Two loudspeakers s1 and s2 are 3.0m apart, emir the same single-frequency time i

Question

Two loudspeakers s1 and s2 are 3.0m apart, emir the same single-frequency time in phase at the speakers. A listener L directly in front of speaker S1 notices that the intensity is a minimum when she is 4.9. From that speaker. What is the lowest possible frequency of the emitted tone, when the speed of sound in air is 340 m/s? Two loudspeakers s1 and s2 are 3.0m apart, emir the same single-frequency time in phase at the speakers. A listener L directly in front of speaker S1 notices that the intensity is a minimum when she is 4.9. From that speaker. What is the lowest possible frequency of the emitted tone, when the speed of sound in air is 340 m/s?

Explanation / Answer

from the given data,

path diffrence of the two sound waves, r2 - r1 = sqrt(4.9^2 + 3^2) - 4.9

= 0.8454 m

we know, for distructive intereference, r2 - r1 = (n/2)*lamda (n = 1,3,5,7...)

0.8454 = (n/2)*lamda

now use threlation, v = lamda*f

==> lamda = v/f

so,

0.8454 = (n/2)*v/f

minum value for n = 1,

0.8454 = (1/2)*v/f_min

so, f_min = (1/2)*340/0.8454

= 201 Hz <<<<<<<<<<----------------Answer

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