Chapter 14, Problem 55P

(a)

To determine

The depth of the well

Answer to Problem 55P

The well has a depth of $21.5\text{m}$.

Explanation of Solution

The well can be considered as a pipe open at one end and closed at the other end. The fundamental modes of vibration of the pipe now show odd harmonics. Assume the depth of the well to be $L$ and the speed of sound to be $v$.

Write the equation for the depth of the well given that the vibrations are odd harmonics.

    $L=\left(2n-1\right)\frac{{\lambda }_1}{4}$        (I)

Here, $n$ is an integer representing the $n^{th}$ mode of vibration and ${\lambda }_1$ is the corresponding wavelength.

Write the equation for the wavelength in the $n^{th}$ mode of vibration.

    ${\lambda }_1=\frac{v}{f_1}$        (II)

Here, $v$ is the speed of sound and $f_1$ is the frequency in the $n^{th}$ mode of vibration.

Substitute equation (II) in equation (I).

    $L=\left(2n-1\right)\frac{v}{4f_1}$        (III)

Write the equation for the depth of the well for the next resonance in succession to the previous.

    $L=\left[2\left(n+1\right)-1\right]\frac{{\lambda }_2}{4}$        (IV)

Here, $n+1$ is an integer representing the ${\left(n+1\right)}^{th}$ mode of vibration and ${\lambda }_2$ is the corresponding wavelength

Write the equation for the wavelength in the ${\left(n+1\right)}^{th}$ mode of vibration.

    ${\lambda }_2=\frac{v}{f_2}$        (V)

Here, $v$ is the speed of sound and $f_2$ is the frequency in the ${\left(n+1\right)}^{th}$ mode of vibration

Substitute equation (V) in equation (IV).

    $L=\left[2\left(n+1\right)-1\right]\frac{v}{4f_2}$        (VI)

Conclusion:

Substitute $343\text{m/s}$ for $v$ and $51.87\text{Hz}$ for $f_1$.in equation (III).

    $L=\left(2n-1\right)\frac{343\text{m/s}}{4\left(51.87\text{Hz}\right)}$        (VI)

Substitute $343\text{m/s}$ for $v$ and $59.85\text{Hz}$ for $f_2$.in equation (VI).

    $L=\left[2\left(n+1\right)-1\right]\frac{343\text{m/s}}{4\left(59.85\text{Hz}\right)}$        (VII)

Compare equation (VI) and equation (VII) as they represent two successive resonances.

    $\begin{array}{c}\left(2n-1\right)\frac{343\text{m/s}}{4\left(51.87\text{Hz}\right)}=\left[2\left(n+1\right)-1\right]\frac{343\text{m/s}}{4\left(59.85\text{Hz}\right)}\\ \frac{2n+1}{59.85}=\frac{2n-1}{51.87}\\ 119.7n-103.74n=111.72\\ n=7\end{array}$

Substitute $7$ for $n$ in equation (VI).

    $\begin{array}{c}L=\left(2\left(7\right)-1\right)\frac{343\text{m/s}}{4\left(51.87\text{Hz}\right)}\\ =\left(13\right)\frac{343\text{m/s}}{207.48{\text{s}}^{\text{-1}}}\\ =21.5\text{m}\end{array}$

Therefore, the depth of the well is $21.5\text{m}$.

(b)

To determine

The number of antinodes

Answer to Problem 55P

There are seven antinodes in the standing wave.

Explanation of Solution

The well is open at one end and closed at the other end thereby showing odd harmonics. Write the equation for the frequency of the first harmonic of the well.

    $f_1=\frac{v}{4L}$

Here, $v$ is the speed of the sound and $L$ is the depth of the well.

Substitute $343\text{m/s}$ for $v$ and $21.5\text{m}$ for $L$.

    $\begin{array}{c}{f}_1=\frac{343\text{m/s}}{4\left(21.5\text{m}\right)}\\ =3.99\text{Hz}\end{array}$        (VIII)

Conclusion:

The pattern of the first harmonic is $AN$ where, $A$ represents an antinode and $N$ represents a node. Similarly, the pattern of the third harmonic is $ANAN$ and that for the fifth harmonic is $ANANAN$. Therefore, a pattern with $n$ antinodes belongs to ${\left(2n-1\right)}^{th}$ harmonic.

The ratio of the frequency of the standing wave to the frequency of the first harmonic tells to which harmonic the frequency of the standing waves belongs.

    $\frac{51.87\text{Hz}}{3.99\text{Hz}}=13$

Therefore, the frequency of the standing wave belongs to ${13}^{th}$ harmonic. A pattern with $n$ antinodes belongs to ${\left(2n-1\right)}^{th}$ harmonic Hence,

    $\begin{array}{c}2n-1=13\\ 2n=14\\ n=7\end{array}$

Therefore, there are $7$ antinodes in the standing wave.