# The Wave Equation %%intermediary diagrams needed%% Remember the definition for the transverse wave - if we had an element on the wave, it'd move perpendicular to the direction the wave is actually travelling. Let's double back to the string element from the [[Waves Foundations|previous doc]] to get an intuition for it. ![[Pasted image 20230908143857.png]] It's time for us to relate this small string to a titan in physics - Newton's Second Law! I can't wait... First, let's take that small mass length in the diagram again. Since the horizontal velocity of the string is constant, there will be no horizontal resultant force on the string - meaning we only have to worry about the vertical component. Take the mass as $dm$ and you get: $F_{y,f} - F_{y,i} = dm a_{y}$ Since there's always going to be a force on the string, we're going to have to subtract the original force from the final force value - which is what we've done above. Remember linear density - we'll be going over it again! We can write the mass $dm$ as a product of the linear density and the length of the string section: $dm = \mu dL$ Like before, this string section has two tension forces acting on it - though these horizontal components balance out, the vertical ones don't. Let's also say that the angle of the string S with the length dx is really, really small - small enough that we can just approximate it as: $dm = \mu dx$ We're back where we left off! Write the acceleration in terms of its displacement - the differential term: $a_{y} = \frac{d^2y}{dt^2}$ Time to look at the forces - the tensional force $T_2$ is almost equal to the magnitude of the string slope at the right end. This means we can equate it by dividing the components: $\frac{T_{f,y}}{T_{f,x}} = S_{f}$ This gives us the vertical and horizontal tensional force components. Using these variables, we can redefine the magnitude of the tensional force: $T_{2} = \sqrt{ T_{f,y}^2 + T_{f,x}^2 }$ As the string's inclination angle is very low the vertical component is negligible - so we remove it, giving us: $T_{f} = T_{2,x}$ Substitute this to get: $T_{f,y} = T_{f}S_{f}$ Now substitute these expressions back into the original equation with the forces, acceleration and mass! We get our expanded equation as: $T_{f}S_{f} - T_{i}S_{i} = (\mu\ dx)(\frac{d^2y}{dt^2})$ The tension stays constant throughout the string - so we can generalise it into $T$. Divide both sides by $dx$ and $T$: $\frac{{S_{f}-S_{i}}}{dx} = \frac{\mu}{T} \frac{d^2y}{dt^2}$ This string element has a mass $dm$ for a reason - it's really, really, small, enough for us to justify ourselves taking the difference of the slope of each adjacent (final and initial) component as equal! The slope is part of [[Calculus Basics (Maths)|calculus]] - it's just the ratio between the change in the $y$ and $x$ axes: $\frac{d(dy/dx)}{dx} = \frac{\mu}{T} \frac{d^2y}{dt^2}$ We need to write this in terms of [[Partial Derivatives (Maths)|partials]] - different variables mean different sensitivities, after all! $\frac{\delta^2y}{\delta x^2} = \frac{\mu}{T} \frac{\delta^2y}{\delta t^2}$ Now remember [[Waves Foundations#Wavespeeds of Strings|the equation for velocity from the previous doc]]. Substituting the $\frac{\mu}{T}$ term into it gives us a neater expression: $\frac{\delta^2y}{\delta x^2} = \frac{1}{v^2} \frac{\delta^2y}{\delta t^2}$ This is our *general wave equation!* Note that this is a partial derivative - the leftmost term dictates the unit change in the height of the wave depending on the displacement, while the left-hand site relates the change in height with the change in time! feedback appreciated i also have a somewhat limited understanding of this will make up for it in the future when i get around to the mandatory waves parts :LiThumbsUp: # Intro - The General Physics yearns for a lack of abstraction, a