Question

Paisley stands in front of two speakers as shown which emit the same tone, in phase. She is 7.00 m from speaker A and 5.00 m fromB speaker B. a. What are the two lowest frequencies for which she hears: i. a loud sound? [171.5 Hz, 343 Hz] ii. a quiet sound [85.8 Hz, 257 Hz] d-5.00m A-7.00m b. The frequency of the speakers is then changed to 686 Hz Does Paisley hear a loud sound or a quiet sound? Support your answer. loud] Question 2 A set of identical speakers emit identical sounds of frequency 478 Hz Amy Audiophile stands 3.70 m in front of the left speaker and hears no sound What is the minimum possible distance between Amy and the speaker to her right? [0.469 ml 3.70 m Question 3 Suzy Soundlover, stands 2.00 m in front of two speakers which are separated by 3.00 m. a. If the speed of sound in air is 343 m/s, determine the lowest frequency for which Suzy i. hears a loud sound [114 Hz] ii. hears no sound [57.2 Hz 3.00 2.00 b. If Suzy moves farther away from the speakers, describe what happens to the sound she hears.

Answer 1

ANS 1

a. i. Paisley will hear loud sound if the path difference
a. i. Paisley will hear loud sound if the path difference p = (7.00 - 5.00)m = 2.00m between the sounds received from both the speakers is integral multiple of wavelength lambda of sound. lambda = n u/v implies v = nu/p v = 343 m/s and for lowest frequencies n = 1, 2 for n = 1 v = 1*343/2.00 = 171.5Hz for n = 2 v = 2*343/2.00 = 343Hz ii. for a quiet sound p = (n + 1/2) lambda, where n = 0, 1, 2,... P = (n + 1/2) u/v v = (n + 1/2) u/p for n = 0 v = (0 + 1/2) 343/2.00 Hz v = 85.8Hz for n = 1 v = (1 + 1/2) 343/2.00 Hz = 257Hz b. Changed frequency v = 686Hz Therefore wavelength, lambda = 343/686 m = 0.5m Therefore, p/lambda = 2.00/0.5 = 4 i.e path difference in this case is integral multiple of lambda. Therefore Paisley will hear loud sound in this case.
between the sounds received from both the speakers is integral multiple of wavelength of sound.

, where v is the velocity and is frequency of sound.

i.e. , where n = 1,2,3,.....

v = 343 m/s and for lowest frequencies n = 1,2

for n=1

for n=2

ii. for a quiet sound   where n = 0,1,2,......

for n = 0

for n = 1

b. Changed frequency

Therefore wavelength, $lambda=rac{343}{686}m=0.5m$

Therefore,   $rac{p}{lambda}=rac{2.00}{0.5}=4$ i.e path difference in this case is integral multiple of . Therefore Paisley will hear loud sound in this case.

ANS 2

Wavelength of sound in this case

For no sound path difference should be odd integral multiple of  

i.e.

maximum possible value of n for which x is positive is 4

ANS 3

a. In this case the path difference between the two sounds received by her is p=3.00 m

i. Minimum frequency for loud sound is:

ii. for no sound

b. As she moves further away from the speakers she will hear loud sound which will keep on decreasing as she moves away till she hear no sound after covering a distance of  , if she continue moving sound will again increase and becomes maximum after covering a further distance equal to half the wavelength and this will continue to happen i.e she will hear alternate loud and quiet sound as she keep moving further away from the speakers.