# Waves on string

1. Jan 7, 2012

### aaaa202

My book derives the result of what the propagation speed of a wave on a string is:
v = √(F/mu)
It did it using the wave equation, so I wonna ask something about this equation. My book started on waves by describing waves that are harmonic, i.e. inwards force line proportional to the the displacement from the equilibrium. Therefrom they derived the wave equation. But I'm pretty sure, that the wave equation is more general and provides a more deep description. Isn't it so? I mean, isn't it applicable for all kinds of waves?
They then derived the result for v by showing that the string can be described by the wave equation. Does this then mean, that the inwards force is indeed proportional to the displacement from the equilibrium line on a string?

2. Jan 8, 2012

### Philip Wood

First: your last sentence... The answer is: yes, provided the displacement is small. You show this by considering components of the string tension which are transverse to the general line of the string.

Next: "Isn't [the wave equation] applicable to all kinds of waves?" Yes, but by definition: a wave can pretty much be defined as a propagating disturbance obeying the wave equation. There's no substitute for showing, in each individual case (e.g longitudinal disturbances in a gas, electric or magnetic field changes) that the wave equation governs the propagation. Only then can you know you have a wave!

3. Jan 8, 2012

### Ken G

That's a pretty complete answer, but let me add one more thing to help you understand why the wave equation, derived the way you mention, is so ubiquitous. At first glance, you might think a restoring force that is proportional to displacement from equilibrium is fairly arbitrary or rare, but it is actually very common indeed. The reason is, if you have an equilibrium, then the force is zero there, and if the force depends on position (fairly common), then you can "Taylor expand" the force as a function of small displacements. A Taylor expansion gives you a series whose first term has no displacement dependence, but here that has to be zero, because we have an equilibrium. The next term is, you guessed it, proportional to displacement, and all higher terms have displacement to some power >1, which will be very small corrections when the displacement is small.