Destructive interference problem

[vkrock]

Homework Statement

Speaker A and B are separated by 5.00m. A listener, C, stands a distance directly in front of speaker B. What is the largest possible distance between speaker B and the listener so that he observes destructive interference? The two speakers vibrate in phase and play identical 125Hz tones, the speed of sound is 343m/s.

The figure is a right triangle with the distance from A to C (AC) as the hypt.

Homework Equations

|Path Diff| = AC - BC

The Attempt at a Solution

I know that for destructive interference, the path difference needs to be n($$\lambda$$/2). The problem is I am not sure how to set up an equation that solves for the maximum distance, rather than just one distance that gives the destructive interference. Can anyone help me figure out how to set this up properly to find a maximum distance?

 

[anathema]

Thank you for your question. To solve for the maximum distance between speaker B and the listener for destructive interference, we can use the equation you provided, |Path Diff| = AC - BC, and set it equal to n(\lambda/2). This will give us the condition for destructive interference. We can then rearrange the equation to solve for the maximum distance, AC, as follows:

AC = BC + n(\lambda/2)

Since we want to find the largest possible distance, we can let n = 1, which will give us the first destructive interference condition. We can then plug in the given values for BC (5.00m), the wavelength (\lambda = v/f = 343m/s / 125Hz = 2.74m), and n = 1, to solve for the maximum distance, AC. This gives us:

AC = 5.00m + (1)(2.74m/2) = 6.87m

Therefore, the largest possible distance between speaker B and the listener for destructive interference is 6.87m. I hope this helps answer your question. Let me know if you have any further inquiries.

 

[Moetasim]

I would approach this problem by first understanding the concept of destructive interference. Destructive interference occurs when two waves with the same frequency and amplitude meet and their crests and troughs line up, causing them to cancel each other out. In this case, the two speakers A and B are producing identical 125Hz tones, and the listener C is standing directly in front of speaker B.

To find the maximum distance between speaker B and the listener C for destructive interference to occur, we need to consider the path difference between the two speakers. The path difference is the difference in distance that the waves travel to reach the listener C. In this case, the path difference needs to be a multiple of half the wavelength (nλ/2) for destructive interference to occur.

Now, let's consider the right triangle formed by the three points A, B, and C. The hypotenuse of this triangle is the distance from A to C (AC), which is the total distance traveled by the sound waves from speaker A to speaker B to reach the listener C. The path difference is then equal to AC - BC, where BC is the distance from speaker B to the listener C.

Using the equation for the path difference, we can set up an equation to find the maximum distance between speaker B and the listener C:

|Path Diff| = AC - BC = n(λ/2)

Since we know the speed of sound (343m/s) and the frequency (125Hz), we can calculate the wavelength (λ) using the formula λ = v/f, where v is the speed of sound and f is the frequency.

Substituting the values, we get:

|Path Diff| = AC - BC = n(343/125)/2

To find the maximum distance, we need to find the value of n that gives the largest possible path difference. Since n can be any integer value, we can set n = 1 to get the maximum value.

Therefore, the maximum distance between speaker B and the listener C for destructive interference to occur is:

BC = AC - n(λ/2) = 5.00m - (1)(343/125)/2 = 5.00m - 1.37m = 3.63m

In conclusion, the maximum distance between speaker B and the listener C for destructive interference to occur is 3.63m. Any distance