Two loudspeakers, 5.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.25 m . Assume the speed of sound is 340 m/s.

Required:
a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.21 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
c.

Question

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Respuesta :

AbsorbingMan AbsorbingMan · Answer 1 · 2021-07-23T05:31:43+02:00

Solution :

Let [tex]$d_1=\frac{5.5}{2}[/tex]

= 2.75 m

[tex]d_2 = 0.21 \ m[/tex]

And [tex]$d=|d_1-d_2|$[/tex]

[tex]$d=(d_1+d_2) - (d_1-d_2)$[/tex]

[tex]$d=(2.75+0.21) - (2.75-0.21)$[/tex]

[tex]$d = 2.96-2.54$[/tex]

[tex]d = 0.42 \ m[/tex]

a). At minimum,

[tex]$d=\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2d$[/tex]

= 2 x 0.42

= 0.84 m

Frequency, [tex]$\nu = \frac{v}{\lambda}$[/tex]

[tex]$=\frac{340}{0.84}$[/tex]

= 404.76 Hz

Therefore, the frequency of he sound, [tex]$\nu$[/tex] = 404.76 Hz

b). At maximum, λ = d = 0.42 m

Therefore, the frequency, [tex]$\nu = \frac{v}{\lambda}[/tex]

[tex]$=\frac{350}{0.42}$[/tex]

= 809.52 Hz