Question
In Fig. 17-36, two speakers separated by distance d! = 2.00m are in phase. Assume the amplitudes of the sound waves from the speakers are approximately the same at the listener's ear at distance d2 = 3.75 m directly in front of one speaker. Consider the full audible range for normal hearing, 20 Hz to Fig. 17-36 Problem 21.
20 kHz. (a) What is the lowest
frequency fmin,! that gives minimum signal (destructive interference)
at the listener's ear? By what number must fmin I be multiplied
to get (b) the second lowest frequency fmin,2 that 'gives minimum
signal and (c) the third lowest frequency fmin,3 that gives
minimum signal? (d) What is the lowest frequency fmaxl that gives
maximum signal (constructive interference) at the list'ener's ear?
By what number must fmax,) be multiplied to get (e) the second
lowest frequency fmax,2 that gives maximum signal and (f) the third
lowest frequency fmax,3 that gives maximum signal?
Explanation / Answer
Given that , Distnce between the speakers , d1 = 2 m distance of the listner from the bottom spealers ,d2 = 3.75m Norml distance : d = [22 + 3.752] = 4.25 m (a)The minimum signal occurs at the destrcutive intereference thus,for the condition phase difference between the sounds from the speakers = (2m+ 1) = 2/ L where :L = d - d2 1/ = (2m+1)/2(d-d2) = (2m+ 1)/2(4.25-3.75) = 2m+ 1 = 1/ (2m+ 1) but ,frquency of the sound is given by the formual f = V / Here , f = 20 K Hz = 20000 Hz , speed of the sound ,V = 343 m/s thus, f = V (2m+ 1) 2m + 1 = f/V so, m = 28.65 thus, m values running form m= 0 to 28 thus, minimum value of m is 0 now ,frquency (Min) = f min = V (2m + 1) = V = 343 Hz (b)for ,m=1 , f ' = 1029 Hz factor : x = f '/ f = 1029 /343 = 3 Multiplication factor is : 3 (c) for ,m = 2 f ''= 1715 Hz multiplication factor :x = f''/ f = 1715 /343 = 5 (d) For constructive interference condition: 1/ = m/(d-d2) 1/ = m/(4.25-3.75) = 1/2m but , f = V/ = 2m V m = f/2V = 20000 /2(343) = 29.1 thus, m values are running form : m = 1 to 29 for ,m= 1 ,f = 2 m V = 2x1x343 = 686 Hz (e) for m= 2 , f = 686x2 = 1372 Hz Hence , Multiflication factor :m = 2 (f) for ,m= 3 , f = 2 (3) (343) = 2058 Hz Hence , Multiplication factor : m = 3