# 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?"