Two loudspeakers, d1 = 2.00 m, are in phase. Assume the amplitudes of the sound from the speakers are approximately the same at the position of a listener, who is d2 = 3.75 m directly in front of one of the speakers. Consider the audible range for normal hearing, 20 Hz to 20 kHz. (a) What is the lowest frequency that gives the minimum signal (destructive interference) at the listener's ear? (b) What is the lowest frequency that gives the maximum signal (constructive interference) at the listener's ear?

Question

contestada

Respuesta :

rejkjavik rejkjavik · Answer 1 · 2019-08-06T10:48:14+02:00

Answer:

Lowest frequency that gives the minimum signal is 343 Hz

lowet frequency for maximum signal 686 Hz

Explanation:

D_1 = 2.00 m

D _2 =3.75 m

Δx = [tex]\sqrt {D_1^2 +D_2^2} - D_2[/tex]

=[tex]\sqrt {2.00^2 +3.75^2} - 3.75[/tex]

= 0.5 m

for minimum distructive interference, the path difference is

Δx = [tex]( n+ \frac{1}{2} \lambda[/tex]

for minimum n = 0

[tex]\lambda = 2\Delta x[/tex]

= 2*0.5 = 1 m

lowest frequency f = [tex]\frac[c}{\lambda}[/tex]

where c is speed of sound in air 343 m/s

[tex]f = \frac{343}{1} = 343 Hz[/tex]

b) lowet frequency for maximum signal

for minimum CONSTRUCTIVE interference, the path difference is

Δx = [tex]n\lambda[/tex]

for minimum n = 1

[tex]\lambda = Delta x[/tex]

= 0.5 m

lowest frequency f =[tex] \frac[c}{\lambda}[/tex]

f = [tex]\frac{343}{0.5} = 686 Hz[/tex]