Q.A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz ?(sound velocity in air = 330 m s$^{-1}$)

The phenomenon of acoustic resonance in a pipe closed at one end dictates that only specific harmonic frequencies, which are odd multiples of the fundamental frequency, can be excited.

1. The fundamental frequency ($f_1$) for a pipe of length $L$ closed at one end is given by the formula $f_1 = \frac{v}{4L}$.

   Given $L = 20 \text{ cm} = 0.2 \text{ m}$ and $v = 330 \text{ m/s}$, the fundamental frequency is:

   $f_1 = \frac{330}{4 \times 0.2} = \frac{330}{0.8} = 412.5 \text{ Hz}$.

2. For a pipe closed at one end, only odd harmonics are possible. The frequencies of these harmonics are $f_n = n f_1$, where $n$ must be an odd integer ($n = 1, 3, 5, \dots$). …

When sound waves travel through a pipe, they reflect from the ends, creating standing waves. For resonance to occur, the length of the pipe must be an integer multiple of half-wavelengths (for open-open or closed-closed pipes) or quarter-wavelengths (for open-closed pipes), satisfying specific boundary conditions at the ends.

For a pipe closed at one end and open at the other:

• At the closed end, the air particles cannot move, so a displacement node (and a pressure antinode) must form.
• At the open end, the air particles are free to move, so a displacement antinode (and a pressure node) must form.

These boundary conditions dictate the possible wavelengths and thus the resonant frequencies.

Let $L$ be the length of the pipe.

1. Fundamental Mode (First Harmonic):

   The simplest standing wave pattern that satisfies the boundary conditions has a node at the closed end and the very next antinode at the open end.

   This means the length of the pipe $L$ corresponds to one-quarter of a wavelength ($\lambda_1$).

   $L = \frac{\lambda_1}{4} \implies \lambda_1 = 4L$

   The fundamental frequency $f_1 = \frac{v}{\lambda_1} = \frac{v}{4L}$.

2. Higher Harmonics:

   For higher resonant modes, additional nodes and antinodes must fit into the pipe, always maintaining a node at the closed end and an antinode at the open end.

   • The next possible mode has a node, then an antinode, then another node, and finally an antinode at the open end. This pattern covers three-quarters of a wavelength ($\lambda_3$). $L = \frac{3\lambda_3}{4} \implies \lambda_3 = \frac{4L}{3}$ The frequency $f_3 = \frac{v}{\lambda_3} = \frac{v}{4L/3} = 3 \frac{v}{4L} = 3f_1$. This is the third harmonic.
   • The next possible mode would correspond to five-quarters of a wavelength ($\lambda_5$). $L = \frac{5\lambda_5}{4} \implies \lambda_5 = \frac{4L}{5}$ The frequency $f_5 = \frac{v}{\lambda_5} = \frac{v}{4L/5} = 5 \frac{v}{4L} = 5f_1$. This is the fifth harmonic.

   Notice a pattern: only odd multiples of the fundamental frequency are possible in a pipe closed at one end. Even harmonics are absent.

Step 1: For a pipe of length $L$ closed at one end, the fundamental frequency is $f_1 = \dfrac{v}{4L}$ (a node at the closed end, an antinode at the open end, quarter-wave fit).

Step 2: Substitute $v=330\ \text{m/s}$, $L=0.20\ \text{m}$:

Step 3: Only odd harmonics exist in a closed pipe: $f_n = n f_1$ with $n=1,3,5,\dots$ …