Chapter 18, Problem 50PQ

(a)

To determine

The wavelengths of fundemental and third harmonic before and after the addition of water.

Answer to Problem 50PQ

Value of fundamental wavelength with and without water will be the same and it is $4.00\text{m}$.

Wavelength of third harmonic with and without water will be the same and it is $1.33\text{m}$.

Explanation of Solution

Write the equation find the wavelength of first harmonic.

    ${\lambda }_1=\frac{4L}{n}n=1,2,\mathrm{3....}$

Here, ${\lambda }_1$ is the wavelength of fundamental and $L$ is the length of pipe.

Note that there is no any term in the above equation which denotes the influence of medium which pipe is filled. It means that fundamental wavelength remain unchanged even the pipe is filled with water.

Write the equation find the wavelength of first harmonic.

    ${\lambda }_3=\frac{4L}{n}n=1,2,\mathrm{3....}$

Here, ${\lambda }_3$ is the third harmonic

The same explanation in the case of fundamental wavelength is applicable to this case also.. Wavelength of third harmonic remain unchanged even the pipe is filled with water.

Conclusion:

Substitute $1.00\text{m}$ for $L$ and $1$ for $n$ in the equation for ${\lambda }_1$.

    $\begin{array}{c}{\lambda }_1=\frac{4\left(1.00\text{m}\right)}{1}\\ =4.00\text{m}\end{array}$

Substitute $1.00\text{m}$ for $L$ and $3$ for $n$ in the equation for ${\lambda }_3$.

    $\begin{array}{c}{\lambda }_1=\frac{4\left(1.00\text{m}\right)}{3}\\ =1.33\text{m}\end{array}$${\lambda }_3=\frac{4L}{n}n=1,2,\mathrm{3....}$

Therefore, value of fundamental wavelength with and without water will be the same and it is $4.00\text{m}$.

Wavelength of third harmonic with and without water will be the same and it is $1.33\text{m}$.

(b)

To determine

The frequencies of fundemental and third harmonic before and after the addition of water.

Answer to Problem 50PQ

Fundamental frequencies in air and water are $85.8\text{Hz}$ and $371\text{Hz}$ respectively.

Third harmonic frequencies in air and water are $371\text{Hz}$ and $1110\text{Hz}$ respectively.

Explanation of Solution

Fundamental and third harmonic frequencies are function of speed of sound and it changes with medium. So, fundamental and third harmonic frequencies will be different in air and water.

Write the equation find the fundamental frequency in air.

    $f_{\text{1,air}}=\frac{n{v}_{air}}{4L}$                                                                                                             (I)

Here, $f_{\text{1,air}}$ is the fundamental frequency in air and $v_{air}$ is the speed of sound in air.

Write the equation find the fundamental frequency in water.

    ${\text{f}}_{\text{1,water}}=\frac{n{v}_{water}}{4L}$                                                                                                         (II)

Here, ${\text{f}}_{\text{1,water}}$ is the fundamental frequency in water and $v_{water}$ is the speed of sound in water

Write the equation find the third harmonic frequency in air.

    ${\text{f}}_{\text{3,water}}=\frac{n{v}_{air}}{4L}$                                                                                                         (III)

Here, $f_{3,\text{water}}$ is the third harmonic frequency in water.

Write the equation find the third harmonic frequency in water.

    ${\text{f}}_{\text{3,water}}=\frac{n{v}_{water}}{4L}$                                                                                                       (IV)

Conclusion:

Substitute $1$ for $n$, $343\text{m}/\text{s}$ for $v_{air}$, and $1.00\text{m}$ for $L$  in equation (I) to find ${\text{f}}_{\text{1,air}}$.

    $\begin{array}{c}{f}_{\text{1,air}}=\frac{\left(1\right)\left(343\text{m}/\text{s}\right)}{4\left(1.00\text{m}\right)}\\ =85.8\text{Hz}\end{array}$

Substitute $1$ for $n$, $1482\text{m}/\text{s}$ for $v_{air}$, and $1.00\text{m}$ for $L$  in equation (I) to find ${\text{f}}_{\text{1,water}}$.

    $\begin{array}{c}{\text{f}}_{\text{1,water}}=\frac{\left(1\right)\left(1482\text{m}/\text{s}\right)}{4\left(1.00\text{m}\right)}\\ =371\text{Hz}\end{array}$

Substitute $3$ for $n$, $343\text{m}/\text{s}$ for $v_{air}$, and $1.00\text{m}$ for $L$  in equation (I) to find ${\text{f}}_{\text{3,air}}$.

    $\begin{array}{c}{f}_{\text{3,air}}=\frac{\left(3\right)\left(343\text{m}/\text{s}\right)}{4\left(1.00\text{m}\right)}\\ =257\text{Hz}\end{array}$

Substitute $3$ for $n$, $343\text{m}/\text{s}$ for $v_{air}$, and $1.00\text{m}$ for $L$  in equation (I) to find ${\text{f}}_{\text{3,water}}$.

    $\begin{array}{c}{f}_{\text{3,water}}=\frac{\left(3\right)\left(1482\text{m}/\text{s}\right)}{4\left(1.00\text{m}\right)}\\ =1110\text{Hz}\end{array}$

Therefore, fundamental frequencies in air and water are $85.8\text{Hz}$ and $371\text{Hz}$ respectively.

Third harmonic frequencies in air and water are $371\text{Hz}$ and $1110\text{Hz}$ respectively.

(c)

To determine

Check whether lower pitched or high pitched sound will hear on tube is filled with water.

Answer to Problem 50PQ

High pitch sound will hear on tube is filled with water.

Explanation of Solution

Higher the pitch, higher will be the frequency of sound. Note that the frequency of sound when tube is filled with water is greater than that of tube filled with air. So the listener should hear higher pitch sound when the tube is filled with water.

Therefore, high pitch sound will hear on tube is filled with water.