A) 90 Hz and 84 Hz
B) 100 Hz and 106 Hz
C) 96 Hz and 90 Hz
D) 206 Hz and 200 Hz
Correct Answer: C
Solution :
We know that \[f=\frac{v}{2l}\] Fundamental frequency of an open organ pipe). Number of beats heard = 6 \[\Rightarrow \] \[{{f}_{P}}-{{f}_{Q}}=6\] or \[\frac{v}{2(0.3)}\frac{v}{2(0.32)}=6\] or \[v=\frac{2\times 0.3\times 0.32\times 6}{0.32-0.3}\] Now, \[{{f}_{P}}=\frac{v}{2(0.3)}=\frac{0.32\times 6}{0.02}=96\,Hz\] and \[{{f}_{Q}}=\frac{v}{2(0.32)}=\frac{0.3\times 6}{0.02}=90\,Hz\]
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