- #1

thomas19981

## Homework Statement

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## 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}##

## Homework Equations

## The Attempt at a Solution

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.