Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 \rm Hz. You are 8.00 \rm m from speaker A. Take the speed of sound in air to be 344 \rm m/s.

What is the closest you can be to speaker B and be at a point of destructive interference?

To determine the closest distance to speaker B where destructive interference occurs, we need to find the path difference between the waves from speakers A and B.

The formula for path difference is:

Δx = r1 - r2

Where Δx is the path difference, r1 is the distance from the reference point (your position) to speaker A, and r2 is the distance from the reference point to speaker B.

In this case, you are given that you are 8.00 m from speaker A. Let's consider this as our reference point. So, r1 = 0 m.

To find the path difference, we need to calculate r2. The total distance from speaker A to B is the distance traveled by sound from speaker A to your position, plus the distance from your position to speaker B.

Calculating the distance traveled by sound from speaker A to your position:

Distance = speed × time

The time it takes for sound to travel from speaker A to your position can be calculated by dividing the distance by the speed of sound:

Time = Distance / Speed

Time = 8.00 m / 344 m/s

Time ≈ 0.0233 s

Now, to find the distance from your position to speaker B, we can use the speed of sound and the frequency of the waves emitted. The wave travels to your position, then from your position to speaker B. So, we are dealing with two times the path difference.

Path difference = 2 × (Speed × Time)

Path difference = 2 × (344 m/s × 0.0233 s)

Path difference ≈ 16.00 m

Now, we have the path difference (Δx = 16.00 m). For destructive interference to occur, the path difference should be equal to a whole number of wavelengths (nλ), where n is an integer (0, 1, 2, etc.) and λ is the wavelength.

Since the waves emitted by the speakers have the same frequency and are in phase, the wavelengths are the same.

Δx = nλ

16.00 m = n × (wavelength)

To find the closest distance you can be to speaker B with destructive interference, we need to find the largest n that satisfies the equation. The smallest value of n will give us the closest distance.

Let's find the wavelength first.

The formula for calculating the wavelength is:

Wavelength = (Speed of Sound) / (Frequency)

Wavelength = 344 m/s / 172 Hz

Wavelength ≈ 2.00 m

Now, substitute the values into the equation:

16.00 m = n × 2.00 m

n = 8

Therefore, the closest you can be to speaker B and be at a point of destructive interference is when the path difference is equal to 16.00 m, which corresponds to the eighth wavelength (n = 8).

In conclusion, you can be 16.00 meters away from speaker B for destructive interference to occur with speaker A.

draw the figure. draw a triangle short side from A, long side from B, other side A to B.

so the difference between the short side and the long side has to be an even multiple of 1/2 wavelength.

start with the wave equation
lambda= v/f
so long side-short side= n(v/2f)

do the degenerate triangle first, where short+long= AB (a point on AB)

nv/2f+nv/2f+v/2f=ab
or v/2f(2n+1)=ab
you know ab as 8m, so solve for n
then solve for short side length.

Now do the same thinking for some triangle .

Use the law of cosines to find short side, then minimize it with differential calculus.
You will find the answer, and discovering something magic about the angle from B, it gets minimized also.