Chapter 14, Problem 20P

(a)

To determine

The violinist’s finger distance from the end of string to play the concert $A$.

Answer to Problem 20P

The violinist’s finger distance from the end of string to play the concert $A$ is $3.82\text{cm}$.

Explanation of Solution

Given info: The frequency of the concert $A$ is $440\text{Hz}$, the frequency of concert $G$ is $392\text{Hz}$, the length of the string in concert $G$ is $0.350\text{m}$ and the half width of the finger is $0.600\text{cm}$.

Formula to calculate length in concert $G$ is,

$\begin{array}{c}{\lambda }_{\text{A}}=2L_{\text{A}}=\frac{v}{f_{\text{A}}}\\ L_{\text{A}}=\frac{v}{2f_{\text{A}}}\end{array}$

Here,

$L_{\text{A}}$ is the length of the string in $A$ concert.

$f_{\text{A}}$ is the frequency of concert $A$.

Formula to calculate the speed of the wave in concert $G$ is,

$\begin{array}{c}{\lambda }_{\text{G}}=2L_{\text{G}}=\frac{v}{f_{\text{G}}}\\ v=2f_{\text{G}}{L}_{\text{G}}\end{array}$

Here,

$f_G$ is the frequency of concert $G$.

$L_{\text{G}}$ is the length of the string in $G$ concert.

$v$ is the speed of wave.

The difference between the length of the concert $G$ and concert $A$ is,

$L_{\text{G}}-L_{\text{A}}$ (1)

Substitute $\frac{v}{2f_{\text{A}}}$ for $L_{\text{A}}$ in equation (1).

$L_{\text{G}}-L_{\text{A}}=L_{\text{G}}-\frac{v}{2f_{\text{A}}}$ (2)

Substitute $2f_{\text{G}}{L}_{\text{G}}$ for $v$ in equation (2).

$\begin{array}{c}{L}_{\text{G}}-L_{\text{A}}=L_{\text{G}}-\frac{2f_{\text{G}}{L}_{\text{G}}}{2f_{\text{A}}}\\ =L_{\text{G}}-\frac{f_{\text{G}}{L}_{\text{G}}}{f_{\text{A}}}\\ =L_{\text{G}}\left(1-\frac{f_{\text{G}}}{f_{\text{A}}}\right)\end{array}$ (3)

Substitute $0.350\text{m}$ for $L_{\text{G}}$, $440\text{Hz}$ for $f_{\text{A}}$ and $392\text{Hz}$ for $f_{\text{A}}$ in equation (3).

$\begin{array}{c}{L}_{\text{G}}-L_{\text{A}}=0.350\text{m}\left(1-\frac{392\text{Hz}}{440\text{Hz}}\right)\\ =0.0382\text{m}\\ \text{=}0.0382\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =3.82\text{cm}\end{array}$

So, the finger must be placed at $3.82\text{cm}$ from the end of the string to play the concert $A$.

Conclusion:

Therefore, the violinist’s finger distance from the end of string to play the concert $A$ is $3.82\text{cm}$.

(b)

To determine

The maximum allowable percentage change in the string tension.

Answer to Problem 20P

The maximum allowable percentage change in the string tension is $3.84\%$.

Explanation of Solution

Given info: The frequency of the concert $A$ is $440\text{Hz}$, the frequency of concert $G$ is $392\text{Hz}$, the length of the string in concert $G$ is $0.350\text{m}$ and the half width of the finger is $0.600\text{cm}$.

The difference between the length of the concert $G$ and concert $A$ is,

$L_{\text{G}}-L_{\text{A}}=3.82\text{cm}$ (4)

Substitute $0.350\text{m}$ for $L_{\text{G}}$ in equation (4) to get the $L_{\text{A}}$.

$\begin{array}{c}0.350\text{m}-L_{\text{A}}=3.82\text{cm}\\ L_{\text{A}}=0.312\text{m}\\ \text{=}0.312\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ =31.2\text{cm}\end{array}$

Formula to calculate the frequency of the wave on the string in concert $A$ is,

$\begin{array}{c}{f}_A=\frac{1}{2L_{\text{A}}}\sqrt{\frac{T}{\mu }}\\ L_{\text{A}}=\frac{1}{2f_A}\sqrt{\frac{T}{\mu }}\end{array}$ (3)

Differentiate the above equation with respect to $T$.

$d{L}_{\text{A}}=\frac{dT}{2f_{\text{A}}\sqrt{T\mu }}$ (4)

Divide the equation (II) by equation (I).

$\begin{array}{c}\frac{d{L}_{\text{A}}}{L_{\text{A}}}=\frac{dT}{2T}\\ \frac{dT}{T}=\frac{2d{L}_{\text{A}}}{L_{\text{A}}}\end{array}$ (5)

• $d{L}_{\text{A}}$ is the half width of the finger.

Convert the equation into percentage.

$\frac{dT}{T}=\frac{2d{L}_{\text{A}}}{L_{\text{A}}}\times 100$ (6)

Substitute $31.18\text{cm}$ for $L_{\text{A}}$ and $0.600\text{cm}$ for $d{L}_{\text{A}}$ in equation (VI) to find the $\frac{dT}{T}$.

$\begin{array}{c}\frac{dT}{T}=\frac{2\times 0.600\text{cm}}{31.2\text{cm}}\times 100\\ =3.85\%\end{array}$

Conclusion:

Therefore, the maximum allowable percentage change in the string tension is $3.85\%$.