What Distance from Speaker A Will a Microphone Detect Minimum Sound Intensity?

[patrickmoloney]

Homework Statement

Two loudspeakers A and B of equal power are separated by a distance of 1.4 m. Both the speakers emit sound waves in phase and of frequency 450 Hz. If a microphone were to be moved from A in a direction perpendicular to AB, at what distances from A will it detect a minimum of sound intensity? ( Take the velocity of sound in air to be 330 m/s )

Homework Equations

c=fλ,

The Attempt at a Solution

I have no idea where to even start. I tried using Pythagoras' theorem and could go no further.

 

[haruspex]

Pythagoras' theorem is certainly relevant.
At a point distance x from A along that line, how far is it from B?
How many wavelengths is it from each? What will the phase difference be at that point?

 

[patrickmoloney]

Distance is sqrt(1.96)+x². How do I calculate the wavelength at each point. lambda=c/f = 330/ 450 = 0.7333 which is all I know. I also know that destructive interference occurs at lamda/2

 

[haruspex]

Distance is sqrt(1.96)+x².

That's not what you meant to write, I hope.

How do I calculate the wavelength at each point.

I did not suggest calculating a wavelength at each point. There is only one wavelength, which you have calculated. I said to calculate the number of wavelengths represented by each of the two distances. But the alternative below may be simpler.

lambda=c/f = 330/ 450 = 0.7333 which is all I know. I also know that destructive interference occurs at lamda/2

For sources in phase, same frequency, completely destructive interference occurs when the difference in the path lengths is lambda/2. What, as a function of x, is the difference in the path lengths? How many wavelengths is that?

 

[Nickie]

Please help!

I would approach this problem by first understanding the concept of sound intensity and how it is affected by distance. Sound intensity is the amount of sound energy passing through a unit area in a unit of time. It is directly proportional to the square of the amplitude of the sound wave and inversely proportional to the square of the distance from the source.

In this problem, we have two speakers emitting sound waves of equal power, which means they have the same amplitude. Since they are in phase, the sound waves will interfere constructively and produce a higher intensity. As we move away from the speakers, the sound waves will start to interfere destructively and the intensity will decrease.

To solve for the distance at which the microphone will detect minimum sound intensity, we can use the equation for sound intensity:

I = P/4πr²

Where I is the sound intensity, P is the power of the speakers, and r is the distance from the source.

We can rearrange this equation to solve for r:

r = √(P/4πI)

Since the power of the speakers and the frequency are given, we can calculate the intensity at any given distance using the equation:

I = (P/4π)(1/r²)

Now, to find the distance at which the intensity will be minimum, we need to find the distance at which the intensity is lowest. This will happen when the denominator, r², is the largest. This means that r should be the largest possible value, which is when the microphone is farthest away from the speakers.

Using the given values, we can calculate the maximum distance using the equation for the speed of sound:

r = vt

Where v is the speed of sound and t is the time it takes for the sound wave to travel from the speakers to the microphone. The time can be calculated using the formula:

t = d/v

Where d is the distance between the speakers (1.4 m). Therefore, the maximum distance at which the microphone will detect minimum sound intensity is:

r = (330 m/s)(1.4 m)/(450 Hz) = 0.91 m

This means that the microphone will detect minimum sound intensity at a distance of 0.91 m from speaker A.