Questions: A 25.0 cm long organ pipe is filled with air and is open at one end and closed at the other. The speed of sound in air at 0°C is 331 m/s. What is the frequency of the fourth mode of vibration? Multiple Choice 1,750 Hz 3,170 Hz 2,720 Hz 1,550 Hz 2,320 Hz

Solution Steps

We have an organ pipe that is open at one end and closed at the other. The length of the pipe is \(25.0 \, \text{cm}\), which is \(0.25 \, \text{m}\). We need to find the frequency of the fourth mode of vibration. The speed of sound in air at \(0^{\circ} \mathrm{C}\) is given as \(331 \, \text{m/s}\).

For a pipe that is open at one end and closed at the other, the resonant frequencies occur at odd harmonics. The \(n\)-th mode of vibration corresponds to the \((2n-1)\)-th harmonic. Therefore, the fourth mode corresponds to the seventh harmonic.

The wavelength \(\lambda\) for the \(n\)-th harmonic in a pipe open at one end and closed at the other is given by:

\[ \lambda_n = \frac{4L}{2n-1} \]

For the fourth mode (\(n = 4\)), the wavelength is:

\[ \lambda_4 = \frac{4 \times 0.25 \, \text{m}}{2 \times 4 - 1} = \frac{1 \, \text{m}}{7} \approx 0.1429 \, \text{m} \]

The frequency \(f\) is related to the speed of sound \(v\) and the wavelength \(\lambda\) by the equation:

\[ f = \frac{v}{\lambda} \]

Substituting the values we have:

\[ f_4 = \frac{331 \, \text{m/s}}{0.1429 \, \text{m}} \approx 2316.8 \, \text{Hz} \]

Final Answer