Chapter 11, Problem 57P

(a)

To determine

The intensity of sound when the waves interfere constructively.

Answer to Problem 57P

The intensity of sound when the waves interfere constructively is $\underset{\_}{0.25{\text{W/m}}^2}$.

Explanation of Solution

Given that the individual intensities of the waves are $0.040{\text{W/m}}^2$, and $0.090{\text{W/m}}^2$.

The intensity of the wave is directly proportional to the square of its amplitude.

Let $A_1$ and $A_2$ be the amplitudes of the two waves and $I_1$ and $I_2$ be their respective intensities.

Write the expression for the ratio of the amplitudes of the two waves.

$\frac{A_1}{A_2}=\sqrt{\frac{I_1}{I_2}}$ (I)

Solve equation (I) for $A_1$.

$A_1=A_2\sqrt{\frac{I_1}{I_2}}$ (II)

Write the expression of the amplitude $A$ of the resultant wave after constructive interference.

$A=A_1+A_2$ (III)

Use equation (II) in (III).

$\begin{array}{c}A=A_2\sqrt{\frac{I_1}{I_2}}+A_2\\ =A_2\left(\sqrt{\frac{I_1}{I_2}}+1\right)\end{array}$ (IV)

Let $I$ be the resultant intensity after constructive interference.

Write the expression for the ratio of the amplitudes of the resultant wave and the wave with amplitude $A$.

$\frac{A}{A_2}=\sqrt{\frac{I}{I_2}}$ (V)

Use equation (IV) in (V) and solve for $I$.

$\begin{array}{c}\frac{A_2\left(\sqrt{\frac{I_1}{I_2}}+1\right)}{A_2}=\sqrt{\frac{I}{I_2}}\\ I=I_2{\left(\sqrt{\frac{I_1}{I_2}}+1\right)}^2\end{array}$ (VI)

Conclusion:

Substitute $0.040{\text{W/m}}^2$ for $I_1$ and $0.090{\text{W/m}}^2$ for $I_2$ in equation (VI) to find $I$.

$\begin{array}{c}I=\left(0.090{\text{W/m}}^2\right){\left(\sqrt{\frac{0.040{\text{W/m}}^2}{0.090{\text{W/m}}^2}}+1\right)}^2\\ =0.25{\text{W/m}}^2\end{array}$

Therefore, the intensity of sound when the waves interfere constructively is $\underset{\_}{0.25{\text{W/m}}^2}$.

(b)

To determine

The intensity of sound when the waves interfere destructively.

Answer to Problem 57P

The intensity of sound when the waves interfere destructively is $\underset{\_}{0.010{\text{W/m}}^2}$.

Explanation of Solution

Write the expression of the amplitude $A$ of the resultant wave after destructive interference.

$A=\left|A_1-A_2\right|$ (VII)

Use equation (II) in (VII).

$\begin{array}{c}A=\left|A_2\sqrt{\frac{I_1}{I_2}}-A_2\right|\\ =\left|A_2\left(\sqrt{\frac{I_1}{I_2}}-1\right)\right|\end{array}$ (VIII)

Use equation (VIII) in (V) and solve for $I$.

$\begin{array}{c}\frac{\left|A_2\left(\sqrt{\frac{I_1}{I_2}}-1\right)\right|}{A_2}=\sqrt{\frac{I}{I_2}}\\ I=I_2{\left|\sqrt{\frac{I_1}{I_2}}-1\right|}^2\end{array}$ (IX)

Conclusion:

Substitute $0.040{\text{W/m}}^2$ for $I_1$ and $0.090{\text{W/m}}^2$ for $I_2$ in equation (IX) to find $I$.

$\begin{array}{c}I=\left(0.090{\text{W/m}}^2\right){\left|\sqrt{\frac{0.040{\text{W/m}}^2}{0.090{\text{W/m}}^2}}-1\right|}^2\\ =0.01{\text{W/m}}^2\end{array}$

Therefore, the intensity of sound when the waves interfere destructively is $\underset{\_}{0.010{\text{W/m}}^2}$.

(c)

To determine

The intensity of sound after interference if the waves are incoherent.

Answer to Problem 57P

The intensity of sound after interference if the waves are incoherent is $\underset{\_}{0.130{\text{W/m}}^2}$.

Explanation of Solution

Write the expression for the resultant intensity when the incoherent waves interfere.

$I=I_1+I_2$ (X)

Conclusion:

Substitute $0.040{\text{W/m}}^2$ for $I_1$ and $0.090{\text{W/m}}^2$ for $I_2$ in equation (X) to find $I$.

$\begin{array}{c}I=0.040{\text{W/m}}^2+0.090{\text{W/m}}^2\\ =0.130{\text{W/m}}^2\end{array}$

Therefore, the intensity of sound after interference if the waves are incoherent is $\underset{\_}{0.130{\text{W/m}}^2}$.