Question

2. Two sources of sound are separated by
a distance 4 m. They both emit sound
with the same amplitude and frequency
(330 Hz), but they are 180° out of phase.
At what points between the two sources,
will the sound intensity be maximum?
[Ans: +0.25, 0.75, +1.25 and
1.75 m from the point at the center]​

Answer 1

Given that,

Distance = 4m

Amplitude1 = Amplitude2 = ?

Frequency1 = Frequency2 = 330Hz

Angle between them = 180°

Velocity of sound = 330 m/s

Let's calculate wave length of the sound wave,

we know the equation,

• velocity = frequency × wavelength

• 330m/s = 330 Hz × λ

• λ = 330 m/s / 330 Hz

• λ = 330 m/s/330s

• λ = 1m

Therefore, wavelength is 1m.

There are four antinode between the two sources of the sound. These measurments are given below.

The positions of antinode are,

• First wave = ±(V/4) = V = 1m = ±(1m/4) = ±0.25m.

• Second wave = ±(V/2+V/4) = ±(1m/2+1m/4) = ±(0.50m+0.25m) = ± 0.75m.

• Third wave = ±(V+V/4) = ±(1m+1m/4) = ±(1m+0.25m) = 1.25m.

• Third wave = ±(3V/2+V/4) = ± (3×1m/2+1m/4) = ±(1.5m+0.25m) = ±1.75m.

Therefore, ±0.25m, ±0.75m, ±1.25m, ±1.75m from the center.

{ Here ± because the sources of sound may be left or right from the center }

Hope this helps you.

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