Chapter 18, Problem 79PQ

(a)

To determine

The possible frequencies for tuning the guitar string.

Answer to Problem 79PQ

The possible frequencies for tuning the guitar string either $192\text{Hz}$ or $200\text{Hz}$.

Explanation of Solution

Write the expression for beat frequency

  $f_{\text{beat}}=\left|f_2-f_1\right|$                                                                                          (I)

Here, $f_1$ is the frequency of the tune in guitar string and $f_2$ is the frequency of the reference tone.

Conclusion:

Substitute $196\text{Hz}$ for $f_2$ and $4.00\text{beats/s}$ for $f_1$ in equation (I) to find $f_{\text{beat}}$.

  $\begin{array}{c}{f}_{\text{beat}}=\left|196\text{Hz}-4.00\text{beats/s}\right|\\ =192\text{Hz}\end{array}$

  $\begin{array}{c}{f}_{\text{beat}}=\left|196\text{Hz}+4.00\text{beats/s}\right|\\ =200\text{Hz}\end{array}$

Therefore, the possible frequencies for tuning the guitar string either $192\text{Hz}$ or $200\text{Hz}$.

(b)

To determine

The frequency of the guitar string, if the musician loosens the guitar string.

Answer to Problem 79PQ

The frequency of the guitar string, if the musician loosens the guitar string is $198\text{Hz}$.

Explanation of Solution

If loosening the string lowers the wave speed and frequency, if the original frequency was $192\text{Hz}$, the difference in frequencies would become larger and the beats would speeds up.

Write the expression for beat frequency, if the musician loosens the guitar string

  ${f^{\prime }}_{\text{beat}}=\left|{f^{\prime }}_2-{f^{\prime }}_1\right|$                                                                                       (II)

Here, ${f^{\prime }}_1$ is the frequency of the loosened guitar string and ${f^{\prime }}_2$ is the original frequency of the guitar string.

Conclusion:

Substitute $200\text{Hz}$ for ${f^{\prime }}_2$ and $2.00\text{beats/s}$ for ${f^{\prime }}_1$ in equation (II) to find ${f^{\prime }}_{\text{beat}}$.

  $\begin{array}{c}{f^{\prime }}_{\text{beat}}=\left|200\text{Hz}-2.00\text{beats/s}\right|\\ =198\text{Hz}\end{array}$

Instead the frequency must have started at $200\text{Hz}$ to become $198\text{Hz}$.

Therefore, frequency of the guitar string, if the musician loosens the guitar string is $198\text{Hz}$

(c)

To determine

The percentage of the tension in the guitar string change for the string to be tuned to $196\text{Hz}$.

Answer to Problem 79PQ

The percentage of the tension in the guitar string change for the string to be tuned to $196\text{Hz}$ is reduced by 2.01%.

Explanation of Solution

Write the expression for fundamental frequency depends on the tension in the string.

  $f_1=\frac{1}{2L}\sqrt{\frac{F_T}{\mu }}$                                                                                                    (III)

Here, $f_1$ is the fundamental frequency, $L$ is the length of the string, $F_T$ is the tension force, and $\mu$ is the linear mass density.

The ratio of the two frequencies then depends on the square root of the ratio of tensions.

  $\frac{f_2}{f_1}=\sqrt{\frac{F_{T1}}{F_{T2}}}$

Here, $f_1$ is the frequency of the guitar string, if the musician loosens the guitar string, $f_2$ is the original frequency of the tone, $F_{T1}$ is the tension due to frequency $f_1$, and $F_{T2}$ is the tension due to frequency $f_2$.

Rearrange the above equation for $F_{T2}$.

  $\begin{array}{c}{\left(\frac{f_2}{f_1}\right)}^2=\frac{F_{T1}}{F_{T2}}\\ F_{T2}={\left(\frac{f_2}{f_1}\right)}^2{F}_{T1}\end{array}$                                                                                          (IV)

Conclusion:

The fractional change should be made in the tension is,

  $\text{fractional change}=\frac{F_{T2}-F_{T1}}{F_{T1}}$

Rearrange the above equation and then compare with equation (IV).

  $\begin{array}{c}\text{fractional change}=\frac{F_{T2}}{F_{T1}}-1\\ ={\left(\frac{f_2}{f_1}\right)}^2-1\end{array}$

Substitute $198\text{Hz}$ for $f_1$ and $196\text{Hz}$ for $f_2$ in the above equation.

  $\begin{array}{c}\text{fractional change}={\left(\frac{196\text{Hz}}{198\text{Hz}}\right)}^2-1\\ =-0.0201\\ =-2.01\%\end{array}$

Therefore, the current tension should be reduced by a factor of 2.01% in order to bring the string in tune with the reference tone.