Question

Two closed organ pipes, when sounded simultaneously, gave $4$ beats per seconds. If a longer pipe has a length of $1\;m$. Then length of the shorter pipe will be ($v=300m/s$)
\[\begin{align}
  & A.185.5cm \\
 & B.94.9cm \\
 & C.90cm \\
 & D.80cm \\
\end{align}\]

Answer 1

Hint: The beat frequency or beats per second is the difference between frequencies of two nodes which interfere to produce beats. To find the length of the shorter pipe, we need to calculate the frequency of the shorter frequency first, that can be done from the given beat.

Formula used:
$B=\dfrac{v}{4L_{2}}-\dfrac{v}{4L_{1}}$

Complete step-by-step answer:
Beats are defined as the periodic repetition of fluctuating intensities of sound waves. This occurs when two sound waves of similar frequencies interfere with one another. It is characterised by waves whose amplitude varies at a regular rate. The beats oscillate to and fro between amplitude zero and maximum amplitude. The beat frequency or beats per second is the difference between frequencies of two nodes which interfere to produce beats.
We know that frequency of a node produced by the closed organ pipes of length $L$, when $v$ is the speed of air is given by $F=\dfrac{v}{\lambda}=\dfrac{v}{4L}$
Here, we have two closed pipes, $L_{1}=1m$ and $L_{2}$. Also given that, $v=300m/s$
Then the frequency of the pipes is given as $F_{1}=\dfrac{v}{4L_{1}}$ and $F_{2}=\dfrac{v}{4L_{2}}$
We know that beat is between frequencies of two nodes which interfere constructively or destructively, and here, $B=4$
Then, $B=F_{2}-F_{2}$
$\implies B=\dfrac{v}{4L_{2}}-\dfrac{v}{4L_{1}}$
$\implies B=\dfrac{v}{4}\left(\dfrac{1}{L_{2}}-\dfrac{1}{L_{1}}\right)$
Substituting the values we have,
 $\implies 4=\dfrac{300}{4}\left(\dfrac{1}{L_{2}}-\dfrac{1}{1}\right)$
$\implies \dfrac{16}{300}+1=\dfrac{1}{L_{2}}$
$\implies \dfrac{1}{L_{2}} =\dfrac{316}{300}$
$\implies L_{2}=\dfrac{300}{316}=0.949m$
Hence the answer is \[B.94.9cm\]

So, the correct answer is “Option B”.

Note: Beat is the periodic repetition of fluctuating intensities. The positive amplitude is called crest and the negative amplitude is called trough. A loud sound is heard when the waves interfere constructively. This happens when two crests or two troughs interfere. Similarly, no sound is heard when the waves interfere destructively. This happens when one crest and one trough interferes.