Answer to Question #87775 in Physics for Sridhar

Question

Two identical loud speakers are placed facing each other, horizontally separated by a distance of 2 m. Each of them emit sound waves of wavelength 80 cm, driven by the same oscillator.
If a listener standing midway between the two speakers walks a distance ‘x’ towards one of the speakers, when the first minimum in the sound intensity is listened, the distance ‘x’ is
(1) 20 cm (2) 10 cm
(3) –10 cm (4) 15 cm

Answer 1

Every Source (speakers) generates sin waves. Following superposition principle,

$$a(t) = A\sin(kx_1-\omega t) + A \sin(k x_2 -\omega t) = 2 A \sin(k\frac{x_1+x_2}{2}-\omega t)\cos(k\frac{x_1-x_2}{2})$$

where k is wavenumber, $k = \frac{2\pi}{\lambda},\, \lambda = 80 \, \text{cm}$ , $a(t)$ is sound intensity function of time t, and x_1, x_2 - distances to speakers.

Therefore, the first minimum whould be

$$a(t)=0 \Rightarrow \cos\left(k\frac{x_1-x_2}{2}\right)=0 \Rightarrow k\frac{x_1-x_2}{2}=\frac{\pi}{2}$$

$$x_1-x_2 = \lambda / 2 = 40 \, \text{cm}$$

Obviously that $x_1+x_2=L = 2\, \text{m}$, hence displacement

$$x = L/2-x_1 = \lambda / 4 = 20 \, \text{cm}$$

Answer: (1) 20 cm