Two identical loudspeakers 10.0 m apart are driven by the same oscillator with a

Question

Two identical loudspeakers 10.0 m apart are driven by the same oscillator with a frequency of f= 21.5 Hz (figure below) in an area where the speed of sound is 344 m/s. Show that a receiver at point A records a minimum in sound intensity from the two speakers. If the receiver is moved in the plane of the speakers, show that the path it should take so that the intensity remains at a minimum is along the hyperbola 9x^2- 16y^2 = 144 (shown in red-brown in the figure above) Can the receiver remain at a minimum and move very far away from the two sources? Yes No If so, determine the limiting form of the path it must take. If not, explain how far it can go.

Explanation / Answer

a)d = 10 m ; f = 21.5 Hz ; vs = 344 m/s

we know that,

lambda = v/f = 344/21.5 = 16 m

first speaker from is A is da = 9 m and distance from B is db = 10 - 9 = 1 m

da - db = 9 - 1 = 8 m

we see that, lambda/2 = (da - db)

Hence, A is the point of minimum intensity

b) distance of first one from A

da = sqrt [(x + 5)^2 + y^2] ; db = sqrt [(x - 5)^2 + y^2]

in order to have minimum intensity at A

d1 - d2 = (2k+a)lambda/2

sqrt [(x + 5)^2 + y^2] - sqrt [(x - 5)^2 + y^2] = lambda/d

sqrt [(x + 5)^2 + y^2] = lambda/2 + sqrt [(x - 5)^2 + y^2]

squaring both sides and solving further we get

[(x + 5)^2 + y^2] = lambda^2/4 + [(x - 5)^2 + y^2] + l[(x - 5)^2 + y^2]

x^2 + 10x + 25 = lambda^2/4 - 10x + 25 + lambda x sqrt [(x - 5)^2 + y^2]

20 x - lambda^2/4 = lambda x sqrt [ (x - 5)^2 + y^2]

again squaring both sides and simplifyinf we get

400x^2 + lambda^4/16 - 10 x lambda^2 = lambda^2 [ (x - 5)^2 + y^2]

9x^2 - 16y^2 = 144

c)Yes

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