Chapter 12, Problem 40P

(a)

To determine

The tension in the string.

Answer to Problem 40P

The tension in the string is $85.6\text{N}$.

Explanation of Solution

Write the equation for the speed of transverse waves on a string.

$v=\sqrt{\frac{F}{\mu }}$ (I)

Here, $v$ is the speed of transverse waves on a string, $F$ is the restoring force, and $\mu$ is the linear mass density.

Write the equation for linear mass density.

$\mu =\frac{m}{L}$ (II)

Here, $m$ is the mass of the string and $L$ is the length of the string,

Write the equation for the first harmonic frequency of standing waves.

$f_1=\frac{v}{2L}$ (III)

Conclusion:

Substitute equation (II) in (I).

$v=\sqrt{\frac{FL}{m}}$ (IV)

Rewrite the equation (III) for $v$.

$v=2L{f}_1$ (V)

Compare the equation (IV) and (V) and rearrange for $F$

$\begin{array}{c}\sqrt{\frac{FL}{m}}=2L{f}_1\\ \frac{FL}{m}=4L^2{f}_1^2\\ F=4mL{f}_1^2\end{array}$

Substitute $0.300\text{g}$ for $m$, $65.5\text{cm}$ for $L$, and $330.0\text{Hz}$ for $f_1$ in the above equation.

$\begin{array}{c}F=4\left(0.300\text{g}\right)\left(\frac{0.001\text{kg}}{1\text{g}}\right)\left(65.5\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right){\left(330.0\text{Hz}\right)}^2\\ =85.6\text{N}\end{array}$

Therefore, the tension in the string is $85.6\text{N}$.

(b)

To determine

The speed of the waves travel on the string.

Answer to Problem 40P

The speed of the waves travel on the string is $432\text{m/s}$.

Explanation of Solution

Write the relation between waves travel on the string with a speed.

$v=2L{f}_1$ (VI)

Here, $v$ is the speed of the waves travel on the string.

Conclusion:

Substitute $65.5\text{cm}$ for $L$ and $330.0\text{Hz}$ for $f_1$ in the equation (VI).

$\begin{array}{c}v=2\left(65.5\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\left(330.0\text{Hz}\right)\\ =432.3\text{m/s}\\ =432\text{m/s}\end{array}$

Therefore, the speed of the waves travel on the string is $432\text{m/s}$.

(c)

To determine

What is the fundamental frequency of the slide whistle?

Answer to Problem 40P

The fundamental frequency of the slide whistle is $335\text{Hz}$.

Explanation of Solution

Since the other musician has been lowering the frequency of whistle.

The frequency of the musician playes with slide whistle is.

$f=f_1+f_2$ (VII)

Here, $f$ is the frequency of the musician playes with slide whistle and $f_2$ is the beat frequency.

Conclusion:

Substitute $5\text{Hz}$ for $f_2$ and $330.0\text{Hz}$ for $f_1$ in the equation (VII).

$\begin{array}{c}f=330.0\text{Hz}+5\text{Hz}\\ =335\text{Hz}\end{array}$

Therefore, the fundamental frequency of the slide whistle is $335\text{Hz}$.

(d)

To determine

The length of the open tube in the slide whistle.

Answer to Problem 40P

The length of the open tube in the slide whistle is $0.256\text{m}$.

Explanation of Solution

Write the equation for the second harmonic frequency of standing waves.

$f=\frac{v}{4L}$ (VIII)

Rewrite the equation (VIII) for the length of the slide whistle.

$L=\frac{v}{4f}$ (IX)

Here, $v$ is the speed of the sound waves.

Conclusion:

Substitute $335\text{Hz}$ for $f$ and $343\text{m/s}$ for $v$ in the equation (IX).

$\begin{array}{c}L=\frac{343\text{m/s}}{4\left(335\text{Hz}\right)}\\ =0.256\text{m}\end{array}$

Therefore, the length of the open tube in the slide whistle is $0.256\text{m}$.