Why does the wave equation support wave motion?

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The wave equation indicates that objects obeying it will exhibit wave-like behavior, as demonstrated by solutions such as y = Asin(2π/λ)(x - vt), which represent sinusoidal waves. The second-order differential equation v²(d²y/dx²) = d²y/dt² reveals that the second derivative relates to the wave's concavity, affecting how disturbances propagate. For instance, a sinusoidal bump on a string accelerates downwards at its peak and upwards at its edges, causing the disturbance to spread. While the term "wave" may seem misleading, the critical aspect is the function's dependency on x - vt, which ensures propagation of features in space and time. Understanding these principles enhances intuition about wave motion and the nature of wave equations.
DunWorry
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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
 
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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.
 
You could just memorize it. You see it enough that it will seem intuitive after a while.
 
DunWorry said:
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?
That would be nice. Solving partial differential equations intuitively. I certainly don't have that kind of intuition.
 
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).
 
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