Question

A pipe closed at one end produces a fundamental note of $412\text{Hz}$. It is cut into two pieces of equal length. The fundamental notes produced by the two pieces are

• A. $824\text{Hz},1648\text{Hz}$
• B. $412\text{Hz},824\text{Hz}$
• C. $206\text{Hz},412\text{Hz}$
• D. $206\text{Hz},824\text{Hz}$

Answer 1

The correct option is A $824\text{Hz},1648\text{Hz}$

$f=\frac{v}{4l}=412\text{Hz}$

$f_1=\frac{v}{2\left(\frac{l}{2}\right)}=\frac{v}{l}=4\times 412=1648\text{Hz}$
$f_2=\frac{v}{4\left(\frac{l}{2}\right)}=\frac{v}{2l}=2\times 412=824\text{Hz}$