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

The distance between these two points is 2.29 m

Step-by-step 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=(2\pi* d)/(\lambda)`

`\phi=(2\pi* d)/((v)/(f))`

Put the value into the formula

`5.1=(2\pi* d)/((340)/(120))`

`d=(5.1*2.833)/(2\pi)`

`d=2.29\ m`

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