Chapter 12, Problem 74GP

To determine

To Find:The mass per unit length of each string relative to that of the lowest string

Answer to Problem 74GP

The mass per unit length of each string relative to that of the lowest string are ${\mu }_B=0.44{\mu }_A,{\mu }_C=0.20{\mu }_A,\text{and}{\mu }_D=0.039{\mu }_A$ .

Explanation of Solution

Given:

Each string frequency is one and half times the neighbour string

Tension is constant for all the strings

Formula used:

The fundamental frequency is given as,

  $f=\frac{v}{2L}=\frac{1}{2L}\sqrt{\frac{F_t}{\mu }}$

Where, $\mu$ is the mass per unit length, L is the string length and ${\text{F}}_{\text{T}}$ is the tension force

Calculation:

Considered the frequencies of the four violin’s strings with $f_A,f_B,f_C,f_D$ and $f_A$ is the frequency of lowest pitch. $\mu$ be the mass per unit length. All strings are of the same length and tension. The frequencies are calculated by using the above formula such that

  $\begin{array}{l}{f}_B=1.5f_A=1.5\times \frac{1}{2L}\sqrt{\frac{F_t}{\mu }}\\ {\mu }_B=\frac{{\mu }_A}{{\left(1.5\right)}^2}=0.44{\mu }_A\end{array}$

Similarly, the other frequencies are

  $\begin{array}{l}{f}_C=1.5f_B={\left(1.5\right)}^2{f}_A\\ f_C={\left(1.5\right)}^2\times \frac{1}{2L}\sqrt{\frac{F_T}{{\mu }_A}}\\ {\mu }_C=\frac{{\mu }_A}{{\left(1.5\right)}^4}=0.20{\mu }_A\end{array}$

  $\begin{array}{l}{f}_D=1.5f_C\\ f_D={\left(1.5\right)}^3{f}_A={\left(1.5\right)}^3\times \frac{1}{2L}\sqrt{\frac{F_t}{{\mu }_A}}\\ {\mu }_D=\frac{{\mu }_A}{{\left(1.5\right)}^6}=0.09{\mu }_A\end{array}$

Conclusion:

Thus, the mass per unit length of each string relative to that of the lowest string for each case is ${\mu }_B=0.44{\mu }_A,{\mu }_C=0.20{\mu }_A,\text{and}{\mu }_D=0.09{\mu }_A$ .