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

• patrickmoloney
In summary, the problem involves two loudspeakers A and B emitting sound waves of frequency 450 Hz in phase. A microphone moving perpendicular to the line AB will detect a minimum of sound intensity at distances where the difference in path lengths is equal to half a wavelength. This can be calculated by finding the difference in path lengths as a function of distance x and determining how many wavelengths it represents.
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 )

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.

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?

1 person
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

patrickmoloney said:
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?

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.

1. What is intensity of sound waves?

Intensity of sound waves is a measure of the amount of energy carried by the sound wave per unit area. It is directly proportional to the amplitude of the wave and is measured in decibels (dB).

2. How is intensity of sound waves related to loudness?

Intensity and loudness are closely related, as a higher intensity sound wave will generally be perceived as louder. However, loudness is also affected by other factors such as frequency and duration of the sound.

3. How is intensity of sound waves measured?

Intensity of sound waves is typically measured using a sound level meter, which detects the pressure variations in the air caused by the sound wave. The results are then converted to decibels (dB) to determine the intensity.

4. What is the difference between intensity and amplitude of sound waves?

Intensity and amplitude are related but not the same. Amplitude refers to the maximum displacement of the particles in the medium caused by the sound wave, while intensity measures the amount of energy carried by the wave per unit area.

5. How does distance affect the intensity of sound waves?

As sound waves travel through a medium, their intensity decreases as they spread out over a larger area. This is known as the inverse square law, which states that the intensity is inversely proportional to the square of the distance from the source.

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