Question

A loudspeaker at the origin emits a 120Hz Hz tone on a day when the speed of sound is 340 m/sm/s. The phase difference between two points on the x-axis is 5.1 rad.What is the distance between these two points?

Answer 1

Answer:

The distance between these two points is 2.29 m

Explanation:

Given that,

Frequency = 120 hz

Speed of sound = 340 m/s

Phase difference = 5.1 rad

We need to calculate the distance between these two points

Using formula of phase difference

$\phi=\dfrac{2\pi\times d}{\lambda}$

$\phi=\dfrac{2\pi\times d}{\dfrac{v}{f}}$

Put the value into the formula

$5.1=\dfrac{2\pi\times d}{\dfrac{340}{120}}$

$d=\dfrac{5.1\times2.833}{2\pi}$

$d=2.29\ m$

Hence, The distance between these two points is 2.29 m.