Question

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.

Answer 1

Let
`$d_1=(5.5)/(2)`

= 2.75 m

`d_2 = 0.21 \ m`

And
`$d=|d_1-d_2|$`

`$d=(d_1+d_2) - (d_1-d_2)$`

`$d=(2.75+0.21) - (2.75-0.21)$`

`$d = 2.96-2.54$`

`d = 0.42 \ m`

a). At minimum,

`$d=(\lambda)/(2)$`

`$\lambda = 2d$`

= 2 x 0.42

= 0.84 m

Frequency,
`$\\u = (v)/(\lambda)$`

`$=(340)/(0.84)$`

= 404.76 Hz

Therefore, the frequency of he sound,
`$\\u$` = 404.76 Hz

b). At maximum, λ = d = 0.42 m

Therefore, the frequency,
`$\\u = (v)/(\lambda)`

`$=(350)/(0.42)$`

= 809.52 Hz