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Please help! I've been stuck on this for a very very long time .

Two loudspeakers, separated by a distance of d1 = 1.60 m, are in phase. Assume the amplitudes of the sound from the speakers are approximately the same at the position of a listener, who is d2 = 3.90 m directly in front of one of the speakers.

See Diagram: http://www.webassign.net/hrw/18_32alt.gif" [Broken]

(a) For what frequencies in the audible range (20-20,000 Hz) does the listener hear a minimum signal? (Give the lowest, second lowest and highest possible frequencies in the audible range.)

I JUST FIGURED THIS ONE OUT:(b) For what frequencies is the signal a maximum? (Give the lowest, second lowest and highest possible frequencies in the audible range.)

KEY: d = delta, f = frequency, v = velocity of air, dL = path length difference, m = nodes I assume?

If not, what's m in the equation (seen below)?

dL = (m+.5)(v/f)

v = 343 m/s

d3 = squareroot(1.6^2 + 3.9^2) = 4.22 m

I was able to derive this: f = (m+.5)(v/dL) from the dL equation given.

Phase difference at the person is = 2pi(d3-d2)/lamda

For minimum intensity at the listener, phasediff = (2m+1)pi

and lamda = 2pi(d3-d2)/phasediff

freq = v/lamda = [(2m+1)v]/[2(d3-d2)]

and that leaves me with (2m+1)[v/(2(d3-d2))]

[v/(2(d3-d2))] = frequency that I get

For me, it's 543.67Hz.

And 20000Hz/543.67Hz = 37.3

2m+1 = 37

So m needs to be between 1 and 18 I believe...

****

I figured out part b.:rofl:

lamda = (1/n)(d3-d2)

so f = v/lamda = nv/(d3-d2) = n(1088.89Hz)

20,000Hz/1088.89Hz = 18.37

n = 1 through 18

Plug in 1, 2, and 18 for frequencies...

****

But I'm still wrong for part a... :grumpy: :surprised :yuck: :zzz:

**1. Homework Statement**Two loudspeakers, separated by a distance of d1 = 1.60 m, are in phase. Assume the amplitudes of the sound from the speakers are approximately the same at the position of a listener, who is d2 = 3.90 m directly in front of one of the speakers.

See Diagram: http://www.webassign.net/hrw/18_32alt.gif" [Broken]

(a) For what frequencies in the audible range (20-20,000 Hz) does the listener hear a minimum signal? (Give the lowest, second lowest and highest possible frequencies in the audible range.)

I JUST FIGURED THIS ONE OUT:(b) For what frequencies is the signal a maximum? (Give the lowest, second lowest and highest possible frequencies in the audible range.)

KEY: d = delta, f = frequency, v = velocity of air, dL = path length difference, m = nodes I assume?

If not, what's m in the equation (seen below)?

**2. Homework Equations**dL = (m+.5)(v/f)

v = 343 m/s

**3. The Attempt at a Solution**d3 = squareroot(1.6^2 + 3.9^2) = 4.22 m

I was able to derive this: f = (m+.5)(v/dL) from the dL equation given.

Phase difference at the person is = 2pi(d3-d2)/lamda

For minimum intensity at the listener, phasediff = (2m+1)pi

and lamda = 2pi(d3-d2)/phasediff

freq = v/lamda = [(2m+1)v]/[2(d3-d2)]

and that leaves me with (2m+1)[v/(2(d3-d2))]

[v/(2(d3-d2))] = frequency that I get

For me, it's 543.67Hz.

And 20000Hz/543.67Hz = 37.3

2m+1 = 37

So m needs to be between 1 and 18 I believe...

****

I figured out part b.:rofl:

lamda = (1/n)(d3-d2)

so f = v/lamda = nv/(d3-d2) = n(1088.89Hz)

20,000Hz/1088.89Hz = 18.37

n = 1 through 18

Plug in 1, 2, and 18 for frequencies...

****

But I'm still wrong for part a... :grumpy: :surprised :yuck: :zzz:

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