Chapter 12, Problem 46P

(a)

To determine

The lowest frequency.

Answer to Problem 46P

  $\mathrm{f}=343\text{ Hz}\text{.}$

Explanation of Solution

Given:

Distance between loudspeakers $\mathrm{d} =1.8\text{ m}$

Temperature, $\left(\mathrm{T}\right)$

  $=20°\text{C}$

The distance between one speaker to person ${\mathrm{d}}_0=3.00\text{m}$

The distance between the second speaker to person ${\mathrm{d}}_1=3.50\text{m}$ .

Formula used:

Write the expression for the destructive interference.

  ${\mathrm{d}}_1-{\mathrm{d}}_0=\left(2\mathrm{m}+1\right)\frac{\mathrm{\lambda }}{2}$

Here, ${\mathrm{d}}_0$ is the distance between one speaker to person,

  ${\mathrm{d}}_1$ is the distance between the second speaker to person.

  $\mathrm{m}$ is the order of interference.

  $\mathrm{\lambda }$ is the wavelength.

Write the expression of the frequency.

  $\mathrm{f}=\frac{\mathrm{v}}{\mathrm{\lambda }}\text{ Hz}\text{.}$

Where,

  $\mathrm{f}$ = frequency, Hz

  $\mathrm{v}$ = speed of wave, m/s

  $\mathrm{\lambda }$ = Wavelength, m

  $$

Calculation:

Write the expression for the destructive interference.

  ${\mathrm{d}}_1-{\mathrm{d}}_0=\left(2\mathrm{m}+1\right)\frac{{\mathrm{\lambda }}_1}{2}$

Here $\mathrm{m}=0$

Substitute the values in the above expression.

  $\begin{array}{c}3.5-3.0=\left(2(0)+1\right)\frac{{\mathrm{\lambda }}_1}{2}\\ \frac{{\mathrm{\lambda }}_1}{2}=0.5\\ {\mathrm{\lambda }}_1=1\end{array}$

Write the expression of frequency.

  ${\mathrm{f}}_1=\frac{\mathrm{v}}{{\mathrm{\lambda }}_1}\text{ Hz}\text{.}$

Speed of the wave is = 343 m/s

  ${\mathrm{f}}_1=\frac{343}{1}\text{ Hz}$

  ${\mathrm{f}}_1=343\text{ Hz}\text{.}$

The lowest frequency is $343\text{ Hz}.$

Conclusion:

The lowest frequency is $343\text{ Hz}.$

(b)

To determine

To Calculate: Two other frequencies that will result in destructive interference at the same point.

Answer to Problem 46P

1040 Hz and 1715 Hz. at $\mathrm{n}=3$ and $\mathrm{n}=5$ respectively.

Explanation of Solution

Given:

Loudspeaker distance $\left(\mathrm{D}\right) =1.8\text{ m}$$$

Temperature, $\left(\mathrm{T}\right)$

  $={20}^0\text{C}$

Formula used:

Write the expression for the destructive interference.

  ${\mathrm{d}}_1-{\mathrm{d}}_0=\left(2\mathrm{m}+1\right)\frac{\mathrm{\lambda }}{2}$

Here, ${\mathrm{d}}_0$ is the distance between one speaker to person,

  ${\mathrm{d}}_1$ is the distance between the second speaker to person.

  $\mathrm{m}$ is the order of interference.

  $\mathrm{\lambda }$ is the wavelength.

Write the expression of the frequency.

  $\mathrm{f}=\frac{\mathrm{v}}{\mathrm{\lambda }}\text{ Hz}\text{.}$

Where,

  $\mathrm{f}$ = frequency, Hz

  $\mathrm{v}$ = speed of wave, m/s

  $\mathrm{\lambda }$ = Wavelength, m

  $$

Where,

  $\mathrm{f}$ = frequency, Hz

  $\mathrm{v}$ = speed of wave, m/s

  $\mathrm{\lambda }$ = Wavelength, m

Calculation:

Write the expression for the destructive interference.

  ${\mathrm{d}}_1-{\mathrm{d}}_0=\left(2\mathrm{m}+1\right)\frac{{\mathrm{\lambda }}_2}{2}$

Here $\mathrm{m}=1$

Substitute the values in the above expression.

  $\begin{array}{c}3.5-3.0=\left(2(1)+1\right)\frac{{\mathrm{\lambda }}_2}{2}\\ \frac{3{\mathrm{\lambda }}_2}{2}=0.5\\ {\mathrm{\lambda }}_2=0.33\text{m}\end{array}$

Write the expression for the frequency.

  ${\mathrm{f}}_2=\frac{\mathrm{v}}{{\mathrm{\lambda }}_2}=\frac{343}{0.33}\text{ Hz}$

  ${\mathrm{f}}_2=1040\text{ Hz}$

For third wave, Calculate the wavelength,

Write the expression for the destructive interference.

  ${\mathrm{d}}_1-{\mathrm{d}}_0=\left(2\mathrm{m}+1\right)\frac{{\mathrm{\lambda }}_2}{2}$

Here $\mathrm{m}=2$

Substitute the values in the above expression.

  $\begin{array}{c}3.5-3.0=\left(2(2)+1\right)\frac{{\mathrm{\lambda }}_2}{2}\\ \frac{5{\mathrm{\lambda }}_2}{2}=0.5\\ {\mathrm{\lambda }}_2=0.2\text{m}\end{array}$

Write the expression for the frequency.

  ${\mathrm{f}}_3=\frac{\mathrm{v}}{{\mathrm{\lambda }}_3}=\frac{343}{0.2}\text{ Hz}$

  ${\mathrm{f}}_3=1715\text{ Hz}$

Conclusion:

The other two frequencies are $1040\text{ Hz}$ and $1715\text{ Hz}.$