Expressing a wave on a string using Fourier series

Question

Where does time function of wave on string go when expressed in the Fourier series?

A Standing wave on string of length $L,$ fixed at its ends $x=0$ and $x=L$ is: $\quad y(x, t)=A \sin (k x) \cos \left(\omega t+\phi_{0}\right) \quad$

Where: $k=\frac{n \pi}{L}$

A periodic function $f(x)$ with period $P$ is represented by the Fourier series: $f(x)=\frac{1}{2} a_{0}+\sum_{n=1}^{\infty} a_{n} \cos \left(n \frac{2 \pi}{p} x\right)+\sum_{n=1}^{\infty} b_{n} \sin \left(n \frac{2 \pi}{p} x\right)$

Where: $$ a_{0}=\frac{2}{p} \int_{-P / 2}^{P / 2} f(x) d x \quad a_{n}=\frac{2}{p} \int_{-P / 2}^{P / 2} f(x) \cos \left(\frac{2 \pi}{p} n x\right) d x \quad b_{n}=\frac{2}{p} \int_{-P / 2}^{P / 2} f(x) \sin \left(\frac{2 \pi}{P} n x\right) d x $$ For question where a guitar is played and the string is put into motion by plucking it. If we want to write $y(x)$ as a sum of the basis function, $y_{n}(x)$ we write: $$ y(x, 0)=\sum_{n=1}^{\infty} a_{n} \sin \left(k_{n} x\right) \quad \rightarrow \quad y(x, t)=\sum_{n=1}^{\infty} a_{n} \sin \left(k_{n} x\right) \cos \left(\omega_{n} t\right) $$ [since the wave function is usually odd, so the $a_n$ function will be eliminated)

Also in the case where the wave is neither an odd or even function when we have values for $a_{0}, a_{n}, b_{n}$ (not just 0).

And the periodic function is given by $f(x)=\frac{1}{2} a_{0}+\sum_{n=1}^{\infty} a_{n} \cos \left(n \frac{2 \pi}{p} x\right)+\sum_{n=1}^{\infty} b_{n} \sin \left(n \frac{2 \pi}{p} x\right)$.

Where do we add the $\cos \left(\omega_{n} t\right)$ part?

Does the equation look like this: $y(x, t)=\frac{1}{2} a_{0} \cos \left(\omega_{n} t\right)+\sum_{n=1}^{\infty} a_{n} \cos \left(n \frac{2 \pi}{p} x\right) \cos \left(\omega_{n} t\right)+\sum_{n=1}^{\infty} b_{n} \sin \left(n \frac{2 \pi}{p} x\right) \cos \left(\omega_{n} t\right)$

Answer 1

Let's start from the equations of motion for a guitar string (with damping). Let $A(x,t)$ be the amplitude of the wave at a point $x$ along the string at time $t$. Then \begin{align} \partial_t^2 A + b\partial_t A - \partial_x^2 A = S(x,t)\,, \end{align} where $b$ is the damping coefficient and $S$ is the source term (representing the pluck). Let's assume that the string is length $L$ and the string is fixed with $A(0,t) = A(L,t) = 0$. The "normal modes" of the string are the eigenfunctions of the operator \begin{align} D = \partial_t^2 + b\partial_t - \partial_x^2\,. \end{align} It is easy to see that the eigenfunctions that satisfy the boundary conditions are of the form \begin{align} f_n(\omega,x,t) = \sin\left(\frac{\pi n}{ L }x\right) e^{{\rm i}\omega t}\,. \end{align} Thus, we can decompose \begin{align} A(x,t) = \sum_{n = -\infty}^\infty \int_{-\infty}^\infty\frac{{\rm d}\omega}{2\pi} A_n(\omega) f_n(\omega,x,t)\,. \end{align} We can now solve for $A_n(\omega)$, \begin{align} A_n(\omega) = \sum_{n = -\infty}^\infty\int_{-\infty}^\infty\frac{{\rm d}\omega}{2\pi}\frac{f_n(\omega,x,t)}{\lambda_n(\omega)}\int_0^L{\rm d}x\int_{-\infty}^\infty{\rm d}t S(x,t)f_n^*(\omega,x,t) \end{align} where $\lambda_n(\omega)$ are the eigenvalues \begin{align} D f_n(\omega, x,t) = \lambda_n(\omega)f_n(\omega,x,t)\,. \end{align}

Answer 2

For each time $t$, there is a different Fourier series. The $t$-dependence is incorporated via the Fourier coefficients. For a function $y(x,t)$ that is always zero at $x = 0$ and at $x = L$, the Fourier series is \begin{equation} \sum_{n=1}^{\infty}b_n(t)\sin\left(\frac{2\pi}{L}x\right). \end{equation} There are no cosine terms because of the boundary conditions. More generally, the Fourier series would be \begin{equation} \frac{1}{2}a_o(t) + \sum_{n=1}^{\infty}\left(a_n(t)\cos\left(\frac{2\pi}{L}x\right) + b_n(t)\sin\left(\frac{2\pi}{L}x\right)\right). \end{equation}

If your function of $t$ and $x$ is \begin{equation} y(x,t) = A\sin\left(\frac{2\pi}{L}x\right)\cos(\omega t + \phi), \end{equation} then your Fourier expansion with $t$-dependent Fourier coefficients is \begin{equation} \underbrace{A\cos(\omega t + \phi)}_{b_n(t)}\sin\left(\frac{2\pi}{L}x\right). \end{equation} All $a_m(t)$ and all other $b_m(t)$ ($m\neq n$) are identically 0.

Answer 3

1. $\omega_0$ depends on the tension of the string, then $\omega_n=n\omega_0$. You have no $a_0$ in your Fourier since $f=0$ at $x=0$, and no cosine terms for the same reason.
2. For a real string you should not go more than $n=4$ or maybe six, and to every wavelength you have the time term as factor.