Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A person stands 5.0 m away from one speaker and 6.2 m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be 330 m/ s. A) 183 Hz B) 275 Hz C) 413 Hz D) 137 Hz E) 550 Hz

C) 413 Hz

Explanation:

For destructive interference, the path difference ΔL = (n + 1/2)λ where ΔL = L₂ - L₁ where L₁ = person's distance from one speaker (the closer one) = 5.0m and L₂ = person's distance from other speaker (the farther one) = 6.2 m and λ = wavelength = v/f where v = speed of sound = 330 m/s and f = frequency

So, ΔL = (n + 1/2)λ

L₂ - L₁  = (n + 1/2)v/f

f = (n + 1/2)v/(L₂ - L₁)

At the second lowest frequency that results in destructive interference at the point where the person is standing, n = 1.

So,

f = (1 + 1/2)v/(L₂ - L₁)

f = 3v/2(L₂ - L₁)

Substituting the values of the variables into the equation, we have

f = 3v/2(L₂ - L₁)

f = 3(330 m/s)/2(6.2 m - 5.0 m)

f = 3(330 m/s)/2(1.2 m)

f = 990 m/s ÷ 2.4 m)

f = 412.5 Hz

f ≅ 413 Hz

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be