In Fig. 17-35, two loudspeakers, separated by a distance of d 1 = 2.14 m, are in
ID: 1726619 • Letter: I
Question
In Fig. 17-35, two loudspeakers, separated by a distance ofd1 = 2.14 m, are in phase. Assume theamplitudes of the sound from the speakers are approximately thesame at the position of a listener, who is d2 =4.54 m directly in front of one of the speakers. Consider theaudible range for normal hearing, 20 Hz to 20 kHz.(a) What is the lowest frequency that gives theminimum signal (destructive interference) at the listener's ear?(b) What is the lowest frequency that gives themaximum signal (constructive interference) at the listener's ear?(Take the speed of sound to be 343 m/s.)
Fig. 17-35
Problem 17.
Explanation / Answer
v = 343 Hz
in the given problem let L1 be thedistance from the closer speaker to the listner
the distance from the other speaker to thelistner will be
L2 = (L12+ d2)
where d is the distance between the speakers
the phase difference at the listner will be
= 2 (L2 - L1)/
where is the wavelength
for minimum intensity at the listner we get
= (2 n + 1)
where n is an integer so that
= 2 (L2 - L1) / (2 n+ 1)
the frequency will be
f = v /
= (2 n + 1) v /2 ((L12 + d2)-L1)
= (2 n + 1) .......Hz
now we can see that
20,000 / 343 = 58.3
so (2n + 1) must range from 0 to 57 for thefrequency to be in the audible range
(a)
the lowest frequency that gives minimum signal isn = 0
fmin,1 = ......... Hz
(b)
the second lowest frequency is n= 1
fmin,2 = ......... Hz
(c)
the third lowest frequency is n= 2
fmin,3 = ......... Hz
for maximum intensity at the listner = 2 n where n is a positive integer
= (1 / n)((L12 + d2)- L1)and
f = v /
= n v / ((L12+ d2)- L1)
= n (........ Hz)
now we can see that
20,000 / (........ Hz) = .........
so n must range from 1 to ........ forthe frequency to be in the audible range
(d)
the lowest frequency that gives maximumsignal is n = 1
fmax,1 = ......... Hz
in the similar manner solve for the others