Question

A tube open at only one end is cut into two tubes of non-equal lengths. The piece open at both the ends has a fundamental frequency of $450\text{Hz}$ and the other piece has a fundamental frequency of $675\text{Hz}$. What is the first overtone frequency of the original tube?

• A. $506.25\text{Hz}$
• B. $606.25\text{Hz}$
• C. $406.25\text{Hz}$
• D. $306.25\text{Hz}$

Answer 1

The correct option is A $506.25\text{Hz}$

For the piece of tube open at both ends,
Fundamental frequency,
$f_{o1}=450\text{Hz}$
$\Rightarrow \frac{v}{2l_1}=450$
$\Rightarrow l_1=\frac{v}{900}$ ............(1)
For the other piece of tube open at one end,
Fundamental frequency,
$f_{o2}=675\text{Hz}$
$\Rightarrow \frac{v}{4l_2}=675$
$\Rightarrow l_2=\frac{v}{2700}$ ............(2)
So, length of original tube,
$l=l_1+l_2$
$\Rightarrow l=\frac{v}{900}+\frac{v}{2700}=\frac{v}{675}$
[from (1) and (2)]
Now, first overtone of original tube $(n=3)$
$f_1=\frac{3v}{4l}=\frac{3v}{4\times \frac{v}{675}}=506.25\text{Hz}$