The first overtone in a closed pipe has a fre | vedclass

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(D) For a closed pipe of length $l$,the fundamental frequency is $f_0 = \frac{v}{4l}$. The first overtone is the next harmonic,which is $f_1 = \frac{3

Vedclass · Accepted Answer

None of the above

Vedclass · Answer

(D) For a closed pipe of length $l$,the fundamental frequency is $f_0 = \frac{v}{4l}$. The first overtone is the next harmonic,which is $f_1 = \frac{3v}{4l}$.For an open pipe of the same length $l$,the fundamental frequency is $f_{open,0} = \frac{v}{2l}$.The first overtone of an open pipe is $f_{open,1} = \frac{2v}{2l} = \frac{v}{l}$.Comparing $f_1 = \frac{3v}{4l}$ with the options:- Option $A$: $f_{open,0} = \frac{v}{2l} = \frac{2v}{4l} 
eq \frac{3v}{4l}$.- Option $B$: $2 	imes f_{open,0} = 2 	imes \frac{v}{2l} = \frac{v}{l} = \frac{4v}{4l} 
eq \frac{3v}{4l}$.- Option $C$: $f_{open,1} = \frac{v}{l} = \frac{4v}{4l} 
eq \frac{3v}{4l}$.Therefore,none of the given options are correct.

The first overtone in a closed pipe has a frequency:

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