Chapter 18, Problem 53P

(a)

To determine

Find the depth of the well.

Answer to Problem 53P

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

Explanation of Solution

Here, the well acts like a pipe open at one end and closed at the other end. The normal mode of vibration of the pipe is odd harmonics.

Write the expression for depth of the well.

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

Here, $L$ is the depth of the well, $n$ is the number of modes of vibration, and ${\lambda }_1$ is the wavelength at first instant.

Write the relation between frequency, wavelength and speed.

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

Here, $v$ is the speed of the sound in air and $f_1$ is the resonance frequency at first instant.

Write the relation between frequency, wavelength and speed at next instant resonance.

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

Here, ${\lambda }_2$ is the wavelength of the wave for next instant resonance and $f_2$ is the resonance frequency at other instant.

The expression for the next resonance instant, depth of the well is,

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

Conclusion:

Substitute equation (II) in equation (I).

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

Substitute equation (III) in equation (IV).

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

Solve the equation (V) and (VI).

  $\left(2n-1\right)\frac{v}{4f_1}=\left(2n+1\right)\frac{v}{4f_2}$

Substitute $343\text{m/s}$ for $v$, $51.87\text{Hz}$ for $f_1$, and $59.85\text{Hz}$ for $f_2$ in the above equation.

  $\begin{array}{c}\left(2n-1\right)\frac{343\text{m/s}}{4\left(51.87{\text{s}}^{-1}\right)}=\left(2n+1\right)\frac{343\text{m/s}}{4\left(59.85{\text{s}}^{-1}\right)}\\ \frac{\left(2n-1\right)}{51.87}=\frac{\left(2n+1\right)}{59.85}\\ 59.85\left(2n-1\right)=51.87\left(2n+1\right)\end{array}$

Solve the above equation for $n$.

  $\begin{array}{c}119.7n-59.85=103.74n+51.87\\ 119.7n-103.74n=51.87+59.85\\ 15.96n=111.72\\ n=7\end{array}$

Substitute $7$ for $n$, $343\text{m/s}$ for $v$, and $51.87\text{Hz}$ for $f_1$ in equation (V) to find $L$.

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

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

(b)

To determine

The number of antinodes are formed in standing wave at frequency $51.87\text{Hz}$.

Answer to Problem 53P

The number of 7 antinodes are formed in standing wave at frequency $51.87\text{Hz}$.

Explanation of Solution

Write the expression for fundamental frequency for first harmonic.

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

Here, $f_1$ is the fundamental frequency.

Write the expression to calculate the number of harmonics.

  $n^{\prime }=\frac{f_2}{f_1}$                                                                                            (VIII)

Here, $n^{\prime }$ is the number of wave pattern formed and $f_2$ is the standing wave frequency at first resonance instant.

Conclusion:

Substitute $343\text{m/s}$ for $v$ and $21.5\text{m}$ for $L$ in equation (VI) to find $f_1$.

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

The pattern for above frequency is AN, similarly the 3^rd harmonic pattern is ANAN, and the 5^th harmonic is ANANAN etc.

Substitute $3.99\text{Hz}$ for $f_1$ and $51.87\text{Hz}$ for $f_2$ in the equation (VIII) to find $n^{\prime }$.

  $\begin{array}{c}{n}^{\prime }=\frac{51.87\text{Hz}}{3.99\text{Hz}}\\ =13\end{array}$

The pattern with number of antinodes is calculated as,

  $n^{\prime }=\left(2n-1\right)$

Rewrite the above relation for $n$.

  $\begin{array}{c}{n}^{\prime }+1=2n\\ n=\frac{n^{\prime }+1}{2}\end{array}$

Substitute 13 for $n^{\prime }$ in the above relation.

  $\begin{array}{c}n=\frac{13+1}{2}\\ =7\end{array}$

Therefore, the number of 7 antinodes are formed in standing wave at frequency $51.87\text{Hz}$.