A certain pipe has resonant frequencies of \(234 \mathrm{Hz}\) $390 \mathrm{Hz},\( and \)546 \mathrm{Hz},$ with no other resonant frequencies between these values. (a) Is this a pipe open at both ends or closed at one end? (b) What is the fundamental frequency of this pipe? (c) How long is this pipe?

For an open pipe at both ends, resonant frequencies are given by: \(f_n = \frac{2nL}{v}\), where \(n\) is an integer value (1, 2, 3, ...), \(L\) is the length of the pipe, and \(v\) is the speed of sound in air. For a closed pipe with one end open, resonant frequencies are given by: \(f_n = \frac{(2n-1)L}{4v}\), where \(n\) is an integer value (1, 2, 3, ...), \(L\) is the length of the pipe, and \(v\) is the speed of sound in air.

We are given that the pipe has resonant frequencies of \(234 \mathrm{Hz}, 390 \mathrm{Hz},\) and \(546 \mathrm{Hz}\) with no other resonant frequencies between these values. If it is an open pipe, both \(234 Hz\) and \(390 Hz\) should be multiples of the fundamental frequency (the first resonant frequency when n=1). Finding the difference between the two frequencies: Difference = \(390 Hz - 234 Hz = 156 Hz\) Now, check if \(234 Hz\) is a multiple of \(156 Hz\). It's not. However, if it is a closed pipe, the fundamental frequency will only be available in every odd multiple, i.e., \(1f, 3f, 5f, ...\) Let's check if the difference between \(234 Hz\) and \(390 Hz\) fits this scenario: Difference = \(390 Hz - 234 Hz = 156 Hz\) Now, check if \(234 Hz\) is a multiple of \(\frac{156 Hz}{2}\) Fundamental frequency: \(f = \frac{156}{2} = 78 Hz\) Thus, the given frequencies fit the scenario of a closed pipe: \(1f = 78 Hz\), \(3f=234 Hz\), \(5f=390 Hz\), \(7f=546 Hz\). (a) The pipe is closed at one end. (b) The fundamental frequency of this pipe is \(78 Hz\).

Now that we know the pipe is closed at one end, and the fundamental frequency is \(78 Hz\), we can calculate the length of the pipe using the formula for the resonant frequency of a closed pipe: \(f_n = \frac{(2n-1)L}{4v}\) For the fundamental frequency, n = 1: \(f_1 = \frac{(2(1)-1)L}{4v} = \frac{1L}{4v}\) We know the speed of sound in air, \(v\), is approximately \(343 \mathrm{m/s}\). Now, we solve for L: \(78 Hz = \frac{1L}{4(343 m/s)}\) \(L = 78 Hz \times 4(343 m/s)\) \(L = 106.68 m\) (c) The length of the pipe is approximately \(106.68 m\).