Two small stereo speakers A and B that are 1.40m apart are sending out sound of wavelength 33.5cm in all directions and all in phase. A person at point P starts out equidistant from both speakers and walks (see the figure (Figure 1) ) so that he is always 1.50m from speaker B. Limit your solutions to the cases where x≤1.50m.

1. Give the five positive values of x for which the sound will be maximally reinforced. Begin at the central maximum (m = 0.)

Enter your answers in decreasing order separating them with commas.

2. Give four positive values of x for which the sound will be cancelled?

Begin with the first minimum that follows the central maximum. Enter your answers in decreasing order separating them with commas.

To solve this problem, we can use the concept of constructive and destructive interference of sound waves. Constructive interference occurs when the waves from two sources (speakers) are in phase and reinforce each other, resulting in a maximum amplitude. Destructive interference occurs when the waves are out of phase and cancel each other out.

1. Finding the values of x for maximum reinforcement (constructive interference):

Here, we are given the distance between the speakers (1.40m) and the wavelength of the sound (33.5cm). We need to find the values of x that maximize the reinforcement.

The condition for constructive interference is that the path difference between the waves from the two sources must be a whole number multiple of the wavelength.

Path difference = distance traveled by wave from speaker B - distance traveled by wave from speaker A

We know that the total distance traveled by the wave from speaker B is always 1.50m. So, the path difference can be calculated as:
Path difference = 1.50m - x

For constructive interference, the path difference should be an integer multiple of the wavelength:
Path difference = n * wavelength, where n is an integer.

Substituting the values, we get:
1.50m - x = n * 33.5cm, where n is an integer.

Now, find the values of x that satisfy the equation. Since we are looking for positive values of x, we'll start with n = 0, which corresponds to the central maximum.

When n = 0:
1.50m - x = 0
x = 1.50m

When n = 1:
1.50m - x = 33.5cm
x = 1.50m - 33.5cm = 1.1665m

When n = 2:
1.50m - x = 2 * 33.5cm
x = 1.50m - 2 * 33.5cm = 0.8335m

When n = 3:
1.50m - x = 3 * 33.5cm
x = 1.50m - 3 * 33.5cm = 0.5m

When n = 4:
1.50m - x = 4 * 33.5cm
x = 1.50m - 4 * 33.5cm = 0.1665m

Therefore, the five positive values of x for maximum reinforcement are: 1.50m, 1.1665m, 0.8335m, 0.5m, and 0.1665m.

2. Finding the values of x for sound cancellation (destructive interference):

The condition for destructive interference is that the path difference should be a half-integer multiple of the wavelength:
Path difference = (n + 0.5) * wavelength, where n is an integer.

Using the same formula as before (1.50m - x = (n + 0.5) * 33.5cm), we can find the values of x for which the sound will be canceled.

Starting with n = 0 (the first minimum after the central maximum):
1.50m - x = 0.5 * 33.5cm
x = 1.50m - 0.5 * 33.5cm = 1.3335m

When n = 1:
1.50m - x = 1.5 * 33.5cm
x = 1.50m - 1.5 * 33.5cm = 1.0m

When n = 2:
1.50m - x = 2.5 * 33.5cm
x = 1.50m - 2.5 * 33.5cm = 0.6665m

When n = 3:
1.50m - x = 3.5 * 33.5cm
x = 1.50m - 3.5 * 33.5cm = 0.3335m

Therefore, the four positive values of x for sound cancellation are: 1.3335m, 1.0m, 0.6665m, and 0.3335m.

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