Two speakers are driven by the same oscillator with frequency of 172 Hz. They are located 3.72 m apart on a vertical pole. A man really far away walks straight toward the lower speaker in a direction perpendicular to the pole.

How far is he from the pole at the first moment he hears a minimum in sound intensity (short of infinity) if the speed on sound is 330 m/s?

To find the minimum in sound intensity, we need to consider the interference pattern created by the two speakers.

The distance between the speakers is 3.72 m, which is half of a wavelength for the 172 Hz frequency. This means that at certain points, the sound waves from the two speakers will interfere constructively, resulting in a maximum in sound intensity. At other points, the waves will interfere destructively, resulting in a minimum in sound intensity.

Since the man walks perpendicular to the pole, we can consider the path difference between the sound waves from the two speakers to the man. The path difference will determine whether constructive or destructive interference occurs at a particular point.

Let's assume that the man is initially at a point where destructive interference occurs. At this point, the path difference between the two speakers should be an odd multiple of half a wavelength.

The wavelength of the sound wave can be calculated using the formula:

wavelength = speed of sound / frequency

wavelength = 330 m/s / 172 Hz ≈ 1.918 m

For destructive interference, the path difference should be an odd multiple of half a wavelength. Therefore, the path difference between the two speakers and the man's position should be:

path difference = (2n + 1) * (wavelength/2)

where n is an integer representing the number of half-wavelength shifts.

To find the minimum distance from the pole, we need to find the smallest value of n that satisfies the path difference condition.

Let's consider the first few values of n to find the smallest distance:

n = 0:
path difference = (2*0 + 1) * (1.918 m/2) = 0.959 m

n = 1:
path difference = (2*1 + 1) * (1.918 m/2) ≈ 2.877 m

n = 2:
path difference = (2*2 + 1) * (1.918 m/2) ≈ 4.795 m

From the calculations above, we can see that the first moment of minimum sound intensity occurs when the path difference is approximately 0.959 m.

Therefore, the man is about 0.959 m away from the pole at the first moment he hears a minimum in sound intensity.

To find the distance at which the man hears a minimum in sound intensity, we need to consider the concept of interference between the sound waves produced by the two speakers.

Interference occurs when two waves interact with each other, resulting in the amplification or cancellation of certain parts of the wave. In this case, the waves from the two speakers will interfere with each other as they propagate towards the man.

Step 1: Determine the wavelength of sound
The frequency of the oscillator is given as 172 Hz. The speed of sound in air is 330 m/s. We can use the formula v = λf, where v is the wave speed, λ is the wavelength, and f is the frequency, to find the wavelength of sound.

Wavelength (λ) = v / f
= 330 m/s / 172 Hz
≈ 1.918 m

Step 2: Determine the distance between the speakers
The speakers are located 3.72 m apart on a vertical pole.

Step 3: Analyze interference patterns
When two waves interfere constructively, their crests align with each other, resulting in regions of increased intensity. When they interfere destructively, their crests align with the troughs, reducing the sound intensity.

In this case, as the man walks towards the lower speaker perpendicularly to the pole, the distance he travels will cause a phase difference between the waves from the two speakers. At certain points, this phase difference will result in destructive interference and a minimum in sound intensity.

The condition for destructive interference is when the path difference between the two waves is equal to an odd multiple of half wavelengths.

Step 4: Estimate the distance of the first minimum
To find the distance at which the first minimum occurs, we need to determine the path difference that corresponds to one-half wavelength.

Path difference = λ/2

Now, we can set up the equation to find the distance (D) to the man at the first minimum:
D = √((3.72/2)^2 + (Path difference)^2)

Plugging in the values, we get:
D = √((1.86)^2 + (1.918/2)^2)
≈ √(3.4596 + 0.4623)
≈ √3.9219
≈ 1.98 m

Therefore, the man would be approximately 1.98 m away from the pole at the first moment he hears a minimum in sound intensity.

First compute the sound wavelength,

L = Vsound/frequency = 330/172 = 1.919 m

Require that the difference between the distances to the speakers be one-half wavelength, so that destructive interference occurs.

If R is the distance from observer to lower speaker and d is the speaker separation,
sqrt(R^2 + d^2) -R = L/2

Use the fact that d/R <<1, so that
sqrt (R^2 + d^2) = R sqrt(1 + (d/R)^2]
= R (1 + (1/2)(d/R)2 + ...)

sqrt(R^2 + d^2) -R = (1/2) d^2/R = L/2
R = d^2/L = 7.2 m