Question

What are the four lowest standing wave frequencies for an organ pipe that is \(4.80 \mathrm{m}\) long and closed at one end?

Short answer

Answer: The four lowest standing wave frequencies for the given organ pipe are approximately 35.7 Hz, 107.1 Hz, 178.5 Hz, and 249.9 Hz.

Step-by-step solution

Recall the formula for the frequency of nth harmonic for a closed pipe

The formula for the frequency of nth harmonic for a pipe closed at one end is: \(f_n = \frac{2n - 1}{4L} \cdot v\) where: - \(f_n\) is the frequency of the nth harmonic (Hz), - \(n\) is the harmonic number (1 for the fundamental frequency, 2 for the second harmonic, etc.), - \(L\) is the length of the pipe (in meters), - \(v\) is the speed of sound in air (approximately \(343 \mathrm{m/s}\) at room temperature).

Plug in the given values and simplify

We are given the length of the pipe as \(L = 4.80 \mathrm{m}\). We will use \(v = 343 \mathrm{m/s}\) as an approximation for the speed of sound. Now we can plug in these values into the formula and simplify: \(f_n = \frac{2n - 1}{4(4.80)} \cdot 343\)

Find the frequencies of the first 4 harmonics

Now, we will calculate the frequencies of the first 4 harmonics by plugging in \(n = 1, 2, 3, 4\) into the simplified formula: 1. For \(n = 1\) (Fundamental frequency): \(f_1 = \frac{2(1) - 1}{4(4.80)} \cdot 343 = \frac{1}{9.6} \cdot 343 \approx 35.7 \mathrm{Hz}\) 2. For \(n = 2\) (Second harmonic): \(f_2 = \frac{2(2) - 1}{4(4.80)} \cdot 343 = \frac{3}{9.6} \cdot 343 \approx 107.1 \mathrm{Hz}\) 3. For \(n = 3\) (Third harmonic): \(f_3 = \frac{2(3) - 1}{4(4.80)} \cdot 343 = \frac{5}{9.6} \cdot 343 \approx 178.5 \mathrm{Hz}\) 4. For \(n = 4\) (Fourth harmonic): \(f_4 = \frac{2(4) - 1}{4(4.80)} \cdot 343 = \frac{7}{9.6} \cdot 343 \approx 249.9 \mathrm{Hz}\) The four lowest standing wave frequencies for the given organ pipe are approximately \(35.7 \mathrm{Hz}\), \(107.1 \mathrm{Hz}\), \(178.5 \mathrm{Hz}\), and \(249.9 \mathrm{Hz}\).