The frequency of the third harmonic if the pipe is closed at one end is = 102.93 Hz.
To find the frequency of the third harmonic if the pipe is closed at one end, we use the formula,
[tex]f_{n} =nf_{1}[/tex]=[tex]n\frac{v}{4L}[/tex] , in this case n= 1,3,5,7,9.....
Here we are given,
L= The length of the pipe = 4.137 m.
v= speed of the sound at 16.164°C .
[tex]v_{0}[/tex]= speed of the sound at 0°C.=331 m/s
Before finding the frequency we have to find the speed of the sound at 16.164°C. To find v, we are using the formula,
[tex]v=v_{0 } \sqrt{\frac{T_{2} }{T_{1} } }[/tex]=[tex]331\times \sqrt{\frac{(273+16.164) }{273 }[/tex]=340.65 m/s.
We have to find the frequency of the third harmonic, so according to the question, n=5
Now we put the known values in the first equation, we can find that
[tex]f_{n}=n\frac{v}{4L}[/tex]
Or,[tex]f_{n} = 5\times\frac{340.65}{4\times 4.137}[/tex]=102.93 Hz.
Therefore, from the above calculation we can conclude that if the pipe is closed at one end the frequency of the third harmonic is 102.93 Hz.
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