# Why does the wave equation support wave motion?

1. Aug 25, 2013

### DunWorry

If motion of an object obeys the wave equation, then it will display wave like behaviour. If you solve the wave equation, you get things like y = Asin $\frac{2∏}{\lambda}$(x - vt) which is a sinosodial wave. But from the second order differential equation v$^{2}$$\frac{d^{2}y}{dx^{2}}$ = $\frac{d^{2}}{dt^{2}}$ how can you tell intuitively that it describes something like a wave?

Thanks

2. Aug 25, 2013

### chingel

For example you could consider an initial condition like a single small sinusoidal bump in the middle of a long string and then try to understand why it would start propagating. The second derivative gives the change of the slope and is therefore related to the concavity. If it is concave the second derivative is negative, and vice versa. The second derivative with respect to time is the acceleration.
Now the middle of the bump is concave so it would start accelerating down, but toward the end of the bump where it is straightening out it is convex so it will start accelerating up. The middle part coming down and the sides going up would correspond to the disturbance spreading out.

3. Aug 25, 2013

### Khashishi

You could just memorize it. You see it enough that it will seem intuitive after a while.

4. Aug 25, 2013

### Staff: Mentor

That would be nice. Solving partial differential equations intuitively. I certainly don't have that kind of intuition.

5. Aug 25, 2013

### voko

A wave is something that propagates. Calling $y = a \sin k(x - v t)$ a wave is a bit of terminological abuse, because it does not really propagate anything (you cannot use pure sinusoidal 'waves' to transmit any signal).

However, it does have a property that any proper wave has: it is a function of $x - v t$. Take any function $f(z)$, and $f(x - v t)$ is a wave. The dependence on $x - v t$ is important, because if $f(z)$ has some feature at $z = z_0$, then this feature will be propagating in space and time with velocity $v$. Using the chain rule, you can see that any such function will satisfy the wave equation ($f(x + v t)$ will satisfy it, too).