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course: 2 × ? Solnwrap ? ? P1 11 S 181??? Diverging × i Reflectir × 1 Converg × L Determin "S Algebraic × A Facebook x ? (--> d ? Secure | https://coursespaces.uvíc.ca/pluginfile.php/1 325607/mod resource/content/0/P111S18M2Soln.pdf Bay P111S18M2Soln.pdf () M. Laidlaw P111 Spring 2018 4 A sound source which emits a frequency of 510Hz is placed dj where d -6.0m and a second sound source is placed at at the origin. The two sources emit sound in phase. A person stands at location ri to monitor the total sound, at distances di and d2 from the two microphones. The speed of sound in air is 340m This is illustrated in the figure below: d2 11:17 AM Type here to search 0? 4/13/2018

Explanation / Answer

Wavelength= speed of sound/frequency= 340/510= 2/3

Here destructive interference is required,so, the difference in distance travelled by sound, which is d1-d2, should be a multiple of half of wavelength but shouldn't be a multiple of the wavelength . So, d1-d2= a multiple of 1/3

When d2=3.4, d1=(6^2+3.4^2)^1/2= 6.9 ,and d1-d2= 6.9-3.4= 3.5 not a multiple of 1/3

Similarly, at d2= 3.1m, d1= (6^2+3.1^2)^1/2= 6.75 and

d1-d2= 3.65 , not a multiple of 1/3

Similarly, at d2= 2.5m, d1= (6^2+2.5^2)^1/2= 6.5 and

d1-d2= 6.5-2.5=4 ,

Also, 4/(1/3)=12,so it is a multiple of 1/3.

So, next value of x for which there is destructive interference is at x = (c)2.5m

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