Question

One closed organ pipe has a length of 2.24 meters. When a second pipe is played at the same time, a beat note with a frequency of 1.1 hertz is heard. By how much is the second pipe too long? Include units in your answer.

Answer 1

0.07 m

Step-by-step explanation

First, we have to find the frequency that the first pipe would play,

`\lambda=4L=4\cdot2.24m=8.96m`

The frequency is,

`f=(v)/(\lambda)=(343m/s)/(8.96m)=38.28125Hz`

The frequency played by the second pipe is,

`f=38.25125Hz-1.1Hz=37.18125Hz`

The wavelength of this note is,

`\lambda=(v)/(f)=(343m/s)/(37.18125s^(-1))\approx9.2251m`

So the length of the second pipe is,

`L=(\lambda)/(4)=(9.2251m)/(4)\approx2.31m`

The difference between the pipes' length is,

`2.31m-2.24m=0.07m`

Hence, the second pipe is 0.07 meters too long