Question

question 4 (4 points) listen a wire with a mass of 1.28 g and a length of 1.28 m is placed under tension on a sounding board. the distance between the two frets of the sounding board is 1.43 m. the string is plucked and the sound pulse is sent through a frequency analyzer. two of the harmonics are 404 hz and 455 hz. assuming these two harmonics are consecutive (no other harmonic frequencies between them), what is the tension in the wire? your answer:

Explanation:

Step1: Find the linear mass density

The linear mass density \(\mu=\frac{m}{L}\). Given \(m = 1.28\times10^{- 3}\space kg\) and \(L = 1.28\space m\), then \(\mu=\frac{1.28\times 10^{-3}}{1.28}=1\times10^{-3}\space kg/m\)

Step2: Use the formula for the frequency of a vibrating string

The frequency of the \(n\) - th harmonic of a vibrating string is \(f_{n}=\frac{n}{2L}\sqrt{\frac{T}{\mu}}\), and for the \((n + 1)\) - th harmonic \(f_{n + 1}=\frac{n + 1}{2L}\sqrt{\frac{T}{\mu}}\)

The difference between consecutive harmonics \(\Delta f=f_{n + 1}-f_{n}\)

Since \(f_{n}=\frac{n}{2L}\sqrt{\frac{T}{\mu}}\) and \(f_{n + 1}=\frac{n + 1}{2L}\sqrt{\frac{T}{\mu}}\), then \(\Delta f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}\)

We are given \(\Delta f=455 - 404=51\space Hz\), \(L = 1.43\space m\) and \(\mu = 1\times10^{-3}\space kg/m\)

From \(\Delta f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}\), we can solve for \(T\)

First, re - arrange the formula for \(T\): \(\sqrt{\frac{T}{\mu}}=2L\Delta f\), then \(T = 4L^{2}\Delta f^{2}\mu\)

Substitute \(L = 1.43\space m\), \(\Delta f=51\space Hz\) and \(\mu = 1\times10^{-3}\space kg/m\)

\(T=4\times(1.43)^{2}\times(51)^{2}\times1\times10^{-3}\)

\(T = 4\times2.0449\times2601\times1\times10^{-3}\)

\(T=4\times2.0449\times2.601\)

\(T = 4\times5.319\)

\(T = 21.276\space N\approx21.3\space N\)

Answer:

\(21.3\space N\)