The Wave Equation (vibrating string)

AI Thread Summary
The wave equation for a vibrating string is expressed as a second-order partial differential equation involving mass per unit length (μ) and tension (T). The general solution is proposed as y(x,t) = f(x - ct), indicating a wave traveling at speed c = √(T/μ). The discussion highlights a misunderstanding regarding the equality of the left-hand side and right-hand side when substituting sinusoidal functions into the equation. For part (ii), a standing wave is correctly identified as the superposition of two waves traveling in opposite directions, resulting in the expression 2A sin(kx) cos(kct). The importance of using the chain rule correctly in derivatives is emphasized for both parts of the problem.
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Homework Statement



The differential equation describing the motion of a stretched string can be written

\frac{\partial ^2 y}{\partial x^2} = \frac{\mu}{T} \frac{\partial^2 y}{\partial t^2}​

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 y(x,t) = f (x -ct), where f describes a pulse of any shape moving at speed c=\sqrt{T/ \mu} 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:

y(x, t) = Asin [k(x -ct)+ \phi] (for some ϕ)

I substituted this in the DE to show that it satisfies it. The LHS becomes:

\frac{\partial ^2 (Asin (kx - kct))}{\partial x^2} = -Ak^2 \sin (kx-ckt)

And the right hand side:

\frac{\mu}{T} \frac{\partial ^2 (Asin (kx - kct))}{\partial t^2} = \frac{\mu}{T} AC^2 \sin (kx-ckt) = A \sin (kx-ckt)

So, why are the LHS and RHS not equal? :confused: 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:

A \ sin(kx-kct)+A \ sin(kx +kct)=2 \ A \sin(kx) \cos(kct)

Is this correct? Any helps would be greatly appreciated.
 
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When taking the time derivatives of the sine function, you didn't quite get it right. Make sure you're using the chain rule of calculus properly.

However, in part (i) I'm not sure that you are meant to assume a sinusoidal solution. The problem seems to want the most general solution. As you noted, this would be a superposition of an arbitrary function of x - ct, f(x-ct), and a different arbitrary function of x + ct, g(x+ct). You then need to show that this superposition satisfies the wave equation. This will require careful use of the chain rule.

Part (ii) looks good to me.
 
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