# Find the frequency of the first two harmonics of a vibrating string

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?

• 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. Jul 1 at 12:33

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 $$\nu=\omega/(2\,\pi)\,.$$