one point (much as a person's finger holds down a string on a violin or guitar), forcing the presence of a "node" there. This means that the only standing waves that can still occur are those having a node at L/4 . The only wavelengths for which that can happen are those with n equal to a multiple of 4 . (The entire length of the string vibrates, but the points which now must be stationary are x = 0, x = L/4, and x = L .)

The permitted waves are thus $$\lambda_{4n} = \frac{2L}{4n} = \frac{L}{2n} , n = 1, 2, 3, ...$$. If we call the fundamental frequency of the string $$f_0 = \frac{v}{2L}$$, then the permitted frequencies of the clamped string are

$$f'_n = \frac{v}{\lambda_{4n}} = \frac{v}{\frac{L}{2n}} = \frac{2nv}{L} , n = 1, 2, 3, ...$$

or, in terms of the fundamental frequency,

$$\frac{f'_n}{f_0} = \frac{\frac{2nv}{L} }{\frac{v}{2L}} = 4n , n = 1, 2, 3, ...$$ .

So the possible frequencies which can be heard from this clamped string are

$$f'_n = 4n \cdot f_0 , n = 1, 2, 3, ...$$.

5. Aug 20, 2008

Pranas Juozas

Thanks for help dynamicsolo.

I don't understnad one thing in your solution. You say that the nodes must be at points x = 0 and x = L although the string is not fixed at the ends (it may sound strange but it is a theoretical problem). In my last solution I thought that at these points there should be anti-nodes because they can oscillate freely.

Am I confusing the issue again?

6. Aug 20, 2008

dynamicsolo

I assumed that the ends of the string are fixed, since a length is specified; problems involving vibrating strings generally require this. This is usually what is intended in a "stretched string" problem. Problems involving standing waves in columns of gas (organ pipes, wind instruments, etc.) are ones which usually allow one or both ends to be open.

If the string is free to vibrate at both ends (a little hard to arrange in practice), then you would be correct in saying that the fundamental would have nodes at x = L/4 and x = 3L/4 , making the fundamental wavelength L. (This is why, however, I assumed this was the usual string with fixed ends: you used a fundamental wavelength of 2L in your velocity relation to find fo.)