Two loudspeakers, 5.5 m apart and facing each other, play identical sounds of th

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 0.35 m . Assume the speed of sound is 340 m/s.

1)What is the frequency of the sound?

2)If the frequency is then increased while you remain 0.35 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Explanation / Answer

Calculates the speed of sound 340 m / s as if.
(a) L1+L2= 5.5m
Ld= |L1-L2|= (2.75+0.35)-(2.75-0.35)= 0.7m
Ld=/2
=2Ld =2*0.7= 1.4m
f=v/ =340/1.4= 242.86[Hz]
(b) 0.7=
f=v/ = 340/0.7= 485.71[Hz]

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