Finding minimum and maximum intensity of waves

[Troi Jones]

Homework Statement

Two loudspeakers are placed 3.00 m apart horizontally.They emit 425 Hz sounds, in phase. A microphone is placed d= 3.20 mid distant from a point midway between the two speakers, where an intensity maximum is recorded.

(a) How far must the microphone be moved to the right to find the first intensity minimum?

(b) Suppose the speakers are reconnected so that the 455 Hz sounds they emit are exactly out of phase. At what positions (how far must the microphone be moved to the right) are the intensity maximum and minimum now?

Homework Equations

S1 - S2= wavelength/2

The Attempt at a Solution

I solved for lengths S_2 and S1 through the Pythagorean theorem which is sqrt(1.5^2 +3.2^2)= 3.534 m. But that's all I know so far. I do not have any other idea of where to go from here. Any type of guidance would be useful, understanding waves is my weak point.[/B]

 

[mfb]

This is just geometry.

If you move the microphone a distance x to the right, what are the new distances? How much do they differ? Which x satisfies the equation you posted?

 

[Charles Link]

The equation you wrote in part 2 is correct and quite necessary to find the position $x$ for which you get the first minimum. Can you write $s_1$ and $s_2$ as a function of $x$ and solve this equation for $x$? @mfb has it correct when he says in post 2, "This is just geometry", but it is still geometry you need to solve to get the correct answer.

 

[Troi Jones]

The equation you wrote in part 2 is correct and quite necessary to find the position $x$ for which you get the first minimum. Can you write $s_1$ and $s_2$ as a function of $x$ and solve this equation for $x$? @mfb has it correct when he says in post 2, "This is just geometry", but it is still geometry you need to solve to get the correct answer.

So should my equation for S2 be sqrt(3.05^2 + (3/2 + x)^2) and S1 be sqrt(3.05^2 +(3/2-x)^2) to solve for x?

 

[Troi Jones]

So should my equation for S2 be sqrt(3.05^2 + (3/2 + x)^2) and S1 be sqrt(3.05^2 +(3/2-x)^2) to solve for x?

After solving for x in S_1 - S_2, I get answer of 6x i think. But I'm not sure what to do next?

 

[Charles Link]

So should my equation for S2 be sqrt(3.05^2 + (3/2 + x)^2) and S1 be sqrt(3.05^2 +(3/2-x)^2) to solve for x?

$d=3.20$, (not 3.05), but otherwise correct.

 

[Troi Jones]

$d=3.20$, (not 3.05), but otherwise correct.

Oops my mistake but ok 6x is correct. So what do I use 6x for? Do i substitute for S1-S2 in S1-S2= wavelength/2?

 

[Charles Link]

Oops my mistake but ok 6x is correct. So what do I use 6x for? Do i substitute for S1-S2 in S1-S2= wavelength/2?

You have $f=425 \,Hz$, and you need to use the value for the speed of sound to convert this to wavelength $\lambda$. Once you solve for $\lambda/2$, you then have a somewhat difficult, but not impossible algebraic expression to solve for $x$: $|s_1-s_2|=\lambda/2$. $\\$ One suggestion is to try solving it numerically by graphing the left side of the equation as a function of $x$, if your algebra is not real good.

 

[Troi Jones]

You have $f=425 \,Hz$, and you need to use the value for the speed of sound to convert this to wavelength $\lambda$. Once you solve for $\lambda/2$, you then have a somewhat difficult, but not impossible algebraic expression to solve for $x$: $|s_1-s_2|=\lambda/2$. One suggestion is to try solving it numerically by graphing the left side as a function of $x$, if your algebra is not real good.

Ok. So the speed of sound is 343 m/s. The wavelength formula equals velocity/ frequency. So wavelength should equal 343/425 which equals about 0.8071 m. Since I still have wavelength/2, that means I divide 0.8071/2 equalling 0.4035. So for my equation to solve for x= x|S1 - S2| = wavelength equals
x= 0.4035/ |S1-S2|?

 

[Charles Link]

Ok. So the speed of sound is 343 m/s. The wavelength formula equals velocity/ frequency. So wavelength should equal 343/425 which equals about 0.8071 m. Since I still have wavelength/2, that means I divide 0.8071/2 equalling 0.4035. So for my equation to solve for x= x|S1 - S2| = wavelength equals
x= 0.4035/ |S1-S2|?

No. $|s_1-s_2|=\lambda/2$, where $s_1=s_1(x)$ , and $s_2=s_2(x)$ which you wrote the expressions for (in post 4=(but use d=3.20, see post 6)), using the Pythagorean theorem. You are still far from being able to write $x=...$. I'm suggesting a numerical solution, to save you from trying to solve for $x$ with something that may be algebraically somewhat difficult. Graphing the left side of the equation vs. x and seeing where it equals the right side of the equation is a feasible numerical approach. $\\$ If you are good with algebra, you should be able to solve for $x$ and get the same answer that you get numerically.