# Tension required to tune a string to the note $E_2$

Hi I'm having trouble with this homework question on wave mechanics. The question is as follows:

"State the boundary condition which must be met at a point where the string of question $2$ is fixed.

Hence find the real standing wave solutions to the wave equation, and determine the allowed oscillation frequencies, when such a string of length $L$ is fixed at its ends.

If the bottom string of a guitar has a mass of $5.4 g m^{-1}$ and its length is determined by the distance $0.648 m$ from the bridge to the nut, find the tension required to tune the string to the note known as $E_2$ (a frequency of $82.4 Hz$)."

Question 2 describes a wave on a guitar string with the wave equation:

$\frac {\partial^2 \psi}{\partial t^2}=\frac WM \frac {\partial^2 \psi}{\partial x^2}$

Where $W$ is the tension, $M$ is the mass per unit length and $\psi$ is the displacement. It has been shown that the velocity of the wave is equal to $\sqrt{W/M}$

My attempt:

So for the first part I stated that $\psi$ must be $0$ at the points where the sting is fixed as it has no displacement.

Next I stated that the real standing wave solution is given as follows: $\psi=\psi_0cos(kx+\phi_x)cos(\omega t+\phi_t)$.

For the allowed oscillation frequencies I did the following:

$\psi=0$ at $x=0$ and $x=L$

Hence, $cos(kx+\phi_t)=0$ at these two points.

Setting $x=0$ yields that $\phi_x=\pi/2$

Setting $x=L$ gives $k=n\pi/L$

Since $k$ is the wavenumber $\lambda=2\pi/k$ so $\lambda=2L/n$

$v=f\lambda$ gives the allowed oscillation frequencies as $f= \frac{n}{2L}\sqrt{W/M}$.

When I come to try to calculate the tension however I rearrange f above to get: $W=\frac{4L^2f^2}{n^2}M$ and I've been given all the numbers to calculate $W$ except I don't know what to pick for the value of $n$. If I had to guess I would say $n=2$ just because in the question it asks for $E_2$ but I have no clue.

Any explanation or help would be very much appreciated.