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Two loudspeakers are located 2.59 m apart on an outdoor stage. A listener is 21.6 m from one and 22.7 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, one that gives minimum signal (destructive interference) at the listener's location

1 Answer

5 votes

Answer:

Frequency
f_(min,1)=155.90\ Hz

Step-by-step explanation:

Given data:

The distance between the speakers, d = 2.59 m

The distance between the listeners, ΔL = 22.7 - 21.6 = 1.1 m

Now, For a destructive interference, we know that


(\Delta L)/(\lambda)=0.5,1.5,2.5,.........

where, λ = wavelength

thus,

frequency
f_(min,n)=((n-0.5)v)/(\Delta L)

where,

v = speed of sound = 343 m/s

for n = 1

we get

frequency
f_(min,1)=((1-0.5)* 343)/(1.1)

or

Frequency
f_(min,1)=155.90\ Hz

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