Physics Friday 17

Suppose we have an ideal string under tension: the string has uniform length density (mass per unit of length) λ is perfectly elastic and flexible, with no resistance to bending; and is under tension large enough, compared to it’s density, that the effects of gravity can be neglected. Let the string undergo small deformations in a plane, moving transversely. Thus, we can describe the string with the function . What differential equation must this function obey?

We consider a short segment of the string, from x to x+Δx. The mass is thus λΔx. At the x end we have a tension force T1 at an angle θ1 from the horizontal; at the x+Δx end we have a tension force T2 at an angle θ2 from the horizontal.
Figure 1

Now, since the motion is entirely transverse, we have a constant horizontal component of tension: .
Looking at the transverse forces, we have:

The acceleration in this direction is , and so

Now, our horizontal equation gives us and . Plugging these into the above equation of motion, we obtain:


Now, note that the angle θ1 of the force at x is equal to the angle of the string there, and trigonometry tells us that the tangent of this angle is equal to the slope of the string. Thus:

and

Thus, our equation of motion becomes:

.
Now, as , the left hand side becomes the partial derivative with respect to x of , and so:


This is the one-dimensional wave equation with wave speed .

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One Response to “Physics Friday 17”

  1. Monday Math 17 « Twisted One 151’s Weblog Says:

    […] the new coordinates is thus: This is the one-dimensional wave equation with speed c (see also this ‘Physics Friday’ post). Now recall that we found the general solution to our original equation to be . Now, inverting […]

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