Water interference physics help

In summary: Good work :)In summary, the conversation discussed the observation of two narrow gaps in the breakwater, located 9.0 m apart and parallel to the shore. The speaker went to the midpoint between the gaps and noticed that at a distance of 1.7 m from the starting point, no waves were reaching them. Beyond the breakwater, 10 wave crests were observed in 18 seconds. The conversation also mentioned the use of the formula r2-r1 = (m+1/2)lambda to find the path difference and the significance of the given frequency, which was not used in the calculations. It was later discovered that the online answer was incorrect and the speaker's answer was correct.
  • #1
jkom329
4
0
While walking along the shore at your beachfront home, you notice that there are two narrow gaps in the breakwater, the wall that protects the shore from the waves. These gaps are 9.0 m apart and the breakwater is 12.0 m from the shore and parallel to it. You go to the shore directly opposite the midpoint between these gaps. As you walk along the shore, the first point where no waves reach you is 1.7 m from your starting point. Out beyond the breakwater you observe that there are ten wave crests in 18 s.

To me, this seems to be a straightforward application of the formula r2-r1 = (m+1/2)lambda. I found the path difference (r2-r1) by simple geometry (1.19m), then I set m=0 since 1.7m is the first point of destructive interference. The answer I have is lambda = 2.38m, and it is wrong for some reason. I also do not understand the significance of the given frequency, and I have not used it in my calculations. I would love to know what I'm missing here. Thanks!
 
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  • #2
jkom329 said:
While walking along the shore at your beachfront home, you notice that there are two narrow gaps in the breakwater, the wall that protects the shore from the waves. These gaps are 9.0 m apart and the breakwater is 12.0 m from the shore and parallel to it. You go to the shore directly opposite the midpoint between these gaps. As you walk along the shore, the first point where no waves reach you is 1.7 m from your starting point. Out beyond the breakwater you observe that there are ten wave crests in 18 s.

To me, this seems to be a straightforward application of the formula r2-r1 = (m+1/2)lambda. I found the path difference (r2-r1) by simple geometry (1.19m), then I set m=0 since 1.7m is the first point of destructive interference. The answer I have is lambda = 2.38m, and it is wrong for some reason. I also do not understand the significance of the given frequency, and I have not used it in my calculations. I would love to know what I'm missing here. Thanks!

What are you being asked to find?
 
  • #3
How far apart are the wave crests?

Forgot to paste it in. Thanks for your help.
 
  • #4
jkom329 said:
How far apart are the wave crests?

Forgot to paste it in. Thanks for your help.

Something here does not compute. You calculated lambda and said you got the wrong answer. Lambda is the distance between wave crests, and your result looked OK to me. The only reason to give the frequency would be to then ask you to find velocity. I am assuming the waves are arriving with crests parallel to the breakwater and that the midpoint between the gaps is a wave maximum, but when you said you had the wrong lambda it made me wonder.
 
  • #5
bump... I still do not see what I'm missing.
 
  • #6
Conclusion: I am not missing anything. I found out today that the online answer is incorrect, and my answer is correct :biggrin:
 
  • #7
jkom329 said:
Conclusion: I am not missing anything. I found out today that the online answer is incorrect, and my answer is correct :biggrin:

That will do it. I was imagining various possibilities, like the slowing if the waves as they approach the shore reducing the wavelength, but clearly the problem did not give enough information to treat that complication, and the fact that they were asking for the wavelength dismissed that possibility.
 

Related to Water interference physics help

1. What is water interference physics?

Water interference physics is the study of the interactions between water and other materials, such as light or sound waves. It involves understanding how water molecules behave and how they can affect the propagation of waves through water.

2. How does water interference affect the movement of waves?

Water interference can cause waves to change direction, speed, or shape. When two or more waves intersect, they can either amplify or cancel each other out, depending on the phase difference between them. This phenomenon is known as constructive and destructive interference, respectively.

3. What is the relationship between wavelength and interference in water?

The wavelength of a wave is directly related to its frequency and speed. In water, waves with longer wavelengths tend to experience less interference because they have a lower frequency and are less likely to interact with other waves. On the other hand, waves with shorter wavelengths are more likely to experience interference and may even be completely canceled out.

4. How does the depth of water affect interference?

The depth of water can significantly impact interference. In shallow water, the waves are more likely to interact with the bottom surface, causing changes in their speed and direction. In deeper water, the waves can travel with less interference, resulting in longer wavelengths and smoother surfaces.

5. What are some practical applications of water interference physics?

Water interference physics has many practical applications, such as in oceanography, underwater acoustics, and optics. It is also essential in understanding the behavior of waves in bodies of water, which is crucial for activities like surfing, sailing, and underwater exploration.

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