Question

A musician plays a string on a guitar that has a fundamental frequency of \(330.0 \mathrm{Hz}\). The string is \(65.5 \mathrm{cm}\) long and has a mass of \(0.300 \mathrm{g} .\) (a) What is the tension in the string? (b) At what speed do the waves travel on the string? (c) While the guitar string is still being plucked, another musician plays a slide whistle that is closed at one end and open at the other. He starts at a very high frequency and slowly lowers the frequency until beats, with a frequency of \(5 \mathrm{Hz}\), are heard with the guitar. What is the fundamental frequency of the slide whistle with the slide in this position? (d) How long is the open tube in the slide whistle for this frequency?

Short answer

Answer: The tension in the guitar string is 98.3 N, the wave speed on the string is 458 m/s, the fundamental frequency of the slide whistle is 325 Hz, and the length of the open tube in the slide whistle is 0.528 m.

Step-by-step solution

a) Tension in the string

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 \(98.3\,\mathrm{N}\).

b) Wave speed on the string

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 \(458\,\mathrm{m/s}\).

c) Fundamental frequency of the slide whistle

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 \(325\,\mathrm{Hz}\).

d) Length of the open tube in the slide whistle

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 \(0.528\,\mathrm{m}\).