Question

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound

Answer 1

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, $d = 1.8 \ m$

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point $D$, the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

$AD-BD = \frac{\lambda}{2}$

$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$

$\lambda = 2 \times 0.4985$

$\lambda = 0.99714 \ m$

Thus the frequency is :

$f=\frac{v}{\lambda}$

$f=\frac{340}{0.99714}$

$f=340.9744$ Hz