Wave Pulse - Finding Average Transverse Acceleration of Segment

1. Sep 26, 2013

ArtVandelay

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
∂^2y/∂x^2= 1/v2 * (∂^2y/∂t2^2)

3. 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

2. Sep 26, 2013

voko

I believe you are supposed to approximate $y(x, t)$ and $\frac {\partial^2 y} {\partial x^2}$ as a power series in the vicinity the crest and then use the given values to estimate whatever coefficients you will end up with.

3. Sep 26, 2013

nasu

You know that dy/dx varies from 0 to tan(17) over a distance Δx=3 mm.
You can approximate the derivative of f(x)=dy/dx as Δf/Δx. And this will be the left term of your equation.

4. Sep 26, 2013

ArtVandelay

Thanks guys! Approximating the derivative of f(x)=dy/dx as Δf/Δx did the trick.

Also, for reference purposes if anyone is reading this thread in search of an answer to a similar problem, I forgot to explicitly mention in my OP that the problem question states: "What is the (average) transverse acceleration of this string segment?"