Chapter 18, Problem 46PQ

(a)

To determine

Whether the three frequencies are the fundamental frequency of the respective bottles.

Answer to Problem 46PQ

Yes, the three frequencies are the fundamental frequency of the respective bottles.

Explanation of Solution

Fundamental frequency as the name itself defines is the lowest frequency of wave generated in an instrument. Ezra can play the beginning notes of the song using the bottles. The notes that he play are B, A, and G. The first bottle is empty but the other two has still some water left inside. The first bottle resonates at $f_{\text{B}}=493.88\text{Hz}$, second bottle at $f_{\text{A}}=440.00\text{Hz}$, and third bottle at $f_{\text{G}}=392.00\text{Hz}$.

All the frequencies are close to each other. If the frequencies were of higher order harmonics, they would be the multiples of the frequency from the empty bottle. As they are very close to each other, they are the fundamental frequency of their respective bottles.

Conclusion:

Therefore, all the three frequencies are the fundamental frequencies of the respective bottles.

(b)

To determine

The bottle which is empty.

Answer to Problem 46PQ

Bottle which produces the G note is empty.

Explanation of Solution

Write the general expression for the fundamental frequency of sound.

  $f_1=\frac{v}{4L}$                                                                                                        (I)

Here, $f_1$ is the fundamental frequency of the sound, $v$ is the velocity of sound, and $L$ is the length of the bottle.

The largest effective length of the vibrating column causes the lowest frequency. That is from the above equation, if the length is maximum, then the resulting frequency will be the minimum. Out of the three frequencies, G note is having the least value of frequency. So the bottle which produces $f_{\text{G}}$ is the empty one.

Conclusion:

Therefore, the bottle which produces the G note is empty.

(c)

To determine

The height of each bottle.

Answer to Problem 46PQ

The height of the empty bottle is $\underset{\_}{0.219\text{m}}$, height of bottle producing $f_{\text{B}}$ is $\underset{\_}{0.174\text{m}}$, and the height of the bottle producing $f_{\text{A}}$ is $\underset{\_}{0.195\text{m}}$.

Explanation of Solution

Solve equation (I) for $L$.

  $L=\frac{v}{4f_1}$                                                                                                   (II)

Use expression (II) for finding the height of the bottle producing G note.

  $L_{\text{G}}=\frac{v}{4f_{\text{G}}}$                                                                                                   (III)

Here, $f_{\text{G}}$ is the fundamental frequency of the bottle producing note G, $v$ is the velocity of sound, and $L_{\text{G}}$ is the height of the bottle producing G note.

Use expression (II) for finding the height of the bottle producing A note.

  $L_{\text{A}}=\frac{v}{4f_{\text{A}}}$                                                                                                 (IV)

Here, $L_{\text{A}}$ is the height of the bottle producing A note, and $f_{\text{A}}$ is the frequency of A note.

Use expression (II) for finding the height of the bottle producing B note.

  $L_{\text{B}}=\frac{v}{4f_{\text{B}}}$                                                                                                       (V)

Here, $L_{\text{B}}$ is the height of the bottle producing B note, and $f_{\text{B}}$ is the frequency of B note.

Conclusion:

Substitute $343\text{m}/\text{s}$ for $v$, and $392.00\text{Hz}$ for $f_{\text{G}}$ in equation (III) to find $L_{\text{G}}$.

  $\begin{array}{c}{L}_{\text{G}}=\frac{343\text{m}/\text{s}}{4\left(392.00\text{Hz}\right)}\\ =0.219\text{m}\end{array}$

Substitute $343\text{m}/\text{s}$ for $v$, and $440.00\text{Hz}$ for $f_{\text{A}}$ in equation (IV) to find $L_{\text{A}}$.

  $\begin{array}{c}{L}_{\text{A}}=\frac{343\text{m}/\text{s}}{4\left(440.00\text{Hz}\right)}\\ =0.195\text{m}\end{array}$

Substitute $343\text{m}/\text{s}$ for $v$, and $493.88\text{Hz}$ for $f_{\text{B}}$ in equation (V) to find $L_{\text{B}}$.

  $\begin{array}{c}{L}_{\text{B}}=\frac{343\text{m}/\text{s}}{4\left(493.88\text{Hz}\right)}\\ =0.174\text{m}\end{array}$

Therefore, height of the empty bottle is $\underset{\_}{0.219\text{m}}$, height of bottle producing $f_{\text{B}}$ is $\underset{\_}{0.174\text{m}}$, and the height of the bottle producing $f_{\text{A}}$ is $\underset{\_}{0.195\text{m}}$.

(d)

To determine

Height of the liquid in the partially filled bottles.

Answer to Problem 46PQ

Height of the liquid in the bottle producing B note is $\underset{\_}{4.5\text{cm}}$, and the height of the liquid in the producing A note is $\underset{\_}{2.4\text{cm}}$.

Explanation of Solution

The height of the liquid inside the bottles is equal to the difference between the height of the empty bottle and the length of the other two bottles with liquid.

Write the expression for the height of the liquid in bottle producing $f_{\text{B}}$,

  $h_{\text{B}}=L_{\text{G}}-L_{\text{B}}$                                                                                             (VI)

Here, $h_{\text{B}}$ is the height of the liquid in bottle producing $f_{\text{B}}$.

Write the expression for the height of the liquid in bottle producing $f_{\text{A}}$,

  $h_{\text{A}}=L_{\text{G}}-L_{\text{A}}$                                                                                             (VII)

Here, $h_{\text{A}}$ is the height of the liquid in bottle producing $f_{\text{A}}$.

Conclusion:

Substitute $0.219\text{m}$ for $L_{\text{G}}$, and $0.174\text{m}$ for $L_{\text{B}}$ in equation (VI) to find $h_{\text{B}}$.

  $\begin{array}{c}{h}_{\text{B}}=0.219\text{m}-0.174\text{m}\\ =\text{0}\text{.045}\text{m}\times \frac{100\text{cm}}{1\text{m}}\\ =4.5\text{cm}\end{array}$

Substitute $0.219\text{m}$ for $L_{\text{G}}$, and $0.195\text{m}$ for $L_{\text{A}}$ in equation (VII) to find $h_{\text{A}}$.

  $\begin{array}{c}{h}_{\text{A}}=0.219\text{m}-0.195\text{m}\\ =\text{0}\text{.024}\text{m}\times \frac{100\text{cm}}{1\text{m}}\\ =2.4\text{cm}\end{array}$

Therefore, the height of the liquid in the bottle producing B note is $\underset{\_}{4.5\text{cm}}$, and the height of the liquid in the producing A note is $\underset{\_}{2.4\text{cm}}$.