Chapter 3.8, Problem 28E

(a)

(i)

To determine

To find: The rate of change of the frequency with respect to the length (When $T$ and $\rho$ are constant)

Answer to Problem 28E

The rate of change of frequency with respect to length is $-\frac{1}{2L^2}\sqrt{\frac{T}{P}}$.

Explanation of Solution

Given:

The frequency of vibrations of a vibrating violin string is as given below.

$f=\frac{1}{2L}\sqrt{\frac{T}{\rho }}$ (1)

Calculation:

Calculate the rate of change of frequency with respect to length when $T$ and $\rho$ are constant.

Differentiate the equation (1) with respect to $L$.

$\begin{array}{c}\frac{df}{dL}=\frac{d}{dL}\left(\frac{1}{2L}\sqrt{\frac{T}{\rho }}\right)\\ =\frac{1}{2}\sqrt{\frac{T}{\rho }}\left(\frac{d{\left(L\right)}^{-1}}{dL}\right)\\ =\frac{1}{2}\sqrt{\frac{T}{\rho }}\left(-L^{-2}\right)\\ =\frac{1}{2}\sqrt{\frac{T}{\rho }}\left(\frac{-1}{L^2}\right)\end{array}$

$\frac{df}{dL}=-\frac{1}{2L^2}\sqrt{\frac{T}{\rho }}$ (2)

Thus, the rate of change of frequency with respect to length is $\frac{df}{dL}=-\frac{1}{2L^2}\sqrt{\frac{T}{\rho }}$.

(i)

To determine

To find: The rate of change of the frequency with respect to the length (When $T$ and $\rho$ are constant)

Answer to Problem 28E

The rate of change of frequency with respect to length is $-\frac{1}{2L^2}\sqrt{\frac{T}{P}}$.

Explanation of Solution

Given:

The frequency of vibrations of a vibrating violin string is as given below.

$f=\frac{1}{2L}\sqrt{\frac{T}{\rho }}$ (1)

Calculation:

Calculate the rate of change of frequency with respect to length when $T$ and $\rho$ are constant.

Differentiate the equation (1) with respect to $L$.

$\begin{array}{c}\frac{df}{dL}=\frac{d}{dL}\left(\frac{1}{2L}\sqrt{\frac{T}{\rho }}\right)\\ =\frac{1}{2}\sqrt{\frac{T}{\rho }}\left(\frac{d{\left(L\right)}^{-1}}{dL}\right)\\ =\frac{1}{2}\sqrt{\frac{T}{\rho }}\left(-L^{-2}\right)\\ =\frac{1}{2}\sqrt{\frac{T}{\rho }}\left(\frac{-1}{L^2}\right)\end{array}$

$\frac{df}{dL}=-\frac{1}{2L^2}\sqrt{\frac{T}{\rho }}$ (2)

Thus, the rate of change of frequency with respect to length is $\frac{df}{dL}=-\frac{1}{2L^2}\sqrt{\frac{T}{\rho }}$.

(ii)

To determine

To find: The rate of change of the frequency with respect to the tension (When $L$ and $\rho$ are constant)

Answer to Problem 28E

The rate of change of frequency with respect to tension is $\frac{1}{4L\sqrt{T\rho }}$.

Explanation of Solution

Calculate the rate of change of frequency with respect to tension when $L$ and $\rho$ are constant.

Differentiate the equation (1) with respect to $T$.

$\begin{array}{c}\frac{df}{dT}=\frac{d}{dT}\left(\frac{1}{2L}\sqrt{\frac{T}{\rho }}\right)\\ =\frac{1}{2L\sqrt{\rho }}\left(\frac{d\left(\sqrt{T}\right)}{dT}\right)\\ =\frac{1}{2L\sqrt{\rho }}\left(\frac{d\left(T^{1/2}\right)}{dT}\right)\end{array}$

$\begin{array}{c}\frac{df}{dT}=\frac{1}{2L\sqrt{\rho }}\left(\frac{T^{-1/2}}{2}\right)\\ =\frac{1}{2L\sqrt{\rho }}\left(\frac{1}{2\sqrt{T}}\right)\end{array}$

$\frac{df}{dT}=\frac{1}{4L\sqrt{T\rho }}$ (3)

Thus, the rate of change of frequency with respect to tension is $\frac{df}{dT}=\frac{1}{4L\sqrt{T\rho }}$.

(iii)

To determine

To find: The rate of change of the frequency with respect to the linear density (When $L$ and $T$ are constant)

Answer to Problem 28E

The rate of change of frequency with respect to linear density is $\frac{-\sqrt{T}}{4L{\rho }^{3/2}}$.

Explanation of Solution

Calculate the rate of change of frequency with respect to linear density when $T$ and $L$ are constant.

Differentiate the equation (1) with respect to $\rho$.

$\begin{array}{c}\frac{df}{d\rho }=\frac{d}{d\rho }\left(\frac{1}{2L}\sqrt{\frac{T}{\rho }}\right)\\ =\frac{\sqrt{T}}{2L}\left(\frac{d\left({\rho }^{-1/2}\right)}{d\rho }\right)\\ =\frac{\sqrt{T}}{2L}\left(\frac{-1}{2}{\rho }^{-3/2}\right)\end{array}$

$\frac{df}{d\rho }=\frac{-\sqrt{T}}{4L{\rho }^{3/2}}$ (4)

Thus, the rate of change of frequency with respect to linear density is $\frac{df}{d\rho }=\frac{-\sqrt{T}}{4L{\rho }^{3/2}}$.

(b)

(i)

To determine

To find: To determine the behavior of pitch of the note “when the effective length of a string is decreased by placing a finger on the string is decreased by placing a finger on the string so a shorter portion of the string vibrates”.

Answer to Problem 28E

The pitch of the note is higher.

Explanation of Solution

Determine the behavior of the pitch note when the effective length of a string is decreased.

Sketch the curve.

Calculate the value of $f$ using the relation.

$f=\frac{k}{L}$

Here $f$ is the frequency of vibration and $L$ is the length.

Substitute 1 for $k$ and 2 for $L$ in the above equation.

$\begin{array}{c}f=\frac{1}{2}\\ =0.5\end{array}$

Similarly, calculate the remaining values of $f$.

Calculate and tabulate the remaining values of $f$ as shown in table (1).

$k$	$L$	$f=\frac{k}{L}$
1	2	0.5
1	3	0.3
1	4	0.25
1	5	0.2
1	6	0.17
1	7	0.14
1	8	0.12
1	9	0.11
1	10	0.1
1	11	0.09

Graph:

Plot the graph using table (1) as shown in figure (1).

Comments:

Refer the figure (1).

Here the frequency of vibration $f$ is increasing when the length $L$ decreases. Thus, the condition $\frac{df}{dL}<0$ is true. Hence the pitch is in higher note.

(i)

To determine

To find: To determine the behavior of pitch of the note “when the effective length of a string is decreased by placing a finger on the string is decreased by placing a finger on the string so a shorter portion of the string vibrates”.

Answer to Problem 28E

The pitch of the note is higher.

Explanation of Solution

Determine the behavior of the pitch note when the effective length of a string is decreased.

Sketch the curve.

Calculate the value of $f$ using the relation.

$f=\frac{k}{L}$

Here $f$ is the frequency of vibration and $L$ is the length.

Substitute 1 for $k$ and 2 for $L$ in the above equation.

$\begin{array}{c}f=\frac{1}{2}\\ =0.5\end{array}$

Similarly, calculate the remaining values of $f$.

Calculate and tabulate the remaining values of $f$ as shown in table (1).

$k$	$L$	$f=\frac{k}{L}$
1	2	0.5
1	3	0.3
1	4	0.25
1	5	0.2
1	6	0.17
1	7	0.14
1	8	0.12
1	9	0.11
1	10	0.1
1	11	0.09

Graph:

Plot the graph using table (1) as shown in figure (1).

Comments:

Refer the figure (1).

Here the frequency of vibration $f$ is increasing when the length $L$ decreases. Thus, the condition $\frac{df}{dL}<0$ is true. Hence the pitch is in higher note.

(ii)

To determine

To find: To determine the behavior of pitch of the note “When the tension is increased by turning a tuning peg.”.

Answer to Problem 28E

The pitch of the note is higher.

Explanation of Solution

Determine the behavior of the pitch note when the tension is increased by turning a tuning peg.

Sketch the curve.

Calculate the value of $f$ using the relation.

$f=k\sqrt{T}$

Here $f$ is the frequency of vibration and $T$ is the tension.

Substitute 1 for $k$ and 0 for $T$ in the above equation.

$\begin{array}{c}f=k\sqrt{T}\\ =0\end{array}$

Similarly, calculate the remaining values of $f$.

Calculate and tabulate the remaining values of $f$ as shown in table (2).

$k$	$T$	$f=k\sqrt{T}$
1	0.0	0.0
1	0.2	0.4
1	0.4	0.6
1	0.6	0.8
1	0.8	0.9
1	1.0	1.0
1	1.2	1.1
1	1.4	1.2
1	1.6	1.3
1	1.8	1.3
1	2.0	1.4
1	2.2	1.5
1	2.4	1.5
1	2.6	1.6
1	2.8	1.7
1	3.0	1.7
1	3.2	1.8
1	3.4	1.8
1	3.6	1.9
1	3.8	1.9
1	4.0	2.0
1	4.2	2.0
1	4.4	2.1
1	4.6	2.1
1	4.8	2.2
1	5.0	2.2
1	5.2	2.3
1	5.4	2.3
1	5.6	2.4
1	5.8	2.4
1	6.0	2.4
1	6.2	2.5
1	6.4	2.5
1	6.6	2.6
1	6.8	2.6
1	7.0	2.6
1	7.2	2.7
1	7.4	2.7
1	7.6	2.8
1	7.8	2.8
1	8.0	2.8
1	8.2	2.9
1	8.4	2.9
1	8.6	2.9
1	8.8	3.0
1	9.0	3.0
1	9.2	3.0
1	9.4	3.1
1	9.6	3.1
1	9.8	3.1
1	10.0	3.2

Graph:

Plot the graph using table (2) shown in figure (2).

Comments:

Refer the figure (2).

Here the frequency of vibration $f$ is increasing when the tension $T$ decreases. Thus, the condition $\frac{df}{dT}>0$ is true. Hence the pitch is in higher note.

(iii)

To determine

To find: To determine the behavior of pitch of the note “When the linear density is increased by switching to another string”.

Answer to Problem 28E

The pitch of the note is lower.

Explanation of Solution

Determine the behavior of the pitch note when the linear density is increased by switching to another string.

Sketch the curve.

Calculate the value of $f$ using the relation.

$f=\frac{k}{\sqrt{\rho }}$

Here $f$ is the frequency of vibration and $\rho$ is linear density.

Substitute 1 for $k$ and 0 for $T$ in the above equation.

$\begin{array}{c}f=\frac{k}{\sqrt{\rho }}\\ =\frac{1}{\sqrt{1}}\\ =1\end{array}$

Similarly, calculate the remaining values of $f$.

Calculate and tabulate the remaining values of $f$ as shown in table (3).

$k$	$\rho$	$f=\frac{k}{\sqrt{\rho }}$
1	0.2	2.2
1	0.4	1.6
1	0.6	1.3
1	0.8	1.1
1	1.0	1.0
1	1.2	0.9
1	1.4	0.8
1	1.6	0.8
1	1.8	0.7
1	2.0	0.7
1	2.2	0.7
1	2.4	0.6
1	2.6	0.6
1	2.8	0.6
1	3.0	0.6
1	3.2	0.6
1	3.4	0.5
1	3.6	0.5
1	3.8	0.5
1	4.0	0.5
1	4.2	0.5
1	4.4	0.5
1	4.6	0.5
1	4.8	0.5
1	5.0	0.4
1	5.2	0.4
1	5.4	0.4
1	5.6	0.4
1	5.8	0.4
1	6.0	0.4
1	6.2	0.4
1	6.4	0.4
1	6.6	0.4
1	6.8	0.4
1	7.0	0.4
1	7.2	0.4
1	7.4	0.4
1	7.6	0.4
1	7.8	0.4
1	8.0	0.4
1	8.2	0.3
1	8.4	0.3
1	8.6	0.3
1	8.8	0.3
1	9.0	0.3
1	9.2	0.3
1	9.4	0.3
1	9.6	0.3
1	9.8	0.3
1	10.0	0.3

Graph:

Plot the graph using table (3) as shown in figure (3).

Comments:

Refer the figure (3).

Here the frequency of vibration $f$ is decreasing when the linear density $\rho$ increases. Thus, the condition $\frac{df}{d\rho }>0$ is true. Hence the pitch is in lower note.