Chapter 16, Problem 76P

(a)

To determine

The two resonant frequencies.

Explanation of Solution

Given:

The distance between nodes for first case is $6.94\text{cm}$ .

The distance between nodes for second case is $5.40\text{cm}$ .

Formula used:

Write expression for the wavelength of the wave.

  ${\lambda }_n=2d$

Here, d is the distance between two nodes.

Write expression for frequency.

  $f_n=\frac{v}{{\lambda }_n}$

Here, $v$ is the velocity of the sound.

Substitute $2d$ for ${\lambda }_n$ in above expression.

  $f_n=\frac{v}{2d}$........ (1)

Calculation:

Substitute $343\text{m}/\text{s}$ for $v$ and $6.94\text{cm}$ for $d$ in equation (1).

  $\begin{array}{l}{f}_n=\frac{343\text{m}/\text{s}}{2\left(6.94\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ f_n=2471\text{Hz}\end{array}$

Substitute ${f^{\prime }}_n$

  $343\text{m}/\text{s}$ for $v$ and $5.40\text{cm}$ for $d$ in equation (1).

  $\begin{array}{l}{f^{\prime }}_n=\frac{343\text{m}/\text{s}}{2\left(5.40\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ {f^{\prime }}_n=3175\text{Hz}\end{array}$

Conclusion:

Thus, the two frequencies are $2471\text{Hz}$ and $3175\text{Hz}$ .

(b)

To determine

The fundamental frequency of the tube.

Explanation of Solution

Given:

The distance between nodes for first case is $6.94\text{cm}$ .

The distance between nodes for second case is $5.40\text{cm}$ .

Formula used:

Write expression for the wavelength of the wave.

  ${\lambda }_n=2d$

Here, d is the distance between two nodes.

Write expression for frequency.

  $f_n=\frac{v}{{\lambda }_n}$

Here, $v$ is the velocity of the sound.

Substitute $2d$ for ${\lambda }_n$ in above expression.

  $f_n=\frac{v}{2d}$........ (1)

Write expression for frequency of $n\text{th}$ resonance.

  $f_n=n{f}_1$

Rearrange above expression for $n$ .

  $n=\frac{f_n}{f_1}$

Write expression for next resonance.

  $f_{n+2}=\left(n+2\right)f_1$

Substitute $\frac{f_n}{f_1}$ for $n$ in above expression.

  $f_{n+2}=\left(\frac{f_n}{f_1}+2\right)f_1$........ (2)

Calculation:

Substitute $343\text{m}/\text{s}$ for $v$ and $6.94\text{cm}$ for $d$ in equation (1).

  $\begin{array}{l}{f}_n=\frac{343\text{m}/\text{s}}{2\left(6.94\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ f_n=2471\text{Hz}\end{array}$

Substitute $f_{n+2}$ for $f_n$ , $343\text{m}/\text{s}$ for $v$ and $5.40\text{cm}$ for $d$ in equation (1).

  $\begin{array}{l}{f}_{n+2}=\frac{343\text{m}/\text{s}}{2\left(5.40\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ f_{n+2}=3175\text{Hz}\end{array}$

Substitute $2471\text{Hz}$ for $f_n$ and $3175\text{Hz}$ for $f_{n+2}$ and solve for $f_1$ in equation (2).

  $\begin{array}{c}3175\text{Hz}=\left(\frac{2471\text{Hz}}{f_1}+2\right)f_1\\ f_1=353\text{Hz}\end{array}$

Conclusion:

Thus, the fundamental frequency is $353\text{Hz}$ .

(c)

To determine

The value of $n$ and $n+2$ .

Explanation of Solution

Given:

The distance between nodes for first case is $6.94\text{cm}$ .

The distance between nodes for second case is $5.40\text{cm}$ .

Formula used:

Write expression for the wavelength of the wave.

  ${\lambda }_n=2d$

Here, d is the distance between two nodes.

Write expression for frequency.

  $f_n=\frac{v}{{\lambda }_n}$

Here, $v$ is the velocity of the sound.

Substitute $2d$ for ${\lambda }_n$ in above expression.

  $f_n=\frac{v}{2d}$........ (1)

Write expression for frequency of $n\text{th}$ resonance.

  $f_n=n{f}_1$

Rearrange above expression for $n$ .

  $n=\frac{f_n}{f_1}$........ (2)

Write expression for next resonance.

  $f_{n+2}=\left(n+2\right)f_1$

Rearrange above expression for $n+2$ .

  $n+2=\frac{f_{n+2}}{f_n}$........ (3)

Substitute $\frac{f_n}{f_1}$ for $n$ in above expression.

  $f_{n+2}=\left(\frac{f_n}{f_1}+2\right)f_1$........ (4)

Calculation:

Substitute $343\text{m}/\text{s}$ for $v$ and $6.94\text{cm}$ for $d$ in equation (1).

  $\begin{array}{l}{f}_n=\frac{343\text{m}/\text{s}}{2\left(6.94\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ f_n=2471\text{Hz}\end{array}$

Substitute $f_{n+2}$ for $f_n$ , $343\text{m}/\text{s}$ for $v$ and $5.40\text{cm}$ for $d$ in equation (1).

  $\begin{array}{l}{f}_{n+2}=\frac{343\text{m}/\text{s}}{2\left(5.40\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ f_{n+2}=3175\text{Hz}\end{array}$

Substitute $2471\text{Hz}$ for $f_n$ and $3175\text{Hz}$ for $f_{n+2}$ and solve for $f_1$ in equation (4).

  $\begin{array}{c}3175\text{Hz}=\left(\frac{2471\text{Hz}}{f_1}+2\right)f_1\\ f_1=353\text{Hz}\end{array}$

Substitute $353\text{Hz}$ for $f_1$ and $2471\text{Hz}$ for $f_n$ in equation (2).

  $\begin{array}{l}n=\frac{2471\text{Hz}}{353\text{Hz}}\\ n=7\end{array}$

Substitute $353\text{Hz}$ for $f_1$ and $3175\text{Hz}$ for $f_{n+2}$ in equation (3).

  $\begin{array}{l}n+2=\frac{3175\text{Hz}}{353\text{Hz}}\\ n+2=9\end{array}$

Conclusion:

Thus, the value of $n$ is $7$ and $n+2$ is $9$ .