Interference - Frequency Interference -- Frequency

[SuperCass]

Interference -- Frequency

Homework Statement

Two loudspeakers at an outdoor rock concert are located 3.5 meters apart. You are standing 16.1 meters from one of the speakers and 19 from the other. During a sound check, the technician sends the exact same frequency to both speakers while you listen. The technician starts at 20Hz and slowly increases it to 30,000Hz.
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a) What is the lowest frequency where you will hear a minimum signal ?
f = Hz
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b) What is the second lowest frequency where you will hear a minimum signal ?
f = Hz
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c) What is the lowest frequency where you will hear a maximum signal ?
f = Hz
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d) What is the second lowest frequency where you will hear a maximum signal ?
f = Hz

Homework Equations

\omega=2\pif
v=\sqrt{T/\mu}

The Attempt at a Solution

I'm not sure where to start!

 

[hikaru1221]

What is the general condition at which the net amplitude is max/min when two waves of the same frequency (and the same vibrating direction) superimpose? Hint: Something about phase difference.

 

[SuperCass]

When there is no phase difference or the phase difference is divisible by pi?

 

[SuperCass]

When there is no phase difference or the phase difference is divisible by pi?

Okay so what I have done so far is found the path length difference, (\DeltaL = L1 - L2).
I know that \DeltaL/\lambda = \Phi / 2\Pi, but is this the right direction?

Where do I go from here?

 

[hikaru1221]

The problem says nothing about the initial phase difference, so I assume that the initial signals coming out of the loudspeakers are in phase.
\Phi is the phase difference, right? So you're on the right track ;)
1 - Now what would \Phi be if it's maximum? And if it's minimum?
2 - Let's take the sound speed v=340m/s. You have \Delta L. So from the above equation you've just pointed out:f = \frac{v}{\lambda} = \frac{\Phi}{2\pi \Delta L}v
Subtitute \Phi for each case (max/min), you will get f.

 

[SuperCass]

Got it! Thank you so so much!