# Wave equation in classical mechanics!

We represent the wavefunction of any wave on the string as $$y=f(x-vt),$$ where $v$ is velocity of the wave and $x$ is distance from origin and $t$ is time taken to reach the given point and $y$ is the displacement from the the $x$-axis.

but my question is why do we represent wave by this equation (i have clearly understood that the vertical displacement is the function of distance from the mean origin and time), why can't we represent a wave function by something like $$y=f(x^2-v^2t^2)$$ or any other function and why is there negative sign between $x$ and $vt$. i think negative sign doesn't make any sense.

on referring other book i saw wave equation was represented as $$y=f(t-\frac{x}{v}),$$ are they both same?

Please try to give answer with simple explanation, i am not expert in differentiation or differential equations, so i can't understand their meanings!

• Perhaps have a look at physics.stackexchange.com/q/174862/50583 Mar 30, 2016 at 21:35
• @ACuriousMind i edited my question that i don't understand differentiation(i know very basic of it) Mar 30, 2016 at 21:37
• @krotov - Sorry to say, but if you are not following along on the math than explaining the why of it isn't going to be very useful. I understand that you might not have gotten to calculus and differential equations. Fear not, if you pursue physics you will! But, in a very simple way, a wave is a 'something' that travels in a given direction with a given velocity. A simple wave will look the same as it moves along (think of waves on the deep ocean - not on the beach!). Thus it must have the functional form y = f(x-vt) - that function looks the same as it moves in x with velocity v. Mar 30, 2016 at 22:52
• I'm fairly certain I answered a very similar question; if I get the time, I'll look out up later this evening. Mar 30, 2016 at 23:15
• Is the yellow block a quote? From where? Mar 31, 2016 at 0:08

A1: In General, second function is not a solution of wave equation. The general solution is$$y=f(x-vt)+g(x+vt).$$One can show that $$y=f(x^2-v^2 t^2)$$satisfies $x^2 \frac {d^2 y}{dx^2}-t^2\frac {1}{v^2} \frac{d^2 y}{dt^2}$ instead of $\frac {d^2 y}{dx^2}-\frac {1}{v^2} \frac{d^2 y}{dt^2}$ by using Chain rule. Moreover, it doesn't mean propagating wave while $$y=f(x-vt)$$ means propagating wave.
A2: The first representation $$y=f(x-vt)$$ and the third $$y=f (t-\frac {x}{v})$$ is equivalent since it is just a change of variable.
• you didn't say why negative sign is there in the equation $f(x-vt)$ Mar 30, 2016 at 23:56