What is the resultant amplitude at X?

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Homework Help Overview

The problem involves two wave generators producing water waves with a specified wavelength and a detector placed at a certain distance from each generator. The waves have a phase difference, and the task is to determine the resultant amplitude at the detector when both generators operate together.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the phase difference and path difference on wave interference, questioning how these factors relate to the resultant amplitude. Some express confusion about the relationship between phase difference and path difference, while others explore the conditions for constructive and destructive interference.

Discussion Status

The discussion includes various interpretations of the phase differences and their effects on wave interference. Some participants have offered insights into the mathematical relationships governing the interference, while others are seeking clarification on specific points. There is an ongoing exploration of different scenarios, including changes in phase difference.

Contextual Notes

Participants are considering specific phase differences (e.g., π and π/2 radians) and their impact on the resultant amplitude, indicating a focus on understanding the underlying principles of wave interference.

songoku
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Homework Statement


Two wave generators S1 and S2 produce water waves of wavelength 0.5 m. A detector X is placed on the water surface 3 m from S1 and 2 m from S2 as shown in the diagram. Each generator produced a wave of amplitude A at X when operated alone. The generator are operating together and producing waves which have a constant phase difference of [tex]\pi[/tex] radians. What is the resultant amplitude at X?

a. 0
b. A/2
c. A/3
d. A
e. 2 A

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Homework Equations


The Attempt at a Solution


Phase difference [tex]\pi[/tex] means that the two waves are completely out of phase.

The path difference (p) between the two sources = 1 m = 2 [tex]\lambda[/tex]. If p = n[tex]\lambda[/tex], it should produce constructive interference, but I think we have to consider the phase difference...

Now I'm confused how to relate these two informations to obtain the answer. Thanks
 
Last edited:
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D path difference corresponds to 2pi*D/lambda phase difference, and the total phase difference counts in interference.

ehild
 


ehild said:
D path difference corresponds to 2pi*D/lambda phase difference, and the total phase difference counts in interference.

Hi ehild

Sorry but I don't get it...:frown:

You mean : phase difference = 2pi*D/lambda ?
 


Lambda path difference is equivalent to 2pi phase difference, 2lambda path difference + the original phase difference between the sources is 4pi + pi phase difference at the meeting point of the waves.

ehild
 


ehild said:
Lambda path difference is equivalent to 2pi phase difference, 2lambda path difference + the original phase difference between the sources is 4pi + pi phase difference at the meeting point of the waves.

I think I get it. It will be destructive interference.

Thanks ehild ! :smile:
 


OK, you got it!:smile:
 


Hi ehild

I want to ask another condition. What if the generator are operating together and producing waves which have a constant phase difference of pi/2 radians?

The phase difference will be 4.5 pi and I think a complete destructive or constructive interference won't occur. How to determine the amplitude at X?

Thanks
 
Last edited:


songoku said:
The phase difference will be 4.5 pi and I think a complete destructive or constructive interference won't occur. How to determine the amplitude at X?

When two waves meet in a point, and one produces a displacement (Displacement can be a lot of things. Change of pressure if it is sound, electric field strength, if it is an electromagnetic wave) y1=A1sin(wt) and the other y2=A2sin(wt+phi) the resultant will be a sin function again with a new amplitude and phase constant.

y1+y2=Asin(wt+alpha)-->
A1 sin (wt)+
A2[sin(wt)cos(phi)+cos(wt)sin(phi)]=A[sin(wt)cos(alpha)+cos(wt)sin(alpha)]

collect the terms containing sin(wt) and do the same with term containing cos(wt).

sin(wt)[A1+A2cos(phi)-Acos(alpha)]=cos(wt)[Asin(alpha)-A2sin(phi)]

The equation should hold for any t. If sin(wt)=0, the right side is also 0, but cos(wt) is different from 0, so

Asin(alpha)-A2sin(phi)=0

The same holds for the left side: A1+A2cos(phi)-Acos(alpha)=0

From this two equations, you get both A (the new amplitude) and alpha( the new phase)

A sin (alpha)=A2sin(phi)
Acos(alpha)=A1+A2cos(phi)

Take the square of both equations and add them up, you get the square of the new amplitude:

A2=A12+2A1*A2*cos(phi)+A22

Assume that the amplitudes are equal A1=A2 and the phase difference is phi = pi/4 + 2k*pi, cos(phi)=sin(phi)=1/sqrt(2).

A=A1*sqrt[2+sqrt(2)]

ehild
 


Hi ehild
ehild said:
sin(wt)[A1+A2cos(phi)-Acos(alpha)]=cos(wt)[Asin(alpha)-A2sin(phi)]

The equation should hold for any t. If sin(wt)=0, the right side is also 0, but cos(wt) is different from 0, so

Asin(alpha)-A2sin(phi)=0

The same holds for the left side: A1+A2cos(phi)-Acos(alpha)=0

We take cos(wt) = 0 to get A1+A2cos(phi)-Acos(alpha)=0 ?

Thanks
 
  • #10


songoku said:
Hi ehild


We take cos(wt) = 0 to get A1+A2cos(phi)-Acos(alpha)=0 ?

Thanks

We can take cos(wt) anything between -1 and +1, the equation must hold. It must hold for cos(wt)=0 (that is, for wt=pi/2+2kpi) One side of the equation is 0, then the other side must be also 0. The right side is the product of sin(wt) and A1+A2cos(phi)-Acos(alpha). One of them must be 0. But it can not be sin(wt), because the sine is either +1 or -1. Therefore A1+A2cos(phi)-Acos(alpha)=0.


ehild

ehild
 
  • #11


Thanks a lot ehild !
 

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