question_answer

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

A)  \[~824\text{ }Hz,\text{ }1648\text{ }Hz\]

B)  \[~412\text{ }Hz,\text{ }824\text{ }Hz\]

C)  \[~206\text{ }Hz,\text{ }412\text{ }Hz\]

D)  \[216\text{ }Hz,\text{ }824\text{ }Hz\]

Correct Answer: A

Solution :

Key Idea: When the dosed pipe is cut one open and one closed pipe are formed. When pipe is closed at one end                                 \[n=\frac{v}{4l}\]                 Given,   \[n=412\,Hz\]                                 \[412=\frac{v}{4l}\]                                         ?(i) When pipe is cut into two equal halves then length of each is \[\frac{l}{2}.\] \[{{n}_{1}}=\frac{v}{4l}\]      (closed pipe) \[{{n}_{2}}=\frac{v}{2l}\]         (open pipe) Where \[l=\frac{l}{2}\]                                 \[{{n}_{1}}=\frac{v}{4\left( \frac{l}{2} \right)}\] Putting \[v=1648l\]from Eq. (i), we get \[{{n}_{1}}=\frac{1648l}{2l}=824\,Hz\] \[{{n}_{2}}=\frac{v}{2\left( \frac{l}{2} \right)}=\frac{v}{l}\,=\frac{1648\,l}{l}\,=1648\,Hz\] Note: A closed pipe produces only odd harmonics while an open pipe produces both even and odd harmonics.