Two identical speakers separated by 14.9 m are driven by the same oscillation with a frequency of 26.3 Hz. x y b (x, y) b 1 st A 2 nd 14.9 m Find the distance from the first speaker to a receiver at A if it records a minimum in sound intensity. Take the speed of sound to be 344.1 m/s

Explanation:

Given that,

Distance between two speakers

d=14.9m

Frequency =26.3Hz

Speed of sound V=344.1m/s

Formula to calculate path difference is given as

d1-d2=λ/2

Where d1= distance from the first speaker

d2=distance from the second speaker

So let the minimum intensity be at x

Distance of the first speaker from A is x m

Distance of the second speaker from A is (14.9-x)m

The wavelength is given as

V=fλ

λ=V/f

λ=344.1/26.3

λ=13.1m

Using the equation.

d1-d2=λ/2

x-(14.9-x)=13.1/2

x-14.9+x=6.54

2x=6.54+14.9

2x=21.44

x=21.44/2

x=10.72m

The minimum sound intensity can be record at a distance of 10.72m from first speaker

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west