Question

A pipe closed at one end produces a fundamental note of 412 Hz. It is cut into two pieces of equal length. The fundamental frequencies produced by the two pieces are :

• A. 206 Hz, 412 Hz
• B. 824 Hz, 1648 Hz
• C. 412 Hz, 824 Hz
• D. 206 Hz, 824 Hz

Answer 1

Answer: (B)

824 Hz, 1648 Hz

Length of the pipe initially be $L$.
Fundamental frequency in closed pipe ${\nu }_o^{\prime }=\frac{v}{4L}=412Hz$
After cutting, one pipe becomes open at both ends and then other remains as closed end at one end type.
Length of both the pipes $L_f=\frac{L}{2}$
Fundamental frequency of pipe open at both ends ${\nu }_o=\frac{v}{2L_f}$
$\therefore$ ${\nu }_o=\frac{v}{2\left(L/2\right)}=\frac{v}{L}=4\times 412=1648Hz$
Fundamental frequency of pipe closed at one end ${\nu }_o^{\prime \prime }=\frac{v}{4L_f}$
$\therefore$ ${\nu }_o=\frac{v}{4\left(L/2\right)}=\frac{v}{2L}=2\times 412=824Hz$