# What is the resultant amplitude at X?

1. Jan 18, 2010

### songoku

1. The problem statement, all variables and given/known data
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 $$\pi$$ radians. What is the resultant amplitude at X?

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

2. Relevant equations

3. The attempt at a solution
Phase difference $$\pi$$ means that the two waves are completely out of phase.

The path difference (p) between the two sources = 1 m = 2 $$\lambda$$. If p = n$$\lambda$$, 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: Jan 18, 2010
2. Jan 18, 2010

### ehild

Re: wave

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

ehild

3. Jan 18, 2010

### songoku

Re: wave

Hi ehild

Sorry but I don't get it...

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

4. Jan 18, 2010

### ehild

Re: wave

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

5. Jan 19, 2010

### songoku

Re: wave

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

Thanks ehild !!

6. Jan 19, 2010

### ehild

Re: wave

OK, you got it!!

7. Jan 19, 2010

### songoku

Re: wave

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: Jan 20, 2010
8. Jan 20, 2010

### ehild

Re: wave

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

9. Jan 20, 2010

### songoku

Re: wave

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

Thanks

10. Jan 20, 2010

### ehild

Re: wave

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. Jan 23, 2010

### songoku

Re: wave

Thanks a lot ehild !!