A tube open at only one end is cut into two tubes of non equal lengths. The piece open at both ends has of fundamental frequency of 450 Hz and other has fundamental frequency of 675 Hz. What is the 1st overtone frequency of the original tube

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Solution

for tube with both ends open $l_{1} = \frac{λ}{2}$
for tube with only one end open $l_{2} = \frac{λ}{4}$

$f_{1} = \frac{v}{λ} = \frac{v}{2 l_{1}}$
$f_{2} = \frac{v}{4 l_{2}}$
length of the original open tube $l_{1} + l_{2}$
therefore its frequency is $n = \frac{v}{4 (l_{1} + l_{2})} = \frac{v}{4 (\frac{v}{2 f_{1}} + \frac{v}{4 f_{2}})} = \frac{f_{1} f_{2}}{2 f_{2} + f_{1}}$
$n = \frac{450 \times 675}{2 \times 675 + 450} = 168.75 H z$
first overtone for original tube is $3 n_{1} = 2 \times 168.75 = 506.25 H z$

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A tube open at only one end is cut into two tubes of non equal lengths. The piece open at both ends has of fundamental frequency of 450 Hz and other has fundamental frequency of 675 Hz. What is the 1st overtone frequency of the original tube

for tube with both ends open l1=λ2for tube with only one end open l2=λ4f1=vλ=v2l1f2=v4l2length of the original open tube l1+l2therefore its frequency is n=v4(l1+l2)=v4(v2f1+v4f2)=f1f22f2+f1n=450×6752×675+450=168.75Hzfirst overtone for original tube is 3n1=2×168.75=506.25Hz