Superposition wave values problem

[Nghi]

Homework Statement

Two wave pulses on a string approach one another at the time t = 0, as shown in the figure below, except that pulse 2 is inverted so that it is a downward deflection of the string rather than an upward deflection. Each pulse moves with a speed of 1.0 m/s. Assume that the superposition principle holds for these waves, and that the absolute value of the height of each pulse is 1 mm in the figure below. Determine the value of the resultant wave at x = 4.1 m at t = 1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s.

Homework Equations

None? o_o

The Attempt at a Solution

Sorry to bother everyone on the same day again, but everyone is just so helpful on this forum! :) This one makes my heart sad because I don't know how to find the slope of pulse 2, which is essentially a straight line. :(

The only solutions I didn't get were t = 2.0 s and 2.5 s. But I think if I understood how to do 2.0 s, then 2.5 would be manageable.

I understand the idea of superposition, but I don't know how to apply it, I guess. Ha ha. :'( My friend mentioned something about finding the slope of pulse 2 first, but I don't know how to do that. I think it's because I'm underthinking.

 

[chaoseverlasting]

The slope of pulse two may be found using the graph. You know the height and the width of the pulse.

 

[ogb p]

Hello,

Thank you for reaching out for help with your homework problem. I am happy to assist you in understanding the concept of superposition and applying it to this specific problem.

Superposition is a fundamental principle in physics that states that when two or more waves overlap, the resulting wave is the sum of the individual waves. In this case, we have two wave pulses approaching each other on a string. The key to solving this problem is understanding how the two pulses will interact and add together.

First, let's consider the two pulses at t = 0. Pulse 1 is an upward deflection of the string, while pulse 2 is an inverted downward deflection. This means that at x = 4.1 m, the resultant wave will have a value of 1 mm (from pulse 1) minus the value of pulse 2 at that point. To find the value of pulse 2 at x = 4.1 m, we need to determine the slope of the line representing pulse 2. This can be done by calculating the change in y (deflection) over the change in x (distance). In this case, we can see that the change in y is -1 mm (since pulse 2 is inverted) and the change in x is 4.1 m, so the slope of pulse 2 is -1 mm / 4.1 m = -0.24 mm/m.

Now, let's look at the values of the resultant wave at different times. At t = 1.0 s, both pulses will have moved 1.0 m (since they are moving at a speed of 1.0 m/s). This means that the value of pulse 1 at x = 4.1 m will be 0 mm (since it has moved beyond that point), and the value of pulse 2 at x = 4.1 m will be -0.24 mm/m * 1.0 m = -0.24 mm. Therefore, the resultant wave at x = 4.1 m will have a value of 0 mm - (-0.24 mm) = 0.24 mm.

At t = 2.0 s, pulse 1 will have moved 2.0 m and pulse 2 will have moved 1.0 m. This means that the value of pulse 1 at x = 4.1 m will be -