Question

Two open organ pipes of length $60 \, \text{cm}$ and $90 \, \text{cm}$ resonate at $6^\text{th}$ and $5^\text{th}$ harmonics respectively. The difference of frequencies for the given modes is ____ $\text{Hz}$.
(Velocity of sound in air $= 333 \, \text{m/s}$)

Answer 1

Correct Answer: 740

The problem asks for the difference in frequencies between two open organ pipes of different lengths, each resonating at a specific harmonic. We are given the lengths of the pipes, the harmonic numbers, and the velocity of sound in air.

Concept Used:

The solution is based on the formula for the frequencies of harmonics in an open organ pipe. An open organ pipe is open at both ends, which means it has antinodes at both ends. As a result, it can support all integer harmonics (both odd and even).

The frequency of the \(n\)-th harmonic (\(f_n\)) for an open organ pipe of length \(L\) is given by:

\[ f_n = n \left( \frac{v}{2L} \right) \]

where:

• \(n\) is the harmonic number (\(n = 1, 2, 3, \ldots\)).
• \(v\) is the velocity of sound in the medium (air).
• \(L\) is the length of the pipe.

Step-by-Step Solution:

Step 1: List the given values and convert them to SI units.

• Length of the first pipe, \(L_1 = 60 \, \text{cm} = 0.60 \, \text{m}\).
• Harmonic number for the first pipe, \(n_1 = 6\).
• Length of the second pipe, \(L_2 = 90 \, \text{cm} = 0.90 \, \text{m}\).
• Harmonic number for the second pipe, \(n_2 = 5\).
• Velocity of sound in air, \(v = 333 \, \text{m/s}\).

Step 2: Calculate the resonant frequency of the first organ pipe (\(f_1\)).

Using the formula for the \(n\)-th harmonic with \(n_1 = 6\) and \(L_1 = 0.60 \, \text{m}\):

\[ f_1 = n_1 \left( \frac{v}{2L_1} \right) = 6 \left( \frac{333}{2 \times 0.60} \right) \] \[ f_1 = 6 \left( \frac{333}{1.2} \right) = 5 \times 333 = 1665 \, \text{Hz} \]

Step 3: Calculate the resonant frequency of the second organ pipe (\(f_2\)).

Using the formula for the \(n\)-th harmonic with \(n_2 = 5\) and \(L_2 = 0.90 \, \text{m}\):

\[ f_2 = n_2 \left( \frac{v}{2L_2} \right) = 5 \left( \frac{333}{2 \times 0.90} \right) \] \[ f_2 = 5 \left( \frac{333}{1.8} \right) = \frac{1665}{1.8} = 925 \, \text{Hz} \]

Step 4: Calculate the difference between the two frequencies.

The difference in frequencies is \(\Delta f = |f_1 - f_2|\).

\[ \Delta f = |1665 \, \text{Hz} - 925 \, \text{Hz}| \] \[ \Delta f = 740 \, \text{Hz} \]

The difference of frequencies for the given modes is 740 Hz.

Answer 2

The frequency \( f \) of an open organ pipe is given by:
\[f = \frac{nv}{2L}\]
The difference in frequency \( \Delta f \) for the two pipes is:
\[\Delta f = \frac{6v}{2 \times 0.6} - \frac{5v}{2 \times 0.9}\]
Substitute \( v = 333 \, \text{m/s} \):
\[\Delta f = \frac{6 \times 333}{2 \times 0.6} - \frac{5 \times 333}{2 \times 0.9}\]
\[\Delta f = 740 \, \text{Hz}\]