question_answer

Two closed organ pipes 100 cm and 101 cm long gives 16 beats in 20 s, when each pipe is sounded in its fundamental mode. Calculate the velocity of sound:

A) 303 ms-1                              

B) 332 ms-1

C) 323.2 ms-1           

D) 300 ms-1

Correct Answer: C

Solution :

Key Idea: Number of beats per second = difference of the frequencies of sound sources. Let \[l\] be the length of pipe and v is the velocity of sound,     then     the frequency of note emitted from the pipe is \[n=\frac{v}{4l}\] Number of beats in 1 s \[=\frac{16}{20}=\frac{4}{5}\] For a closed organ pipe \[x={{n}_{1}}-{{n}_{2}}=\frac{v}{4}\left( \frac{1}{{{l}_{1}}}-\frac{1}{{{l}_{2}}} \right)\] \[\frac{4}{5}=\frac{v}{4}\left( \frac{1}{1}-\frac{1}{1.01} \right)=\frac{0.01v}{4\times 1\times 1.01}\] \[\Rightarrow \]\[0.01\times 5v=16\times 1.01\] \[\therefore \]   \[v=\frac{16\times 1.01}{0.05}=323.2\,m/s\]