physics help please

Two speakers are driven by a common oscillator at 870 Hz and face each other at a distance of 1.20 m. Locate the points along a line joining the two speakers where relative minima of pressure amplitude would be expected. (Use v = 343 m/s. Choose one speaker as the origin and give your answers in order of increasing distance from this speaker.) im not sure how to go about this problem.

Answers

This is an example of an interference pattern due to sound waves generated from the two speakers. To find the points of minimum pressure amplitude, the distance from one speaker to the other (1.20 m) must be divided by the wavelength associated with a frequency of 870 Hz. To calculate the wavelength, use the equation: \lambda = \frac{v}{f} where \lambda is the wavelength, v is the speed of sound, and f is the frequency of the oscillator. In this case, v = 343 m/s and f = 870 Hz, so the wavelength associated with the oscillator is: \lambda = \frac{343 \; m/s}{870 \, Hz} \approx 0.395 \; m To find the points of minimum pressure amplitude, divide the distance between the two speakers (1.20 m) by the wavelength (0.395 m). The result is 3.03, which means that the points of minimum pressure amplitude will be located 3.03 wavelengths away from the origin speaker. This equivalent to 1.20 m / 0.395 m = 3.03, so the points of minimum pressure amplitude will be located at 1.20 m / 3.03 = 0.398 m, 0.799 m, and 1.20 m away from the origin speaker. Therefore, the points of minimum pressure amplitude are located at 0.398 m, 0

Answered by bennettbrian

We have mentors from

Contact support