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.You need to login to perform this action.
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