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

This problem requires determining the frequency differential between two open organ pipes of differing lengths, each operating at a specified harmonic. The lengths of the pipes, their respective harmonic numbers, and the speed of sound in air are provided.

Underlying Principle:

The resolution relies on the equation governing the harmonic frequencies in an open organ pipe. An open organ pipe, open at both extremities, exhibits antinodes at its ends. Consequently, it sustains all integer harmonics, encompassing both odd and even orders.

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

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

where:

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

Sequential Calculation:

Step 1: Enumerate the given parameters and convert them to SI units.

• Length of pipe one, \(L_1 = 60 \, \text{cm} = 0.60 \, \text{m}\).
• Harmonic order for pipe one, \(n_1 = 6\).
• Length of pipe two, \(L_2 = 90 \, \text{cm} = 0.90 \, \text{m}\).
• Harmonic order for pipe two, \(n_2 = 5\).
• Speed of sound in air, \(v = 333 \, \text{m/s}\).

Step 2: Compute the resonant frequency for the first organ pipe (\(f_1\)).

Applying the \(n\)-th harmonic formula 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: Compute the resonant frequency for the second organ pipe (\(f_2\)).

Applying the \(n\)-th harmonic formula 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 disparity between the two frequencies.

The frequency difference is quantified as \(\Delta f = |f_1 - f_2|\).

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

The frequency difference for the specified modes is 740 Hz.