Solved: Wave Interference Frequency & Phase Constant

In summary: As you can see, this is not possible, which means that there is no integer value of n that satisfies the equation. This means that the points given in the problem are not the nodes. In fact, the nodes are not evenly spaced, as you noticed. The actual nodes are at x = 0.323m, x = 0.686m, and x = 1.049m, which you can find by setting the phase difference to odd multiples of π. So, in summary, the frequency of the two sources is equal to the difference in distance between them (0.375
  • #1
theneedtoknow
176
0

Homework Statement


2 sources emitting radio waves of the same frequency are placed at x=0 meters and x = 1.2 meters. There are nodes at x= 0.344m, x= 0.719m, and x=1.094m. What is the frequency of the 2 sources and the difference in their phase constant


Homework Equations





The Attempt at a Solution


Well i know that maximum destructive interference occurs for every 2pi increase in the difference in phase of the waves
The difference in phase is:
dФ = 2Pi (dr)/wavelength + dФ0

where dr is r2 - r1 = difference in radial distance between each source and the point in question
now here is my problem:
since dФ0 stays the same, it means that there will be a node for every increase in dr equal to one wavelength
however, if i use those points mentioned above to find the radial difference for each point, i get
dr1 = 0.512
dr2 = 0.238
dr3 = 0.988

so each one of these nodes has an increase in phase difference of 2Pi from the previous
so taking the 2 with lower drs :
dФ1 = 2Pi (0.238)/wavelength + dФ0
dФ2 = 2Pi (0.512)/wavelength + dФ0 = dФо+ 2Pi
dФ2 - dФ1 = 2 pi = 2pi (0.512)/wavelength - 2Pi (0.238)/wavelength
wavelength = 0.512 - 0.238

so the difference between 2 consecutive nodes should give me the wavelength
...but these are not evenly spaced, so if i use the 2nd set of points instead i get a different result (0.988 - 0.512 = 0.476) so I don't get it, where is my thinking going wrong?
 
Physics news on Phys.org
  • #2


Hello there, thank you for your post. Your approach to finding the wavelength and phase difference is correct. The issue here is that the points given (0.344m, 0.719m, 1.094m) are not the nodes, but rather the positions where the waves are in phase. This means that at these points, the phase difference between the two sources is a multiple of 2π. In other words, at these points, the difference in phase is an integer multiple of 2π, not just 2π.

To find the nodes, we need to look for points where the waves are in destructive interference, meaning the phase difference is an odd multiple of π. So, we can rewrite your equation for the phase difference as:

dФ = 2π(dr/λ + n) + dФ0

where n is an integer representing the number of wavelengths between the two sources. Now, if we use the points given in the problem, we can solve for n:

dФ1 = 2π(0.344/λ + n) + dФ0
dФ2 = 2π(0.719/λ + n) + dФ0

Since dФ1 and dФ2 represent the same phase difference (2π), we can set them equal to each other and solve for n:

2π(0.344/λ + n) + dФ0 = 2π(0.719/λ + n) + dФ0
0.344/λ + n = 0.719/λ + n
0.344/λ = 0.719/λ
λ = 0.375m

So, the wavelength is 0.375m, which is equal to the difference in distance between the two sources. Now, we can use this value to find the phase difference:

dФ = 2π(1.094/0.375 + n) + dФ0
dФ = 2π(2.917 + n) + dФ0

Since dФ1 and dФ2 represent the same phase difference (2π), we can set them equal to each other and solve for n:

2π(2.917 + n) + dФ0 = 2π(0.719/0.375 + n) + dФ0
2π(2.917 + n) =
 
  • #3


I would approach this problem by first examining the given information and equations to ensure that they are accurate and applicable to the scenario. It is important to understand the underlying principles of wave interference and how they apply to this specific situation.

Next, I would suggest using trigonometric functions and the Pythagorean theorem to calculate the radial distance (dr) between each source and the points of destructive interference (nodes). This will provide more accurate values for dr and allow for a more precise calculation of the wavelength.

Additionally, it may be helpful to graph the data and visually analyze the points of interference to ensure that the calculations are correct. If the points are not evenly spaced, it could indicate that there are other factors at play, such as diffraction or reflection, which may affect the interference pattern.

In terms of finding the frequency and phase constant of the sources, I would use the formula c = fλ, where c is the speed of light and λ is the wavelength. By plugging in the calculated wavelength, the frequency can be determined. The phase constant can then be found using the formula dФ0 = 2π(d1)/λ, where d1 is the distance between the first source and the point of destructive interference.

It is always important to carefully analyze and double-check calculations and assumptions in scientific problem-solving. If there are discrepancies or questions, it may be necessary to revisit the equations and data to ensure accuracy.
 

What is wave interference?

Wave interference is a phenomenon that occurs when two or more waves meet and interact with each other. This can result in either constructive interference, where the waves reinforce each other, or destructive interference, where the waves cancel each other out.

What is frequency?

Frequency is the number of complete cycles or oscillations of a wave that occur in one second. It is measured in Hertz (Hz) and is directly related to the energy and wavelength of a wave.

What is a phase constant?

A phase constant is a measure of the position of a wave at a specific point in time. It is represented by the Greek letter phi (φ) and is used to describe the relationship between two waves that are interfering with each other.

How do you calculate wave interference frequency?

The frequency of wave interference can be calculated by dividing the speed of the wave by the wavelength of the wave. This formula is represented as f= v/λ, where f is the frequency, v is the speed, and λ is the wavelength.

What are some real-world applications of wave interference?

Wave interference has many practical applications, such as in noise-canceling technology, where sound waves are used to cancel out unwanted noise. It is also used in radio and television broadcasting to transmit signals over long distances without losing quality. Additionally, wave interference is essential in medical imaging techniques such as ultrasound and MRI, which use sound waves to create images of internal body structures.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
293
  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
956
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
8K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
3K
Back
Top