Two loudspeakers, 4.5 m apart and facing each other, play identical sounds of th

Question

Two loudspeakers, 4.5 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 0.17 m . Assume the speed of sound is 340 m/s. Part A: What is the frequency of the sound? Part B: If the frequency is then increased while you remain 0.17 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?

Explanation / Answer

Given, speed of sound = 340 m / s

(a) L1+L2= 4.5 m
Ld= |L1-L2|= (2.25+0.17)-(2.25-0.17)= 2.42 - 2.08 = 0.34 m
Ld=/2
=2Ld =2*0.34= 0.68 m
f=v/ =340/0.68 = 500.00 Hz

(b) 0.34 =
f=v/ = 340/0.34 = 1000.00 Hz

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.