# Two in-phase loudspeakers emit identical 1000 Hz sound waves along the x-axis. (a) What distance...

## Question:

Two in-phase loudspeakers emit identical {eq}1000 \ Hz {/eq} sound waves along the {eq}x {/eq} -axis.

(a) What distance should one speaker be placed behind the other for the 'no sound' effect?

## Sound Waves

The sound waves from two sources having their wavelength interfering constructively then the resultant wave gets amplified, however, but when their wavelengths interfere destructively, the resultant wave would be having less or no amplitude based on their path difference.

## Answer and Explanation:

The speed of the sound wave in normal temperature and pressure is 343 m/s (approx.).

Therefore, the wavelength for the wave having frequency 1000 Hz is,

{eq}\lambda = \dfrac{v_{(wave)}}{f}\\ where,\\ \lambda = wavelength\;of\;the\;wave\\ f = frequency\\ v_{(wave)} = velocity\;of\;the\;wave\\ \Rightarrow\\ \lambda = \dfrac{343}{1000}\\ \lambda = 0.343\;m {/eq}

The condition of path difference for destructive interference is,

{eq}Path\;difference = \dfrac{n \lambda}{2}\\ where,\\ n = 1,3,5,7..... {/eq}

For distance to be minimum the n = 1. Therefore we get,

{eq}Path\;difference = \dfrac{(1)(0.343)}{2}\\ Path\;difference = 0.172\;m\;or\; 17.2\;cm {/eq} 