Finding the Longest Wavelength for Interference Patterns

AI Thread Summary
The discussion focuses on calculating the longest wavelengths for constructive and destructive interference from two radio towers positioned 136 meters apart, with an observation point 41 meters from one tower. Constructive interference occurs when the waves are in sync, while destructive interference happens when they are half a wavelength out of sync. The key formulas involve the path length difference and the relationship between wavelength and distance traveled by the waves. The calculated wavelengths for constructive and destructive interference are 136 meters and 272 meters, respectively. The conversation highlights the importance of understanding path length differences and their impact on wave interference patterns.
mrjeffy321
Science Advisor
Messages
876
Reaction score
1
I have a problem in which there are two radio broadcasting towers places a given distance apart (136 m). Each tower is broadcasting at the same frequency as the other one, but the frequency (or rather the wavelength is what I am concerned about) can be adjusted to alter the interference at a point (Q) that is another given distance (41 m) away from one of the towers, making it 177 m distance away from the other tower since it is arranged in a line.

I am looking for the longest wavelength that will cause destructive interference at point Q.
Also, I am looking for the longest wavelength for which there will be constructive interference at point Q.

I know constructive interference occurs when the waves are in sync with each other and thus add, creating a stronger (higher amplitude) wave.
Destructive interference occurs when the waves are 1/2 wavelength out of sync with each other and the waves cancel each other out.

I have a formula to describe the electric field of the waves,
E_1 = A*cos(omega*t + phi)
E_2 = A*cos(omega*t)
where omega is the angular frequency, t is time, and phi is the phase of the wave you begin watching it at.
One can modify the formula using the relationship between omega*t to be,
E = A*cos((2*pi / lamba)x + phi)
where lamba is the wavelength and x is the distance traveled.

So now I am loooking for the [longest] wavelength that can be emmited from tower A, travel 177 m, and either be in sync, or 1/2 cycle out of sync from a wave of the save wavelngth emmited from tower B and only traveling 41 m.

I can't seem to grasp the concept of how to do this and where real numbers fall in.
 
Physics news on Phys.org
You need to calculate the relative phase of each wave at the point of observation. This will depend on the various distances you mentioned.

Claude.
 
That still isn't much help to me.
How would I do that? and what then?
 
Since both towers and the observation point are all collinear, you can easily figure out the path length difference for the towers. That term should ring a bell, but if it doesn't then you should look it up in your book. That section of your book should also explain what the conditions are on the path length difference so that you observe constructive and destructive interference.
 
It just occurred to me the correct answer (136 m for constructive, 272 m for desctructive), and it turns out I am right.
But I still can't show it mathmatically if I had to with all that cosine stuff.
I figured that the smallest wavelength to cause interference would either have to be .5 or 1 wavelengths out of phase of the other wave, not 1.5 or 2, or 2.5 or 3, ... since that would mean the wave length is getting short.
So after I knew that for destructive interference, the wave was.5 wavelengths behind, and I knew that it was traveling an extra 136 meters, ta da, the wave length must be 272. and then the same process for constructive interference.


Funny, while I was typing this last reply, Tom Mattson came along and pointed out the very section in the book that gave me this revelation.
R_2 - R_1 = m*lamba
 
Last edited:
mrjeffy321 said:
But I still can't show it mathmatically if I had to with all that cosine stuff.

You could do it using the following trig identity.

\sin(\alpha)+\sin(\beta)=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top