- #1
roam
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Homework Statement
The differential equation describing the motion of a stretched string can be written
[itex]\frac{\partial ^2 y}{\partial x^2} = \frac{\mu}{T} \frac{\partial^2 y}{\partial t^2}[/itex]
μ is the the mass per unit length, and T is the tension.
(i) Write down the most general solution you can for this wave equation, and show that it satisfies the equation.
(ii) Find an expression for sinusoidal a standing wave that satisfies the equation above.
The Attempt at a Solution
(i) I think the most general solution to that PDE is of the form [itex]y(x,t) = f (x -ct)[/itex], where f describes a pulse of any shape moving at speed [itex]c=\sqrt{T/ \mu}[/itex] in the positive x direction. And it goes to the negative direction if we have "+ct". So if we take the function to be sinusoidal the general solution is:
[itex]y(x, t) = Asin [k(x -ct)+ \phi][/itex] (for some ϕ)
I substituted this in the DE to show that it satisfies it. The LHS becomes:
[itex]\frac{\partial ^2 (Asin (kx - kct))}{\partial x^2} = -Ak^2 \sin (kx-ckt)[/itex]
And the right hand side:
[itex]\frac{\mu}{T} \frac{\partial ^2 (Asin (kx - kct))}{\partial t^2} = \frac{\mu}{T} AC^2 \sin (kx-ckt) = A \sin (kx-ckt)[/itex]
So, why are the LHS and RHS not equal? How can I show that the solution satisfies the wave equation?
(ii) I'm not quite sure how to approach this part. But I think we need an expression which does not contain the "kx-wt", therefore it's not an axpression of a single wave, but the superposition of two identical waves traveling in opposite directions:
[itex]A \ sin(kx-kct)+A \ sin(kx +kct)=2 \ A \sin(kx) \cos(kct)[/itex]
Is this correct? Any helps would be greatly appreciated.