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 Hz, 1648 Hz

B)  412 Hz, 824 Hz

C)  206 Hz, 412 Hz

D)  216 Hz, 824 Hz

Correct Answer: A

Solution :

Key Idea: When the closed 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 \[l=\frac{l}{2}\]                 \[\therefore \] \[{{n}_{1}}=\frac{v}{4\left( \frac{l}{2} \right)}\] Putting \[v=1648\,l\] from Eq. (i), we get                 \[{{n}_{1}}=\frac{1648\,l}{2\,l}=824\,\,Hz\]                 \[{{n}_{2}}=\frac{v}{2\left( \frac{l}{2} \right)}=\frac{v}{l}=\frac{1648\,\,l}{l}=1648\,\,Hz\] Note: A closed pipe produces only odd harmonics while an open pipe produces both even and odd harmonics.