Guitar string

The amplitudes $A_n$ (fourier coefficients) results $\begin{aligned} A_n &= 2 \int_{0}^l u_0(x) \sin\left( \frac{n\pi}{l} x \right) dx \\ &= \frac{4 A_0}{l} \left[ \int_{0}^{l/2} x \sin\left( \frac{n\pi}{l} x \right) dx + \int_{l/2}^{l} (l - x) \sin\left( \frac{n\pi}{l} x \right) dx \right]\\ \Rightarrow \qquad A_{2k} &= 0\,, \\ A_{2k+1} &= \frac{8 A_0}{l} \int_{0}^{l/2} x \sin\left( \frac{(2k+1) \pi}{l} x \right) dx \\ &= \frac{8 A_0}{l} \left( \frac{l}{(2k + 1) \pi} \right)^2 \left[\sin(x) - x \cos(x) \right]_0^{(2k + 1)\pi/2} \\ &= \pm \dfrac{8 A_0 l}{(2k + 1)^2 \pi^2} \end{aligned}$ For the even multiples $n = 2k$ the amplitude is zero, because due to the symmetry the two partial integrals cancel each other out. The kinetic energy of the vibration for a string of mass $m$ is $\begin{aligned} E_\text{kin} &= \frac{1}{2} m \cdot \frac{1}{l} \int_{0}^l \left(\frac{\partial u}{\partial t}\right)^2 dx \\ &= \pi^2 m \sum_{n = 1}^\infty (f_n A_n)^2 \sin^2(2 \pi f_n t) \\ &= m \left(\frac{8 A_0 l \nu}{\pi}\right)^2 \sum_{k = 0}^\infty \frac{\sin^2(2 \pi (2 k + 1) \nu t)}{(2 k + 1)^2} \\ \Rightarrow \quad \langle E_\text{kin} \rangle &= \frac{1}{2} m \left(\frac{8 A_0 l \nu}{\pi}\right)^2 \underbrace{\sum_{k = 0}^\infty \frac{1}{(2 k + 1)^2}}_{\pi^2/8} \end{aligned}$ Hereby, $\langle . \rangle$ is the time average. The relative energy contribution of the fundamental mode gives $\frac{\langle E_{\text{kin},k=0} \rangle}{\langle E_\text{kin} \rangle} = \frac{1}{\sum_{k = 1}^\infty \frac{1}{(2 k + 1)^2} } = \frac{8}{\pi^2} \approx 81\,\text{\%}$ Finally, a plot of the energy spectrum of the string vibration:

Here's a quick and dirty solution:

We only need to know the energy carried in each "mode" for the triangle wave and use orthogonality to show that the total energy is just sum of the energies of the modes. (Alternatively, one can just use Parseval's Theorem, but I will show the details here.)

Let's make life easy and denote the time dependent normal modes of vibration as vectors $\Psi_n(x)e^{-i\omega_n t}=\sqrt{\frac{2}{l}}\sin{\left(\frac{n\pi x}{l}\right)}e^{-i\omega_n t}$ .

Since they are normal modes, they satisfy the orthogonal property $\int_0^l\Psi_n(x)\Psi_m(x)dx=\delta_{mn}$ , where $\delta_{mn}$ is the Kronecker delta function.

From this it is easily seen that the kinetic energy for each mode with amplitude $A_n$ is given by integrating the kinetic energy per unit mass over the entire length of the string with mass per unit length $\mu=m/l$ :

Here, the integral we need to compute is the modulus of the complex amplitude:

$KE_n = \frac{1}{2}\mu A_n^2\int_0^l{(-i\omega_n)(+i\omega_n)\Psi_n(x)^2e^{-i\omega_n t}e^{+i\omega_n t}}dx=\frac{1}{2}\mu A_n^2\omega_n^2$

Here's the neat trick, when we compute the energy of the superposition of modes that makes up the triangle wave, we can ignore the cross-terms since they integral to zero by the orthogonality of the modes. Only the self-terms survive. Also, note that $\omega_n = n\omega_1$ .

$KE_{total}=\sum_{n=1}{\frac{1}{2}\mu n^2A_n^2\omega_1^2}$

Thus, once the common factors are cancelled out, the ratio of the fundamental energy to the total energy is just

$\eta_1=\frac{A_1^2}{\sum_{n=1}n^2A_n^2}=\left(\sum_{n=1}{n^2(\frac{A_n}{A_1})^2}\right)^{-1}$ .

Now all that is left is to obtain the $A_n$ 's for the triangle wave.

It turns out that it is easier to obtain the Fourier series for a square wave with 50/50 duty cycle, and take the integral of the entire wave to obtain a triangle wave. The actual amplitudes don't matter, since we just need the ratio to the fundamental. Since all integrals of the sines cancel between 0 to l/2 if the sine wave is even, only the odd integrals survive. It isn't hard to convince yourself that the $A_n$ 's are proportional to $\frac{1}{n}$ where $n$ is odd.

Now, comes the neat trick, since the triangle wave is really just the integral of the square wave, we see that term by term integration of the wave will just get you a set of Fourier coefficients that goes like $\frac{1}{n^2}$ where $n$ is odd. (I'll leave this part for you to check.)

Thus, in computing the energy ratio, we have $n^2(\frac{A_n}{A_1})^2=1/n^2$ where $n$ is odd.

$\eta_1 = \left(\sum_{n=1,odd}\frac{1}{n^2}\right)^{-1}$

Now we compute the sum:

$\sum_{n=1}\frac{1}{n^2}=\sum_{n=1,odd}\frac{1}{n^2}+\sum_{n=2,even}\frac{1}{n^2}$ since the series is absolutely convergent.

It's is not hard to recognize that the second sum is just 1/4 of the original sum. This means the odd sum is just 3/4 of the original sum.

The LHS sum here is a special series and is the Riemann Zeta function for $z=2$ which sums to $\frac{\pi^2}{6}$ .

Thus, our odd terms will add up to $\frac{3}{4}\frac{\pi^2}{6}=\frac{\pi^2}{8}$ .

Finally, we plug this back to get the ratio we were looking for earlier:

$\eta_1 = \frac{8}{\pi^2} \approx 0.81$

Voila, 81% of the energy is in the fundamental.