# Form of the displacement, y(x,t), for the normal modes of a string

1. Apr 19, 2014

### RyanTG

1. The problem statement, all variables and given/known data
The displacement, y(x; t), of a tight string of length, L, satisﬁes the conditions
y(0, t) = $\frac{\delta y}{\delta x}$(L,t) = 0

The wave velocity in the string is v.

a) Explain what is meant by a normal mode. Give the form of the displacement, y(x; t),
for the normal modes of the string and give their frequencies as a function of their
wavenumbers

3. The attempt at a solution

I thought the solution was simply: y(x,t) = Asin(kx)e(-iωnt), where k = $\frac{\pin}{L}$ and ωn=vk. (Subbing the value for k in as well).

But from the mark scheme and looking at later questions, it's obvious that the string is fixed at one point but not at the other. My answer, I think, assumes fixed ends. How can you tell the string is only fixed at one end when the question hasn't wrote in words that is what is going on?

In the mark scheme it has k = $\frac{(2n-1)pi}{2L}$. Which, when substituted into sin(kx) and then simplified, results in: sin[x($\frac{\pi n}{L}$-$\frac{\pi}{2L}$)].

I'm not really sure how to get to the answer they have given. A hint in the right direction would be great, I know there is probably just a gap in my knowledge on this subject.

Edit: Apparently, $\frac{\delta y}{\delta x}$(L,t) = 0, is the boundary condition which describes the string with a massless ring free to move up and down a frictionless pole parallel to the y-direction at one end. Why is this the case?

Last edited: Apr 19, 2014
2. Apr 19, 2014

### dauto

A massless ring must have zero net force acting on it. The horizontal component of the tension along the string is countered by the contact force with the pole at the end of the string but there is nothing to counter the vertical component of the tension since the ring is free to move up and down without friction. That means there is to vertical component of the tension at the end of the string, the tension is horizontal at the end of the string, and therefore the string itself is horizontal at its very end. That means the derivative of the wavefunction is zero there - as stated in the problem.

The string is fixed at the origin so sin(kx) is correct. To find k set x=L in the expression for the derivative which will be zero at x=L. use that fact to find the condition that must be satisfied by k. Than find all values of k that satisfy that condition.

3. Apr 19, 2014

### RyanTG

So I have to differentiate (x,L) = Asin(kL)e(-ivknt)? It seems quite a bit of work for a question like this...

4. Apr 19, 2014

### dauto

The function you'll differentiate is y(x,t) = A sin(kx) eiωt. You only set x = L after the differentiation. It is easy. Since the exponential does not depend on x, it plays as a constant

Last edited: Apr 19, 2014
5. Apr 20, 2014

### RyanTG

Sorry for being so incompetent...

Is this correct?
kAcos(kL)=0

Disregarding the exponential part since it is a constant. What do I do from here to get the answer in my mark scheme?

Using the above I just end up with kL = 1, k = 1/L, surely? Or L = 1/k.

I haven't done a question like this before and I can't find any resources that help, i'm sorry for literally not having a clue what to do.