(a) What is the lowest frequency that givesminimum signal (destructive interfere

Question

(a) What is the lowest frequency that givesminimum signal (destructive interference) at the listener'sear?
1 Hz

(b) What is the second lowest frequency that gives minimumsignal?
2 Hz

(c) What is the third lowest frequency that gives minimumsignal?
3 Hz

(d) What is the lowest frequency that gives maximum signal(constructive interference) at the listener's ear?
4 Hz

(e) What is the second lowest frequency that gives maximumsignal?
5 Hz

(f) What is the third lowest frequency that gives maximumsignal?
6 Hz

Explanation / Answer

   v = 343 Hz    in the given problem let L1 bethe distance 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 thespeakers    the phase difference at the listner willbe     = 2 (L2 -L1) /    where is the wavelength    for minimum intensity at the listner weget     = (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 minimumsignal is n = 0    fmin,1 = ......... Hz (b)    the second lowest frequency is n= 1    fmin,2 = ......... Hz (c)    the third lowest frequency isn = 2    fmin,3 = ......... Hz    for maximum intensity at the listner = 2 n where n is a positive integer    fmin,2 = ......... Hz (c)    the third lowest frequency isn = 2    fmin,3 = ......... Hz    fmin,3 = ......... Hz     = (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 ........for the 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 theothers    20,000 / (........ Hz)= .........    so n must range from 1 to ........for the 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 theothers    fmax,1 = ......... Hz    in the similar manner solve for theothers

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