Answer:
[tex]d=3.9m[/tex]
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:
[tex]f_{n+2}=154Hz\\and\\f_{n+4}=198Hz[/tex]
The nth harmonic is related to the fundamental frequency f₁ as fn=nf₁. Similarly fn+2=(n+2)f₁
Then we have
[tex]f_{n+2}-f_{n}=(n+2)f_{1}-nf_{1}=154-110=44Hz[/tex]
f₁ is common factor on left side.So we have
[tex](n+2-n)f_{1}=44Hz\\2f_{1}=44\\f_{1}=22Hz[/tex]
So the fundamental frequency of vibrations f₁=22Hz
Depth d of well is given by:
[tex]d=\frac{v}{4f_{1}}\\ d=\frac{343m/s}{4*22Hz}\\ d=3.9m[/tex]
