A tube, open at one end and closed at the other end, is cut into three shorter,

Question

A tube, open at one end and closed at the other end, is cut into three shorter, unequal length pieces. The piece that is closed at one end has a fundamental frequency of 225 Hz, while the pieces that are open at both ends have fundamental frequencies of 335 Hz and 445 Hz, respectively. What was the fundamental frequency of the original tube?

Please write clear and show all the steps. Thanks.

A tube, open at one end and closed at the other end, is cut into three shorter, unequal length pieces. The piece that is closed at one end has a fundamental frequency of 225 Hz, while the pieces that are open at both ends have fundamental frequencies of 335 Hz and 445 Hz, respectively. What was the fundamental frequency of the original tube?

Explanation / Answer

see, here

we have

fundamanetal frequency given by f=(1/2l)* velocity for open tube

and

f=(1/4l) for closed

so let lenght of two open tubes be La, Lb , and closed be Lc

so

we have

225=(1/4Lc)*v

335=(1/2La)*v

445=(1/2Lb)*v

v = velocity of sound

so

for that full tube

f=(1/4(La+Lb+Lc))*v

putting values of La,Lb,Lc we have

f=1/(4*(1/(4*225)+1/(2*335)+1/(2*445)))=67.0736 hz

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