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I have a string of length $2 \;\text{m}$ and the wave velocity is $120 \;\text{m/s}$, find the frequency of the first two harmonics.

My attempt, what I must do is to solve the wave equation on the interval $[0,2]$.

So I have the problem $$u_{tt}=u_{xx} $$ $$u(t,0)=0$$ $$u(t,2)=0$$ I am not sure about the initial conditions I must use so for now I will use $$u(0,x)=f(x)$$ $$u_t(0,x)=g(x)$$

Set $$u(t,x)=T(t)X(x)$$ I end up with two equations $$T''=kT$$ $$X''=kX$$ For $k<0$ I have $$X=A\cos \sqrt{k} x +B\sin \sqrt{k} x$$ I end up with $$X=B_n \sin \sqrt{k} x$$ with eigenvalues $$k=\frac{n^2\pi^2 }{4}$$

Also $$T(t)=C_n \cos (n \pi/2) +D_n \sin (n \pi/2)$$

I am not sure of what initial conditions to use, I know that the wave velocity is $v=120 \;\text{m}$ should I use $v$ as an initial velocity?

Can you help?

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  • $\begingroup$ Why are you starting with the wave differential equation? That's like building an internal combustion engine from scratch when you need an automobile in 2021. $\endgroup$
    – Bill N
    Jul 1 at 12:33
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Since your string is fixed at both ends you are better off to look for a standing wave: $$ u(t,x)=\cos(\omega\,t)\,\sin(k\,x)\,, $$ where $2\,k=\pi$ for the longest wave length $\lambda=4$, and in general, $2\,k=n\,\pi$ for wave length $\lambda=4/n\,.$ The phase speed $c=120 m/s$ is related to $\omega$ and $k$ by $c^2k^2=\omega^2\,.$ The frequency of the wave is $\nu=\omega/(2\,\pi)\,.$

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