# Calculating the length of a string from a standing wave

1. Apr 4, 2016

### Thimble

1. The problem statement, all variables and given/known data
A horizontally oriented string with mass per unit length $\mu=0.1kgm^{-1}$ is under tension T=1.0N. The left end of the string moves up and down in a simple harmonic motion with an amplitude A=0.10m and frequency f=3.0Hz. This sets up a sinusoidal wave along the string with the wave function $y(x,t)=Acos(kx-\omega t+\phi)$.
i.Find the wave velocity (v) and the wavelength for the wave ($\lambda$).
ii.y(0,0)=0.071m, find $\phi$.
iii.After travelling for some distance the wave arrives at the right end of the string and is reflected. The reflected wave is described by $y_r=-Acos(kx+\omega t +\phi)$. Show that the superposition of the incoming and reflected waves gives the resultant standing wave: $y_s=2Asin(kx+ \phi)sin(\omega t)$.
iv. Using the information available from (i) to (iii) above, determine the length of the string.

2. Relevant equations
i. $v=\sqrt{\frac{T}{\mu}}$
$\lambda=\frac{v}{f}$
ii. $\phi=cos^{-1}(0.71)$

3. The attempt at a solution
I think i've done parts (i) to (iii) but i'm struggling with part iv. We have what the wave equation for the standing wave is so I think this should just be a case of using a boundary condition for the ends of the string. The one i've been using is that $y_s(x,t)=0$ when $x=L \forall t$ where L is the length of the string. This boundary condition being taken from the wave equation of the reflected wave being inverted and travelling in the opposite direction to the incident wave so it has a fixed end.However, this gives me that $$sin(kL+\phi)=0 \implies L=\frac{n\pi-\phi}{k}$$ where n is a positive integer.

Firstly, is this correct? And secondly is there a way for me to figure out which value of n to use? I recall that for a string with one free end and a fixed end we should only have odd numbered harmonics but i see nothing here excluding the even values of n. I did consider using a second boundary condition for x=0 but it seems like as we have fixed values of A,$\phi$ and k we already know what the amplitude at x=0 has to be.

2. Apr 4, 2016

### BiGyElLoWhAt

Well, the sure fire way to solve this is using the equation for arc length of f(x), with y_s being f(x).
I don't think you can do that. As I'm interpreting what's going on, L is the x component of the displacement along the string. So if you follow the string up and down and to the right, L represents the "right" portion of the distance, and plugging that in gives you the height at some value L for horizontal displacement along the x axis.

I think you need to calculate
$\int_0^{x'} \sqrt{1+f'(x)^2}dx$ to get the arc length. I suppose x' would end up being a multiple, nL, so I don't see a way to get rid on the 'n' in your answer.

3. Apr 4, 2016

### Thimble

Thanks for responding but won't this method give a time varying answer since when the wave is at its full amplitude the integral here will be larger than when the string is completely at equilibrium position? I think the L i'm looking to find is the x coordinate of the end of the string as shown in the picture i've attached (Apologies for the awful picture). Could you elaborate as to why the method i've tried to use is wrong?

4. Apr 4, 2016

### BiGyElLoWhAt

Well, L isn't the length of the string, it's the strings projection onto the x axis. The answer won't be time varying, as it's the same length at every point. So just pick some time, t , such that sin(wt) = 1 to make it easy. Then solve the integral. It's sinusoidal in t, but for any t, the integral should be the same.
I'm not really sure how to explain this, so apologies if I butcher this.
Each point (constant x) oscillates in time via Asin(wt), and the whole wave has a value (for constant t) of A sin(kx - phi).

As to put that in rigorous mathematical terms, I'm not really sure, and this is a very conceptual physicsy type answer, but in short, the string length doesn't change, but the equation from part iii makes it seem like it does.
Perhaps A is supposed to be small enough that the length along the sin wave is the same as the distance L?
I'm not sure, but looking at equation y_s from iii, it appears that way, as the whole wave will be 0 at some t values, namely when wt = npi.
So... yea.
I suppose there are two approaches to this. Assume A is very small, and solve for L (and essentially that the string is elastic). By solving for L, you're solving for the minimum length of the string.
The other approach, which would be solving the max length, is to do what I outlined above, and set sin(wt) = 1, and then use arc length of f(x).

5. Apr 4, 2016

### BiGyElLoWhAt

The more I think about it, with the given information, I think you are supposed to assume that the length is the same at max as it is at min. So then yes, L would be the length, which would be some integer multiple of the wavelength.