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 *T*_{1} at an angle *θ*_{1} from the horizontal; at the *x*+*Δx* end we have a tension force *T*_{2} at an angle *θ*_{2} from the horizontal.

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 *θ*_{1} 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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This entry was posted on April 25, 2008 at 4:13 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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April 28, 2008 at 4:11 am |

[…] 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 […]