Question

A pipe closed at one end and open at the other end resonates with a sound of frequency $135Hz$ and also with $165Hz$, but not any other frequency intermediate between these two, then, the frequency of the fundamental note of the pipe is:
$\begin{align}
  & \text{A}\text{. }60Hz \\
 & \text{B}\text{. }7.5Hz\text{ } \\
 & \text{C}\text{. }30Hz \\
 & \text{D}\text{. }15Hz \\
\end{align}$

Answer 1

- Hint: For calculating the frequency of overtones in a closed organ pipe, we will use the expression for relation between the frequency of an overtone and the fundamental frequency. Closed organ pipes only produce odd harmonics. All even harmonics are absent in a closed organ pipe.

Formula used:
For closed organ pipe,
$\text{Overtone = }\left( 2n+1 \right)\times \text{fundamental frequency}$

Complete step-by-step solution
Organ pipes are the type of musical instruments which are used to produce musical sound by the process of blowing air into the pipe. Organ pipes are of two types: Closed organ pipes, which are closed at one end and Open organ pipes, which are open at both the ends.
For closed organ pipe,
${{f}_{o}}=\dfrac{V}{4L}\text{, }{{f}_{1}}=3{{f}_{o}}\text{, }{{f}_{2}}=5{{f}_{o}}$….and so on.
For open organ pipe,
${{f}_{o}}=\dfrac{V}{2L}\text{, }{{f}_{1}}=2{{f}_{o}}\text{, }{{f}_{2}}=3{{f}_{o}}$….and so on.
The note produced by an open organ pipe is composed of both even and odd harmonics, but the note produced by a closed organ pipe is composed of odd harmonics only. Even harmonics are absent in a closed organ pipe.
Fundamental frequency or the natural frequency of an organ pipe is defined as the lowest frequency of a periodic waveform. Any discrete system with $n$ degrees of freedom can have $n$number of natural frequencies. Similarly, a continuous system of particles can have infinite natural frequencies. The lowest natural frequency of a system is called Fundamental frequency.
For a closed organ pipe, the resonant frequency is the odd multiple of fundamental frequency or, we can say that,
$\text{Overtone = }\left( 2n+1 \right)\times \text{fundamental frequency}$
Or,
\[{{f}_{n}}=\left( 2n+1 \right)\dfrac{V}{4L}\]
Now, we are given that the organ pipe resonates with sound frequency of $135Hz$ and $165Hz$
Therefore,
${{f}_{{{n}_{1}}}}=\left( 2{{n}_{1}}+1 \right)\dfrac{V}{4L}=135Hz$
(Let’s say equation 1)
And,
${{f}_{{{n}_{2}}}}=\left( 2{{n}_{2}}+1 \right)\dfrac{V}{4L}=165Hz$
(Let’s say equation 2)
Dividing equation 1 and equation 2, we get,
$\dfrac{{{f}_{{{n}_{1}}}}}{{{f}_{{{n}_{1}}}}}=\dfrac{\left( 2{{n}_{1}}+1 \right)\dfrac{V}{4L}}{\left( 2{{n}_{2}}+1 \right)\dfrac{V}{4L}}=\dfrac{135}{165}$
Solving above,
$\dfrac{2{{n}_{1}}+1}{2{{n}_{1}}+1}=\dfrac{135}{165}$
Cross multiplying both sides,
$\left( 2{{n}_{1}}+1 \right)165=\left( 2{{n}_{2}}+1 \right)135$
(Let’s say equation 3)
Also,
${{n}_{2}}={{n}_{1}}+1$
Put ${{n}_{2}}={{n}_{1}}+1$ in equation 3,
We get,
\[\begin{align}
  & \left( 2{{n}_{1}}+1 \right)165=\left\{ 2\left( {{n}_{1}}+1 \right)+1 \right\}135 \\
 & 231{{n}_{1}}+165=270{{n}_{1}}+270+135 \\
 & {{n}_{1}}=4 \\
\end{align}\]
And,
$\begin{align}
  & {{n}_{2}}={{n}_{1}}+1 \\
 & {{n}_{2}}=5 \\
\end{align}$
Thus,
\[\begin{align}
  & {{f}_{{{n}_{2}}}}=\left( 2{{n}_{2}}+1 \right)\dfrac{V}{4L} \\
 & {{f}_{{{n}_{2}}}}=\left[ \left( 2\times 5 \right)+1 \right]\dfrac{V}{4L}=165 \\
 & \left( 11 \right)\dfrac{V}{4L}=165 \\
 & \dfrac{V}{4L}=15 \\
\end{align}\]
Fundamental frequency of closed organ pipe is given by,
${{f}_{o}}=\dfrac{V}{4L}$
The fundamental note of the given closed organ pipe is $15Hz$
Hence, the correct option is D.

Note: Students should keep in mind that both odd and even harmonics are present in an open organ pipe while only odd harmonics are present in a closed organ pipe. All even harmonics are absent in a closed organ pipe. The value of any overtone frequency produced by an organ pipe will be less than its fundamental frequency.