Question

Two loudspeakers, 3.5 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 0.15 m . Assume the speed of sound is 340 m/s .
a. What is the frequency of the sound?
b. If the frequency is then increased while you remain 0.15 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Answer 1

For the given question the answer to the first part is that frequency of sound will be 404.76 Hz. and answer to the second part will be that the initial frequency for which that place will have the highest level of sound intensity

Given, the speed of sound 340 m/s

a) L1 + L2 = 3.5 m

and Ld = | L1 - L2 | = ( 1.75 + 0.21 ) - (1.75 - 0.21 ) = 1.96 - 1.54 = 0.42 m

Ld = λ/2

λ = 2Ld = 2×0.42 = 0.84 m

and finally,

f = v/λ

f = 340/0.84

f = 404.76 hertz

Frequency came out to be 404.76 hertz in this case

b) For the first frequency

0.42 = λ

f = v/λ

f = 340 / 0.42

f = 809.52 Hertz

Frequency came out to be 809.52 Hertz in this case.

To conclude with we can say that the Frequency of the sound in case on came out to be 404.76 hertz which is approximately 405 Hz after applying all the concepts and calculations, in second case first frequency for which that location will be a maximum of sound intensity came out to be 809.52 Hertz after applying all the concepts and calculations.

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