Question

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when speaker 2 is at the origin and speaker 1 is at x = 0.540 m . If speaker 1 is slowly moved forward, the sound intensity decreases and then increases, reaching another maximum when speaker 1 is at x = 0.870 m .

What is the frequency of the sound? Assume velocity of sound is 340m/s.
What is the phase difference between the speakers?

Answer 1

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

Answer 2

Frequency of the sound = f = 1,030.3 Hz

Phase difference = Φ = 229.09°

Details required for solving the sound frequency & the phase difference:

A listener in front of both speakers hears a maximum sound intensity when speaker 2 is at the origin and speaker 1 is at x = 0.540 m. And,  speaker 1 is at x = 0.870 m .

Calculation of sound frequency & difference of phase:

The frequency of the sound is

$f = v \div \lambda$

λ = Xfin - Xini

= 0.870 - 0.540 = 0.33

So,

$f = 340 \div 0.33$

f = 1,030.3 Hz

Now the phase difference is

$\phi = (2\pi\div \lambda)\times (Xini - \lambda) = (2\pi\div0.33)\times (0.540-0.33) = 19.04\times 0.21 = 3.9984\phi = 3.9984 rad \times (360^{\circ}\div 2\pi rad)$

Φ = 229.09°

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