Two small loudspeakers are separated by a distance of d = 5.5 cm, as shown in th
ID: 1570219 • Letter: T
Question
Two small loudspeakers are separated by a distance of d = 5.5 cm, as shown in the figure below. The speakers are driven in phase with a sine wave signal of frequency 20 kHz. A small microphone is placed a distance L = 1.60 m away from the speakers on the axis running through the middle of the two speakers, and the microphone is then moved perpendicular to the axis. Where does the microphone record the first minimum and the first maximum of the interference pattern from the speakers? The speed of sound in air is 343 m/s. y_1st min = 37.4 cm y_1st max = cmExplanation / Answer
d = 5.5 cm , f =20 kHz, L =1.6 m,
v =343 m/s
for constructive interference condition: dsin(theta) = m*lamda , m =0,1,2 ..
for destructive interference condition: dsin(theta) = (m+0.5)*lamda , m =0,1,2 ..
v = f*lamda
lamda = v/f = 343/20000 = 0.02144 m
(theta)max =arcsin(lamda/d)
(theta)max =arcsin(0.02144/0.055)
(theta)max = 23 degrees
ymax = Ltan(theta)max = 1.6*tan(23) =0.6792 m
ymax = 67.92 cm
(theta)min =arcsin(lamda/2d)
(theta)min =arcsin(0.02144/(2*0.055))
(theta)max = 11.24 degrees
ymin = Ltan(theta)min = 1.6*tan(11.24) =0.318 m
ymin = 31.8 cm
ymax = 67.92 cm
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.