Plucking Guitar Strings I was given this prompt:
A musician frets a guitar string of length 1.5 m at x = 0.28 m with one finger, and simultaneously plucks the string at x = 0.14 m with another finger (raising it to a height h = 2.8 mm. Both fingers are simultaneously removed from the string, and it is allowed to vibrate. The string has a tension of 86 N, and a linear mass μ = 4.2 g/m. We saw a similar problem last week; the initial position of the string looks something like this:

Compute a) the wave speed v for this string, b) the frequency f of this string, and the period T of the waves on this string. It is intended that you use knowledge from Analytical Physics II to solve this problem.
I tried solving for the velocity using the tension and the linear mass density, then solving for the frequency using v=f*w where w is the wavelength, assumedly being twice the length of the plucked section (so .56m). Then f=1/T which got me the period. However, this was wrong. Is there some other way I'm supposed to find the frequency, or is this not the wavelength? I'm having trouble even visualizing what the string will do once it is released so any help would be appreciated.
 A: A vibrating string satisfies the wave equation $\displaystyle\frac{\partial^{2}f}{\partial t^{2}}=c^{2}\frac{\partial^{2}f}{\partial x^{2}}$, where $\displaystyle c=\sqrt{\frac{T}{\rho}}$. The wave speed for this string is $\displaystyle c=\sqrt{\frac{T}{\rho}}$.
Let the displacement of the string at position x and time t be y(x,t).
The boundary conditions satisfied by the string are: y(0,t)=0, y(l,t)=0 for all t, where l=length of the guitar string.
The initial conditions are: y(x,0)=f(x), and y'(x,0)=0, where y' stands for the velocity of the string.
The vibrating string satisfies the wave equation $\displaystyle\frac{\partial^{2}y}{\partial t^{2}}=c^{2}\frac{\partial^{2}y}{\partial x^{2}}$. Solving this equation by the method of separation of variables, and applying the boundary conditions results in the solution $\displaystyle y_{n}(x,t)=(A_{n}\cos\lambda_{n}t+B_{n}\sin\lambda_{n}t)\sin\frac{n\pi x}{l}$, where $\displaystyle\lambda_{n}=\frac{cn\pi}{l}$, n=1,2,...
In order that the initial conditions be satisfied, the general solution should be $\displaystyle y(x,t)=\sum_{n=1}^{\infty} y_{n}(x,t)$. It can be shown that since the initial velocity of the string is zero, the coefficients $B_{n}$ are zero for n=1,2,... One can see from the final form for y(x,t) that there is not just one frequency of vibration but infinitely many, given by $\displaystyle f_{n}=\frac{c}{\lambda_{n}}$, where $\displaystyle\lambda_{n}=\frac{cn\pi}{l}$, as seen above. The frequency of the fundamental mode is $\displaystyle f_{1}=\frac{c}{\lambda_{1}}$, where $\lambda_{1}=\frac{c\pi}{l}$.
So to summarize, the speed of the wave is $\displaystyle c=\sqrt{\frac{T}{\rho}}$ and the frequencies of vibration are $\displaystyle f_{n}=\frac{c}{\lambda_{n}}$, $\displaystyle\lambda_{n}=\frac{cn\pi}{l}$.
By inserting the appropriate data in the problem the speed of the wave and the frequency of the fundamental mode(n=1) can be obtained.
