Wave Pulse - Finding Average Transverse Acceleration of Segment

[ArtVandelay]

Homework Statement

A traveling wave pulse is shown in figure 1 below, traveling 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.

Homework Equations

∂^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

 

[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.

 

[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.

 

[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?"

 

[Nickie]

either because I am not using the information given in figure 2.

Dear student,

Thank you for your question. It seems like you have a good understanding of the equations involved in finding the average transverse acceleration of a wave pulse on a string. However, it is important to note that the equation ∂^2y/∂x^2= 1/v2 * (∂^2y/∂t2^2) is not applicable in this scenario as it is for a wave propagating in a medium, not a wave pulse on a string.

To find the average transverse acceleration of the segment in figure 2, we need to use the equation a = Δv/Δt, where a is the average acceleration, Δv is the change in velocity, and Δt is the change in time.

In this case, the change in velocity (Δv) can be found by taking the difference between the initial and final velocities of the segment. The initial velocity can be calculated using the slope of the graph at point 1, as you mentioned. The final velocity can be calculated using the slope of the graph at point 2, where the angle of the string is zero.

The change in time (Δt) can be calculated using the given information that the pulse is traveling at a speed of 6 m/s and the distance between point 1 and point 2 is 3 mm. Since we know the speed and distance, we can use the formula v = d/t to solve for the change in time.

Once you have calculated Δv and Δt, you can plug them into the equation a = Δv/Δt to find the average transverse acceleration of the segment.

I hope this helps clarify the approach to solving this problem. Keep up the good work!