Question

Two loudspeakers in a plane are 2.0 m apart and in phase with each other. The speed of sound is 340 m/s. Assume the amplitude of the sound from each speaker is approximately the same at the position of a listener, who is 3.75 m directly in front of one of the speakers. a) (10pts) For what three lowest frequencies will there be a minimum signal (destructive interference)? b) (10pts) For what three lowest frequencies will there be a maximum signal (constructive interference)? rı = 3.75 m 2 & I = 2.00 m

Answer 1

a) DX = Y2 - 87 Axa Mint &? -Y, = UE75'+22 ) – 3:45 Dx = 0.5 For frequencies three consecutive lowest minima 89 = ", 39, 5m Ci do 2x0.5 atm FE * 340 = 340 Hz. so = 30 34= 21 DK d = 2x0.5 = 1/2 Īf340x3 = 1020 Hz

Tri> 0¢ 50 do Raxoys = F 340x 5 = 1700. Hz. Three lowest frequency for minima at listener f = ( 340, 1020, 10o) Hy by for maxima Ags 20, 471 ,617 for at m K d 20 = 27 ok to asm 340 = 680 Hz For so =640 un = 211 DK d = 0.25 = 1360 Hz 340. 0.25

for sp = 60 6rt = 217 Ak d do Diş If= 390x3 = 2040 Hz) 0.5. M4 Three lowest frequency for maxima f = 1680, 1360, 2040) Hz