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.
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
.
Tags: Friday Physics, physics, wave equation, waves
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 [...]