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Homework Statement
I am having a issue understanding this question I have solve the PDE below, but I cant understand where or how you the characteristic frequency, what more confusing is that I don’t know if that lambda is just a constant or a wavelength.
Homework Equations
The Attempt at a Solution
$$U\left(x,t\right)=X\left(x\right)T\left(t\right)$$
$$\frac{X''\left(x\right)}{X\left(x\right)}=\frac{T''\left(t\right)}{c^2T}+\lambda ^2=k^2$$
$$X''\left(x\right)+k^2X\left(x\right)=0$$
using boundary condtions ##U\left(0,t\right)=U\left(l,t\right)=0##, ##X(x)## can be forund to be:
$$X\left(x\right)=C_1sin\left(kx\right),\:X_n\left(x\right)=C_nsin\left(\frac{\pi n}{l}\right)$$
Now solving for T
$$T''\left(t\right)+\left(k^2+\lambda ^2\right)c^2T\left(t\right)=0$$
$$b^2=\left(c^2\left(\lambda ^2+k^2\right)\right)$$
$$T''\left(t\right)+b^2T\left(t\right)=0$$
Solving this I get the following equation
$$T\left(t\right)=C_2cos\left(bt\right)+C_3sin\left(bt\right),\:T_n\left(t\right)=A_ncos\left(c\sqrt{\lambda ^2+k^2}\right)+B_nsin\left(c\sqrt{\lambda \:^2+k^2}\right)$$
$$U_n\left(x,t\right)=\sum \:sin\left(\frac{\pi n}{l}\right)\left(A_ncos\left(c\sqrt{\lambda \:^2+k^2}t\right)+B_nsin\left(c\sqrt{\lambda \:\:^2+k^2}t\right)\right)$$
and this is where I am lost how do I find the characteristic frequency from this? I just cant seem to understand how they have gotten the follwing euqation,
I know that normally the frequency and be workout from ##f_n=\frac{nc}{2l}## where c is the speed of the wave along the string.
Could anyone please give some adice it would be much appreciated
edit: I just thought of something which give me somewhat of the correct answer. But I am not too sure if I can say c is a constant if it is I have the correct ans, but as I mentioned perivous I assume c is the speed of the wave.
Here what I have.
If I have a standing wave then, I have two solutions
$$y_1=Asin\left(kx\omega t\right)$$
and
$$y_2=Asin\left(kx+\omega t\right)$$
as this sinerio gives a standing wave then I get
$$y_1+y_2=2Asin\left(kx\right)cos\left(\omega t\right)$$
so comparing this to what I have I can see that
$$\omega =2\pi f=c\sqrt{\lambda ^2+k^2}=c\sqrt{\lambda ^2+\left(\frac{n\pi }{L}\right)^2}$$
If I reagrange and solve for ##f_n## I get the following
$$f_n=\frac{c}{2}\sqrt{\frac{\lambda ^2}{\pi ^2}+\frac{1}{\pi ^2}\left(\frac{n\pi }{L}\right)^2}$$
which simplifying to the ans required except I have a 1/2 in mine can this be absorbed by c as it a constant or is there a step I am missing?
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