Question

Two closed organ pipes of length 100 cm and 101 cm produce 16 beats in 20 seconds fundamental frequency formula in open when each pipe is sounded in its fundamental frequency mode to calculate the velocity of sound.
A) 303 m/s
B) 332 m/s
C) 323.2 m/s
D) 300 m/s

Answer 1

Hint: In the given question both pipes are open ended and producing sound at their fundamental frequency. Relate it with the formula of an open ended pipe when length and velocity are given and which is resonating in \[{N^{th}}\] harmonic . And we know that the number of beats produced is different between two frequencies.

Complete step by step solution:
Given an open ended pipe and formula for frequency \[f\] when pipe is resonating over \[{N^{th}}\] harmonic is:
\[f = \frac{{nv}}{{4l}}\]
Where:
\[f\] = frequency
\[n\] = \[{N^{th}}\] harmonic
\[v\] = velocity of sound
\[l\] = length of the pipe
Formula for number of beats is :
Number of beats = \[|{f_{}} - {f_2}|\]
Now according to question:
As 16 bests are produced in 20 seconds then
Number of beats per second is = \[\frac{{16}}{{20}} = \frac{4}{5}\]
\[|{f_{}} - {f_2}| = \frac{{16}}{{20}} = \frac{4}{5}\]
\[|{f_{}} - {f_2}| = \frac{v}{{4{l_{}}}} - \frac{v}{{4{l_2}}}\]
\[|{f_{}} - {f_2}| = \frac{v}{4}(\frac{1}{{{l_1}}} - \frac{1}{{{l_2}}})\]
\[{l_1} = 100cm = 1m\]
\[{l_2} = 101cm = 1.01m\]
\[|{f_{}} - {f_2}| = \frac{v}{4}(\frac{1}{1} - \frac{1}{{1.01}})\]
\[\frac{v}{4}(\frac{1}{1} - \frac{1}{{1.01}}) = \frac{4}{5}\]
\[v = \frac{{1.01 \times 16}}{{5 \times 0.01}}\]
\[v = 323.2m/s\]

Hence, option C is correct.

Note: In open end pipes The longest standing wave in a tube of length L with two open ends has displacement antinodes (pressure nodes) at both ends. It is called the fundamental or first harmonic. The next longest standing wave in a tube of length L with two open ends is the second harmonic. It also has displacement antinodes at each end.