A) 57.2 cm and 42.9cm
B) 57. 2 cm and 45.8 cm
C) 42.9 cm and 32.2 cm
D) 42.9 cm and 34.0 cm
Correct Answer: A
Solution :
Given that fundamental frequency of open organ pipe \[(A)=300\,Hz\] \[{{f}_{1}}=300\,Hz=\frac{v}{2{{l}_{1}}}\] ... (i) Frequency of second harmonic-of open organ pipe \[(A)=\frac{v}{{{l}_{1}}}\] and frequency of third harmonic of closed organ pipe \[=\frac{3v}{4{{l}_{2}}}\] Now according to question \[\frac{v}{{{l}_{1}}}=\frac{3v}{4{{l}_{2}}}\] \[\Rightarrow \] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\left( \frac{4}{3} \right)\] ... (ii) Form Eq. (i), we get \[300=\frac{343}{2{{l}_{1}}}\] \[{{l}_{1}}=\frac{343}{600}=52.16=57.2\,\,cm\] Now from Eq. (ii), we get \[{{l}_{2}}=\frac{3}{4}\times {{l}_{1}}=\frac{3}{4}\times 57.2\] = 42.9 cm
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