A student sets up two speakers spaced 2.5 meters apart, each playing the same tone of a single frequency. She stands at point P, at an equal distance from each speaker and hears the tone playing loudly. She walks towards point Q, directly opposite one speaker, and when she gets there she find the tone to be extremely hard to hear (unlike the rest of the area between P and Q). She knows that the speakers emit the sound waves in phase (so that the compressions of the waves are emitted simultaneously from each speaker), and that the speed of sound in the room is 340 m/s. What is the frequency of the sound emitted from the speakers?

well, at a point q it must be 180 out of phase.

I am wondering what you mean directly opposite one speaker...

directly across from one of the speakers

To find the frequency of the sound emitted from the speakers, we need to consider the concept of constructive and destructive interference of sound waves.

Let's assume that the distance between the student (P) and each speaker is 'd'. As the student moves towards point Q, which is directly opposite one of the speakers, the sound waves emitted from the two speakers will reach the student with a slight time delay due to the difference in distance traveled by the sound waves.

When the waves combine at the student's position P, they undergo interference. Constructive interference occurs when the compressions of the waves from both speakers coincide, resulting in an amplified sound. Destructive interference, on the other hand, occurs when the compressions and rarefactions of the waves coincide, resulting in reduced or canceled sound.

We know that the speed of sound in the room is 340 m/s. Therefore, the time delay between the sound waves reaching the student from both speakers can be calculated as follows:

Time delay = (distance traveled by sound wave from one speaker - distance traveled by sound wave from the other speaker) / speed of sound

Given that the speakers are spaced 2.5 meters apart, the distance traveled by the sound wave from one speaker is (d + 2.5) and from the other speaker is (d - 2.5). The time delay can be represented as:

Time delay = [(d + 2.5) - (d - 2.5)] / 340

Simplifying this gives:

Time delay = (5 / 340)

Since the sound waves are known to be in phase, the time delay should be an integer multiple of the period of the sound wave, which is the reciprocal of the frequency.

Therefore, we can write:

Time delay = N * (1 / frequency)

where N is an integer (0, 1, 2, 3, ...)

Substituting the value of time delay, we have:

(5 / 340) = N * (1 / frequency)

To solve for the frequency, we rearrange the equation:

frequency = (1 / N) * (340 / 5)

Now, we need to find the value of N for which the frequency is the highest. Since the student hears the tone playing loudly at point P, we can assume that the interference at P is constructive, leading to the maximum amplitude of sound.

At point Q, however, the student finds the tone extremely hard to hear, indicating that destructive interference is occurring.

Destructive interference occurs when the path length difference between the two speakers is equal to an odd number of half-wavelengths. In other words:

(d + 2.5) - (d - 2.5) = (2n + 1) * (λ / 2)

Simplifying this gives:

5 = (2n + 1) * (λ / 2)

From this equation, we can conclude that the path length difference of 5 meters corresponds to one-half of a wavelength.

Given that the speed of sound is 340 m/s, we can calculate the wavelength:

wavelength = speed of sound / frequency

Substituting the value of speed of sound, we get:

(wavelength) = 340 / frequency

Since the path length difference of 5 meters corresponds to one-half of a wavelength, we have:

5 = (1 / 2) * (wavelength)

Substituting the value of wavelength, we can write:

5 = (1 / 2) * (340 / frequency)

Simplifying this equation gives:

frequency = (1 / 2) * (340 / 5)

Therefore, the frequency of the sound emitted from the speakers is:

frequency = 34 Hz