Both drawings show the same square, each of which has a side of length L=0.75 m. An observer O is stationed at one corner of each square. Two loudspeakers are locate at corners of the square, as in either drawing 1 or drawing 2. The speakers produce the same single-frequency tone in either drawing and are in phase. The speed of sound is 343 m/s. Find the single smallest frequency that will produce both constructive interference in drawing 1 and destructive interference in drawing 2.

In drawing 1, the two speakers are at opposite corners; that is, a line segment between them would be the hypotenuse of a right triangle, and the distance can be calculated to be 1.06m. The observer is at a third corner, 0.75 m from both speakers.
In drawing 2, the two speakers are adjacent to each other. The distance between them is 0.75 m. The observer is 0.75 m from one speaker, and 1.06 from the other.

To find the single smallest frequency that will produce both constructive interference in drawing 1 and destructive interference in drawing 2, we can use the concept of path difference.

Path difference is the difference in distance traveled by two waves from their sources to a particular point. For constructive interference, the path difference should be an integer multiple of the wavelength, while for destructive interference, the path difference should be an odd multiple of half the wavelength.

Let's start with drawing 1. The observer O is located at one corner of the square, which is 0.75 m away from both speakers. The distance between the two speakers is the diagonal of the square, which can be calculated as follows:

Diagonal = √(0.75^2 + 0.75^2) = √0.5625 + 0.5625 = √1.125 = 1.06 m

For constructive interference in drawing 1, the path difference should be an integer multiple of the wavelength. The path difference between the two speakers can be calculated by:

Path difference = Diagonal - Distance between the observer and one speaker
= 1.06 m - 0.75 m
= 0.31 m

Now, let's move on to drawing 2. In this case, the two speakers are adjacent to each other, with a distance of 0.75 m. The observer is located at a distance of 0.75 m from one speaker and 1.06 m from the other.

For destructive interference in drawing 2, the path difference should be an odd multiple of half the wavelength. The path difference between the two speakers can be calculated by:

Path difference = Distance between the observer and one speaker - Distance between the observer and the other speaker
= 0.75 m - 1.06 m
= -0.31 m

Since the path difference in drawing 1 is positive (0.31 m), and the path difference in drawing 2 is negative (-0.31 m), they are equal in magnitude. This means that the total path difference for the same frequency in both cases is 0.62 m.

To find the wavelength corresponding to this path difference, we can use the formula:

Path difference = wavelength * n, where n is an integer

Substituting the values:

0.62 m = wavelength * n

Since we want the smallest frequency, we want the smallest wavelength. This occurs when n = 1 (the smallest integer).

0.62 m = wavelength * 1
wavelength = 0.62 m

Finally, to find the frequency, we can use the formula:

Speed = wavelength * frequency

Rearranging the formula and substituting the values:

343 m/s = 0.62 m * frequency

frequency = 343 m/s / 0.62 m
frequency = 552.42 Hz

Therefore, the single smallest frequency that will produce both constructive interference in drawing 1 and destructive interference in drawing 2 is approximately 552.42 Hz.

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