A travelling wave pulse is shown in figure 1 below, travelling at v=6 m/s across a string. In figure 2, a short segment of the string is shown zoomed in. The angle of this string goes from θ1 = 17o to zero within a small horizontal distance Δx = 3 mm.
∂^2y/∂x^2= 1/v2 * (∂^2y/∂t2^2)
The Attempt at a Solution
I know using that equation I want to solve for the acceleration, which is the second derivative of position (y), so I should solve for (∂^2y/∂t2^2) in terms of v, but when I do that I am still left with an equation in terms of ∂^2y/∂x^2, the concavity, which I am not sure what do do with / what it equals.
So next I tried using my intuition:
|average transverse acceleration| = |Δv|/Δt = initial transverse velocity / Δt.
And since initial transverse velocity = the slope of the graph at point 1 (where Δx begins) = rise/run = tan(θ), and Δt = Δx/v, the above eqn becomes:
( v * tan(θ) ) / Δx
But that isn't right