Question

Two loudspeakers are located 3.85 m apart on an outdoor stage. A listener is 16.2 m from one and 17.3 m from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range (20 Hz to 20 kHz). (a) What is the lowest frequency fmin,1 that gives minimum signal (destructive interference) at the listener's location? By what number must fmin, 1 be multiplied to get (b) the second lowest frequency fmin,2 that gives minimum signal and (c) the third lowest frequency fmin,3 that gives minimum signal? (d) What is the lowest frequency fmax,1 that gives maximum signal (constructive interference) at the listener's location? By what number must fmax,1 be multiplied to get (e) the second lowest frequency fmax, 2 that gives maximum signal and (f) the third lowest frequency fmax,3 that gives maximum signal? (Take the speed of sound to be 343 m/s.) (a) Number (b) Number (c) Number (d) Number (e) Number (f) Number Click if you would like to Show Work for this question: Units The number of significant digits is set to 3; the tolerance is +/-296 Units Units Units Units Open Show Work

Answer 1

Answer:

For destructive interference,

Minimum frequency f_min,n = [(2n-1)v] / 2$\Delta$L where v is the sound velocity = 343 m/s and $\Delta$L is the distance between the two loudspeakers.

Here, $\Delta$L = 17.3 m - 16.2 m = 1.1 m, therefore

f_min,n = (n - 1/2) (v/$\Delta$L) = (n-1/2) (343 m/s)/(1.1 m) = (n-1/2)(311.81 Hz)       

f_min,n = (n-1/2)(311.81 Hz)

(a) The lowest frequency that gives destructive interference is (n = 1)

f_min,1 = (1 - 1/2) (311.81 Hz) = 155.90 Hz.

(b) The second lowest frequency that gives destructive interference is (n = 2)

f_min,2 = (2-1/2)(311.81 Hz) = 467.72 Hz = 3 f_min,1.

So the factor is 3.

(c) The third lowest frequency that gives destructive interference is (n = 3)

f_min,3 = (3-1/2)(311.81 Hz) = 779.52 Hz = 5f_min,1.

So the factor is 5.

For constructive interference f_min,n = nv/$\Delta$L = n (311.8 Hz)

(d) The lowest frequency that gives constructive interference is (n=1).

f_min,1 = 1 (311.81 Hz) = 311.81 Hz.

(e) The second lowest frequency that gives constructive interference is (n=2).

f_min,2 = 2(311.81 Hz) = 623.62 Hz.

(f) The third lowest frequency that gives constructive interference is (n=3).

f_min,3 = 3 (311.81 Hz) = 935.43 Hz.