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

**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) = [itex]\frac{\delta y}{\delta x}[/itex](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 = [itex]\frac{\pin}{L}[/itex] and ω_{n}=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 = [itex]\frac{(2n-1)pi}{2L}[/itex]. Which, when substituted into sin(kx) and then simplified, results in: sin[x([itex]\frac{\pi n}{L}[/itex]-[itex]\frac{\pi}{2L}[/itex])].

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, [itex]\frac{\delta y}{\delta x}[/itex](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?

http://ift.tt/1jkiToY

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