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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

Expert verified
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

01

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).
02

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: \(f_n = \frac{2n - 1}{4(4.80)} \cdot 343\)
03

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 \(249.9 \mathrm{Hz}\).

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