# Expressing a wave on a string using Fourier series

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

• I restored your post; please don't vandalize your posts in the future. Sep 26 '20 at 18:24

## 3 Answers

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}

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 $$$$\sum_{n=1}^{\infty}b_n(t)\sin\left(\frac{2\pi}{L}x\right).$$$$ There are no cosine terms because of the boundary conditions. More generally, the Fourier series would be $$$$\frac{1}{2}a_o(t) + \sum_{n=1}^{\infty}\left(a_n(t)\cos\left(\frac{2\pi}{L}x\right) + b_n(t)\sin\left(\frac{2\pi}{L}x\right)\right).$$$$

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

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.