Two speakers driven by the same force

[aaronb]

Homework Statement

Two loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 800 Hz.

(a) What is the phase difference between the two waves when they reach the observer? (Your answer should be between 0 and 2π.)

(b) What is the frequency closest to 800 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?

Homework Equations

delta r = (phi/2pi)lamda
frequency = velocity/lamd

The Attempt at a Solution

I found the dimensions of the triangle that is formed when the person stands in front of one speaker 4.00m away which is:
a=4,b=1,c=sqrt(17)

 

[Redbelly98]

You'll need to get the difference in the distances to each speaker.

 

[nlightner]

(a) The phase difference between the two waves can be calculated using the equation delta r = (phi/2pi)lamda, where delta r is the difference in distance traveled by the two waves, phi is the phase difference, and lamda is the wavelength. In this case, the wavelength can be calculated using the equation lamda = velocity/frequency, where velocity is the speed of sound and frequency is 800 Hz. So, lamda = (343 m/s)/800 Hz = 0.42875 m.

The difference in distance traveled by the two waves can be calculated using the Pythagorean theorem: delta r = sqrt(17) - 4 = 1.123 m.

Substituting these values into the first equation, we get: 1.123 m = (phi/2pi)0.42875 m

Solving for phi, we get: phi = 2.625 radians or approximately 150.38 degrees.

Therefore, the phase difference between the two waves when they reach the observer is 2.625 radians or 150.38 degrees.

(b) To find the frequency closest to 800 Hz to which the oscillator may be adjusted such that the observer hears minimal sound, we can use the equation frequency = velocity/lamda. We know that the velocity of sound is 343 m/s and the distance between the two speakers is 1 m. So, the wavelength for minimal sound can be calculated as: lamda = 343 m/s/1 m = 343 m.

Substituting this value into the equation for frequency, we get: frequency = 343 m/s/343 m = 1 Hz.

Therefore, the frequency closest to 800 Hz to which the oscillator may be adjusted such that the observer hears minimal sound is 1 Hz. This means that the two speakers would be producing waves that are completely out of phase, cancelling each other out and producing minimal sound for the listener.