A) 10 cm
B) 8 cm
C) 12 cm
D) 5 cm
Correct Answer: D
Solution :
[d]: In an open pipe, the fundamental frequency is \[\frac{v}{2{{L}_{O}}}\]and all harmonics are present. Third overtone means fourth harmonic. \[\therefore \]Frequency of third overtone \[{{\upsilon }_{4}}=\frac{4v}{2{{L}_{O}}}\]where v is the velocity of sound and \[{{L}_{O}}\]is the length of the open pipe. In a closed pipe, the fundamental frequency is \[\frac{v}{4{{L}_{C}}}\] and only odd harmonics are present. Second overtone mean fifth harmonic \[\therefore \]Frequency of second overtone \[{{\upsilon }_{5}}=\frac{5v}{4{{L}_{C}}}\]where \[{{L}_{C}}\] is a length of a closed pipe. As\[{{\upsilon }_{4}}={{\upsilon }_{5}}\] \[\frac{4v}{2{{L}_{O}}}=\frac{5v}{4{{L}_{C}}}\]or\[{{L}_{C}}=\frac{10}{16}{{L}_{O}}=\frac{10}{16}\times 8cm=5cm\]
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