A) 2
B) 6
C) 8
D) 7
Correct Answer: D
Solution :
Given, \[{{f}_{o}}-{{f}_{c}}=2\] ... (i) Frequency of fundamental mode for a closed organ pipe, \[{{f}_{c}}=\frac{v}{4{{L}_{c}}}\] Similarly frequency of fundamental mode for an open organ pipe, \[{{f}_{o}}=\frac{v}{2{{L}_{o}}}\] Given \[{{L}_{c}}={{L}_{o}}\] \[\Rightarrow \] \[{{f}_{o}}=2{{f}_{c}}\] ... (ii) From Eqs. (i) and (ii), we get \[{{f}_{o}}=4\,Hz\] and \[{{f}_{c}}=2\,Hz\] When the length of the open pipe is halved, its frequency of fundamental mode is \[f_{o}^{'}=\frac{v}{2\left[ \frac{{{L}_{o}}}{2} \right]}=2{{f}_{o}}=2\times 4\,Hz=8\,Hz\] When the length of the closed pipe is doubled, its frequency of fundamental mode is \[f_{o}^{'}=\frac{v}{4(2{{L}_{c}})}=\frac{1}{2}\,{{f}_{c}}=\frac{1}{2}\times 2=1\,Hz\] Hence, number of beats produced per second is \[f_{o}^{'}-f_{c}^{'}=8-1=7\]
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