# A loudspeaker at the origin emits a 130 Hz tone on a day when the speed of the sound is 340 m/s....

## Question:

A loudspeaker at the origin emits a 130 Hz tone on a day when the speed of the sound is 340 m/s. The phase difference between two points on the x-axis is 5.4 rad.

What is the distance between these two points?

## Phase and Path Difference

The phase difference between two points on a wave is related to the path difference between these points, by the relation, {eq}\Delta \phi = \dfrac{2 \pi}{\lambda} \Delta p {/eq}, where

• {eq}\lambda {/eq} is the wavelength of the wave
• {eq}\Delta p {/eq} is the path difference
• {eq}\Delta \phi {/eq} is the phase difference

Given :

• The frequency of the wave is, {eq}f = 130 \ Hz {/eq}
• The speed of the sound wave is, {eq}v = 340 \ m/s {/eq}
• The phase difference between two given points is, {eq}\Delta \phi = 5.4 \ rad {/eq}

Let the distance between the given points be, {eq}\Delta p {/eq}

The wave length of the wave produced by the speaker will be,

{eq}\begin{align*} \lambda &= \dfrac{v}{f} \\ &= \dfrac{340}{130} \\ &= 2.6154 \ m \end{align*} {/eq}

Applying the relation between phase difference and path difference, we get,

{eq}\begin{align*} \Delta \phi &= \dfrac{2 \pi}{\lambda} \Delta p \\ \Delta p &= \dfrac{\lambda }{2 \pi } \Delta \phi \\ &= \dfrac{( 2.6154)}{2 \pi}(5.4) \\ &= \boxed{ 2.25 \ m } \end{align*} {/eq}