Answer
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Hint: In order to answer this first we know about the fundamental frequency .In a vibrating object the lowest resonant frequency is known as fundamental frequency.
Complete step-by-step solution:
From this problem we get,
\[v\Rightarrow 350m/s\],
\[l\Rightarrow 42.10\times {{10}^{-2}}m\]
\[d\Rightarrow 3.5cm\Rightarrow 3.5\times {{10}^{-2}}m\]
Here fundamental frequency at one end is given by ,
\[n=\dfrac{V}{4L}\]
\[\Rightarrow \]\[n=\dfrac{350}{4\times 42.10\times {{10}^{-2}}}\]
\[\Rightarrow n=202.78Hz\]
Again the \[Pth\] overtone , \[\Rightarrow np=(2p+1)n\]
Now , the frequency of \[Pth\] overtone is , \[\Rightarrow {{n}_{5}}=(2\times 5+1)202.78\]
\[\Rightarrow {{n}_{5}}=2230.59Hz\]
Therefore , the frequency of \[Pth\] overtone of a vibrating air column is \[2230.59Hz\].
Note: Reverberation is the inclination of a framework to vibrate with expanding amplitudes at certain frequencies of excitation. The resonator may have a crucial recurrence and quite a few sounds.
Complete step-by-step solution:
From this problem we get,
\[v\Rightarrow 350m/s\],
\[l\Rightarrow 42.10\times {{10}^{-2}}m\]
\[d\Rightarrow 3.5cm\Rightarrow 3.5\times {{10}^{-2}}m\]
Here fundamental frequency at one end is given by ,
\[n=\dfrac{V}{4L}\]
\[\Rightarrow \]\[n=\dfrac{350}{4\times 42.10\times {{10}^{-2}}}\]
\[\Rightarrow n=202.78Hz\]
Again the \[Pth\] overtone , \[\Rightarrow np=(2p+1)n\]
Now , the frequency of \[Pth\] overtone is , \[\Rightarrow {{n}_{5}}=(2\times 5+1)202.78\]
\[\Rightarrow {{n}_{5}}=2230.59Hz\]
Therefore , the frequency of \[Pth\] overtone of a vibrating air column is \[2230.59Hz\].
Note: Reverberation is the inclination of a framework to vibrate with expanding amplitudes at certain frequencies of excitation. The resonator may have a crucial recurrence and quite a few sounds.
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