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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

Respuesta :

Answer:

Frequency [tex]f_{min,1}=155.90\ Hz[/tex]

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

[tex]\frac{\Delta L}{\lambda}=0.5,1.5,2.5,.........[/tex]

where, λ = wavelength

thus,

frequency [tex]f_{min,n}=\frac{(n-0.5)v}{\Delta L}[/tex]

where,

v = speed of sound = 343 m/s

for n = 1

we get

frequency [tex]f_{min,1}=\frac{(1-0.5)\times 343}{1.1}[/tex]

or

Frequency [tex]f_{min,1}=155.90\ Hz[/tex]

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