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1. Two loudspeakers are placed 4.5 m apart. They produce the same sounds, in step, at a frequency of 829 Hz. Constructive interference is observed at point P, which is 5.30 m from one loudspeaker and a distance x from the other. Which is a possible value of x? Assume the speed of sound is 340 m/

a.6.60

b. 6.77

c. 6.80

d. 6.88

e. 6.53

2. Two loudspeakers are placed 6.0 m apart. They produce the same sounds, in step, across a frequency range of 252 Hz to 665 Hz. Point P is located 5.10 m from one loudspeaker and 3.60 m from the other. What frequency of sound from the two speakers will produce constructive interference at point P? Assume the speed of sound is 344 m/s.

a. 631
b. 401
c. 459
d. 573
c. 516

Respuesta :

MacNCheese0816

What was the answer? I go to Connexus too and I’m struggling in physics.

#1

speed of sound = 340 m/s

frequency = 829 Hz

wavelength of sound = speed / frequency

[tex]\lambda = \frac{340}{829} = 0.41 m[/tex]

Now for constructive interference we know that

path difference of two waves = integral multiple of wavelength

[tex]x - 5.30 = N\lambda[/tex]

[tex]x = 5.30 + N\lambda[/tex]

if N = 3

[tex] x = 6.53 m[/tex]

#2

speed of sound = 344 m/s

frequency range = 252 Hz - 665 Hz

wavelength of sound = speed / frequency

[tex]\lambda = \frac{344}{f}[/tex]

Now for constructive interference we know that

path difference of two waves = integral multiple of wavelength

[tex]5.10 - 3.60 = N\lambda[/tex]

[tex]N\lambda = 1.5[/tex]

Now we will have

[tex]\lambda = \frac{v}{f}[/tex]

[tex]\frac{1.5}{N} = \frac{344}{f}[/tex]

by rearranging above terms we have

[tex]f = \frac{N(344)}{1.5} = 229.3N[/tex]

[tex]f = 459 Hz[/tex]