91Ó°ÊÓ

For a guitar string, the fundamental frequency is given by: \(f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\) Where \(f_1\) is the fundamental frequency, \(L\) is the length of the string, \(T\) is the tension, and \(\mu\) is the linear mass density. We are given \(f_1 = 330.0\,Hz\), \(L = 65.5\,cm=0.655\,m\), and mass \(m = 0.300\,g = 3 \times 10^{-4}\,kg\). We first need to find the linear mass density \(\mu\), which is calculated by dividing the mass of the string by its length: \(\mu = \frac{m}{L}= \frac{3 \times 10^{-4}\,kg}{0.655\,m} = 4.58 \times 10^{-4}\,\mathrm{kg/m}\) Now, we can rearrange the equation for the fundamental frequency to find the tension \(T\): \(T = (\frac{2L f_1}{\sqrt{\mu}})^2\) Plugging in the values, we get: \(T = (\frac{2(0.655\,m)(330.0\,Hz)}{\sqrt{4.58 \times 10^{-4}\,\mathrm{kg/m}}})^2 = 98.3\,\mathrm{N}\) The tension in the string is \(98.3\,\mathrm{N}\).

To find the wave speed \(v\) on the string, we can use the equation: \(v = \sqrt{\frac{T}{\mu}}\) Plugging in the values we found in part (a), we get: \(v = \sqrt{\frac{98.3\,\mathrm{N}}{4.58 \times 10^{-4}\,\mathrm{kg/m}}} = 458\,\mathrm{m/s}\) The speed of the waves on the guitar string is \(458\,\mathrm{m/s}\).

Let \(f_s\) be the fundamental frequency of the slide whistle, which produces beats of frequency \(5\,\mathrm{Hz}\) with the guitar. Then, we have \(|f_s - f_1| = 5\,\mathrm{Hz}\). Solving for \(f_s\), we get two possible values: \(f_s = f_1 + 5\,\mathrm{Hz} = 335\,\mathrm{Hz}\) or \(f_s = f_1 - 5\,\mathrm{Hz} = 325\,\mathrm{Hz}.\) Now, because the musician lowers the frequency of the slide whistle, we choose the lower value, \(f_s = 325\,\mathrm{Hz}\). The fundamental frequency of the slide whistle is \(325\,\mathrm{Hz}\).

An open tube has a fundamental frequency given by the equation: \(f = \frac{v}{2L}\) Where \(L\) is the length of the tube and \(v\) is the speed of sound in air. The speed of sound in air at room temperature is approximately \(v = 343\,\mathrm{m/s}\). We are given the frequency \(f = 325\,\mathrm{Hz}\). Rearranging the equation to find the length \(L\): \(L = \frac{v}{2f}\) Plugging in the values, we get: \(L = \frac{343\,\mathrm{m/s}}{2(325\,\mathrm{Hz})} = 0.528\,\mathrm{m}\) The length of the open tube in the slide whistle is \(0.528\,\mathrm{m}\).