Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 3.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. Part A What is the lowest possible frequency of sound for which this is possible

1089.74 Hz

Explanation:

Using Pythagoras theorem, we can find the distance from the point to the second speaker.

Thus;

d2 = √(3² + 2.3²)

d2 = √(9 + 5.29)

d2 = √14.29

d2 = 3.78 m

Then, the path distance which is the extra distance travelled would be;

Δd = d2 - d1

Δd = 3.78 - 3

Δd = 0.78 m

Now, the destructive interference condition is given by the formula;

Δd = (m + ½)λ

λ is the wavelength

m is a non - negative integer.

In this case, m = 2

Thus;

0.78 = (2 + ½)λ

λ = 0.78/(2½)

λ = 0.312 m

Now the formula for frequency of a wave is given by;

f = v/λ

Where v is speed of sound.

Thus;

f = 340/0.312

f = 1089.74 Hz