Question

A person hums into the top of a well and finds that standing waves are established at frequencies of 110, 154, and 198 Hz. The frequency of 110 Hz is not necessarily the fundamental frequency. The speed of sound is 343 m/s. How deep is the well?

Answer 1

`d=3.9m`

Step-by-step explanation:

The well is closed at bottom and opened at top.So we will have vibration corresponding to n=1,3,5,7......

Let the frequency of nth harmonic vibration is fn=110Hz.Then we have:

`f_(n+2)=154Hz\\and\\f_(n+4)=198Hz`

The nth harmonic is related to the fundamental frequency f₁ as fn=nf₁. Similarly fn+2=(n+2)f₁

Then we have

`f_(n+2)-f_(n)=(n+2)f_(1)-nf_(1)=154-110=44Hz`

f₁ is common factor on left side.So we have

`(n+2-n)f_(1)=44Hz\\2f_(1)=44\\f_(1)=22Hz`

So the fundamental frequency of vibrations f₁=22Hz

Depth d of well is given by:

`d=(v)/(4f_(1))\\ d=(343m/s)/(4*22Hz)\\ d=3.9m`