Sound from point source S gets to the detector D via two different paths: path 1

Question

Sound from point source S gets to the detector D via two different paths: path 1 and path 2. Path 1 is parallel to the wall, and has length L. The reflection at the wall causes a “phase shift” in the sound wave, effectively lengthening path 2 by ½ wavelength. Distance d is 0.250 m, and distance L is 1.00 m. The highest audible frequency is 20,000 Hz. (a) Determine the equation for the frequencies, n f , as a numerical multiple of n, for which the sound waves along the two paths arrive at D exactly in phase (b) Determine the value of n for the two highest audible frequencies. (c) Determine the two highest audible frequencies. (The speed of sound is 344 m/s.)

Explanation / Answer

a)

path differnce, r2-r1=n*lambda

2*sqrt(d^2+(l/2)^2)-l=n*lambda

2*sqrt((0.25)^2+(0.5)^2)-1=n*lambda

0.118=n*lambda

0.118=n*(v/fn)

fn=n*v/(0.118)

fn=n*(344/0.118)

fn=n*2915.25 Hz

b)

and

fmax=20.000 Hz

for n=1 , f1=1*2915.25 =2915.25Hz

for n=2 , f1=2*2915.25 =5830.51Hz

for n=3 , f1=3*2915.25 =8745.76Hz

for n=4 , f4=4*2915.25 =11661.0 Hz

for n=5, f5=5*2915.25=14576.27 Hz

for n=6   f5=6*2915.25=17491.5 Hz

for n=5,6

c)

for n=5, f5=5*2915.25=14576.27 Hz

for n=6   f5=6*2915.25=17491.5 Hz

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