Question

Two speakers both emit sound of frequency 320 Hz, and are in phase. A receiver sits 2.3 m from one speaker, and 2.9 m from the other. What is the phase difference between the two sounds detected by the receiver?

a. 32.3 λλ
b. 0.60 λλ
c. 0.089 λλ
d. 0.29 λλ
e. 0.56 λλ

Answer 1

The phase difference between the two sounds detected by the receiver is approximately 0.564 wavelengths.

Step-by-step explanation:

When two sound waves are in phase, it means that their crests and troughs line up with each other. In this scenario, the two speakers both emit sound waves of frequency 320 Hz, and are in phase. A receiver is located 2.3 m from one speaker and 2.9 m from the other. To find the phase difference between the two sounds detected by the receiver, we can calculate the path difference between the two speakers and the receiver.

The path difference is given by the equation:

Path Difference = Distance from one speaker - Distance from the other speaker

Path Difference = 2.9 m - 2.3 m = 0.6 m

The phase difference is calculated by dividing the path difference by the wavelength of the sound wave:

Phase Difference = Path Difference / (Wavelength)

Since the frequency of the sound wave is given as 320 Hz, and the speed of sound is approximately 340 m/s, we can calculate the wavelength as:

Wavelength = Speed of sound / Frequency

Wavelength = 340 m/s / 320 Hz = 1.0625 m

Now, we can calculate the phase difference:

Phase Difference = 0.6 m / 1.0625 m = 0.564 (approximately)

Therefore, the phase difference between the two sounds detected by the receiver is approximately 0.564 wavelengths.