Assignment #13 nmentProblemD-57238337 deb PHSX 114- Fall 2015 Resources segnment

Question

Assignment #13 nmentProblemD-57238337 deb PHSX 114- Fall 2015 Resources segnment E3 Problem 12.53 e provious 3 of 12n Problem 12.53 Part A What is the lowest frequency at which destructive inte rference will ccur at this point if the speakers are in phase? Two loudspeakers are 1,.70 m apart. A person stands 3.50 tm from ore speaker and 4.10 m from the other Value H Sulbmit Incorrect; Try Again; 3 attempts remaining Part B Calcuiale two other trequencses that atso result in destructiee interlerence at this point igive the noxt two highest). LetT- 20 C Enter your answers numerically separated by a comma ·214.4.357.3 Ha Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

The path difference between then two waves reaching the person will be ;

r1-r2 = 4.10-3.50 = 0.6 m

No destructive interference is given by ; r1-r2 = (2n+1)y/2

where y is the wavelength.

Also, y = c/f , where c is the speed of sound.

Thus, 0.6 = 340/ (2*f) ------------------------------------------------ since the velocity of sound is 340m/s

Thus, f = 283.33 hz.

For least value, we have taken n=0.

Part B:

For two other destructive interference;

n=1:

0.6 = (2+1)*340/2f

Thus, f = 850 hz.

Also, for n=2:

0.6= ((2*2)+1)*340/2f

Thus, f = 1416.66hz.

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