Interference of rectangular pulse

In summary, the figure shows two rectangular wave pulses traveling toward each other on a stretched string. Each pulse is traveling with a speed of 1.00 mm/s and has a height and width shown in the figure. At t = 4.00 s, the two pulses will meet and the string will be at equilibrium if the pulses were of equal amplitude. However, since the right pulse has an amplitude of 4.00 mm, the string will have an upward pulse with an amplitude of 1.00 mm and a width of 4.00 mm. At t = 6.00 s, the pulses will be 4.00 mm apart, with one pulse traveling in the negative x direction with an amplitude of
  • #1
sapiental
118
0
Question Details:
The figure shows two rectangular wave pulses
traveling toward each other on a stretched string.
Each pulse is traveling with a speed of 1.00 mm/s,
and has the height and width shown in the figure.
If the leading edges of the pulses are 8.00 mm
apart at t = 0, sketch the shape of the string at
t = 4.00 s, t = 6.00 s, and t = 10.0

you guys don't have to sketch the result for me but please be as descriptive as possible.

Im thinking that at t = 4s the 2 pulses will meet. If the pulses were of equal amplitude the string would be at equilibrium. But the right one is 4.00mm so I think it will look something like a upward pulse with 1.00mm amplitude and 4.00mm width?

At t = 6s the pulses will be 4mm apart. One pulse traveling in the negative x direction with an amplitude of 4mm and width 4mm and velocity 1mm/s. The second pulse traveling in the positive x direction with amplitude 3mm, width 4mm, and velocity 1mm.

not sure about t = 10s, but i think the problem wants us to presume that the pulses reflect off the wall..

Any help is much appreciated. And could you guys also explain how you got your answers.. i.e. by what rules etc. tho one I can think of is the principle of superposition (y(x,t) = y1(x,t) + y2(x,t). thanks

Please see the attached pic for the diagram.
 

Attachments

  • untitled.JPG
    untitled.JPG
    7 KB · Views: 1,217
Physics news on Phys.org
  • #2
You have the right idea -- just add them with superposition. Where is the wall you are referring to?
 
  • #3


I would first like to clarify the parameters of the problem. It states that the pulses are traveling with a speed of 1.00 mm/s, but does not specify the direction of this velocity. I will assume that both pulses are traveling in opposite directions, as shown in the diagram.

At t = 4.00 s, the two pulses will meet at the point where the distance between their leading edges is 4.00 mm. The resulting shape of the string at this time can be described by the principle of superposition, which states that the total displacement of a medium at a given point and time is equal to the sum of the displacements caused by each individual wave. In this case, the two pulses will interfere with each other, resulting in a new wave with a higher amplitude.

Since the right pulse has an amplitude of 4.00 mm and the left pulse has an amplitude of 3.00 mm, the resulting wave will have an amplitude of 7.00 mm. The width of the new wave will be equal to the sum of the widths of the two original pulses, which is 8.00 mm. Therefore, at t = 4.00 s, the string will have a new pulse with an amplitude of 7.00 mm, a width of 8.00 mm, and a velocity of 1.00 mm/s in the direction of the right pulse.

At t = 6.00 s, the two pulses will have moved further apart and will be 4.00 mm apart. The left pulse will have traveled a distance of 6.00 mm, while the right pulse will have traveled a distance of 2.00 mm. The resulting shape of the string at this time can be described by the principle of superposition again. The left pulse will now have an amplitude of 2.00 mm, a width of 4.00 mm, and a velocity of 1.00 mm/s in the negative x direction. The right pulse will still have an amplitude of 4.00 mm, a width of 4.00 mm, and a velocity of 1.00 mm/s in the positive x direction.

At t = 10.00 s, the two pulses will have reached the end of the string and will be reflected back in the opposite direction. They will continue to interfere with each other as they travel back towards each other. The resulting shape of the
 

Related to Interference of rectangular pulse

1. What is the definition of interference of rectangular pulse?

Interference of rectangular pulse refers to the phenomenon where two or more rectangular pulses, which are identical in shape but may have different amplitudes, overlap and create a new pulse with a different shape and amplitude.

2. What causes interference of rectangular pulse?

Interference of rectangular pulse is caused by the superposition of two or more pulses. This occurs when the pulses are traveling in the same medium and their paths intersect, resulting in their amplitudes being added together at certain points.

3. How does the amplitude of the resulting pulse in interference of rectangular pulse compare to the individual pulses?

The amplitude of the resulting pulse in interference of rectangular pulse is equal to the sum of the amplitudes of the individual pulses at that point. This is known as constructive interference. However, if the pulses have opposite polarities, they will cancel each other out, resulting in a lower amplitude or even no pulse at all. This is known as destructive interference.

4. What is the role of phase difference in interference of rectangular pulse?

Phase difference plays a crucial role in determining the resulting pulse in interference of rectangular pulse. If the pulses have the same phase, they will constructively interfere, resulting in a larger amplitude. On the other hand, if they have opposite phases, they will destructively interfere, resulting in a lower amplitude or no pulse at all.

5. What are the practical applications of interference of rectangular pulse?

Interference of rectangular pulse is a fundamental concept in signal processing and is used in various applications such as radar, sonar, telecommunications, and medical imaging. It can also be used to manipulate and control light in optics and create interference patterns for experiments in physics.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
11K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
8K
Back
Top