Chapter 18, Problem 18.1OQ

In figure OQ18.1 (page 566), a sound wave of wave-lenght 0.8 m divides into two equal parts that recombine to interfere constructively, with the original difference between their path lengths being |r₂ – r₁| = 0.8 m. Rank the following situations according to the intensity of sound at the receiver from the highest to the lowest. Assume the tube walls absorb no sound energy. Give equal ranks to situations in which the intensity is equal.

(a) From its original position, the sliding section is moved out by 0.1 m. (b) Next it slides out an additional 0.1 m. (c) It slides out still another 0.1 m. (d) It slides out 0.1 m more.

To determine

The ranking of the situations according to the intensity of sound at the receiver in descending order.

Answer to Problem 18.1OQ

The ranking of the situations according to the intensity of sound at the receiver in descending order is $d>a=c>b$.

Explanation of Solution

Given info: The wavelength of the sound wave is $\lambda =0.8\text{m}$ and the difference in the paths is $\left|r_2-r_1\right|=0.8\text{m}$.

Write the expression for the intensity heard by the receiver.

$I=I_0{\cos }^2\frac{\varphi }{2}$ . (1)

Here,

$\varphi$ is the phase difference.

$I$ is the intensity heard by the receiver.

$I_0$ is the initial intensity.

For Case (a);

The section is moved out by $0.1\text{m}$, then the path length $r_2$ is increased by $0.2\text{m}$

The path length $r_2$ is increased so, the path difference is,

$\begin{array}{c}\left|r_2-r_1\right|=0.8\text{m}+\text{0}\text{.2}\text{m}\\ =\text{1}\text{m}\end{array}$

Thus, the value of $\left|r_2-r_1\right|$ is $1\text{m}$.

Write the expression for phase difference,

$\varphi =\left(\frac{2\pi }{\lambda }\right)\left(r_2-r_1\right)$

Substitute $\text{1}\text{m}$ for $\left(r_2-r_1\right)$ and $0.8\text{m}$ for $\lambda$ in above equation.

$\begin{array}{c}\varphi =\left(\frac{2\pi }{0.8\text{m}}\right)\left(\text{1}\text{m}\right)\\ =2.5\pi \end{array}$

Thus, the value of $\varphi$ is $2.5\pi$.

Substitute $2.5\pi$ for $\varphi$ and $I_{\text{a}}$ for $I$ in equation (1)

$\begin{array}{c}{I}_{\text{a}}=I_0{\cos }^2\frac{2.5\pi }{2}\\ =\frac{I_0}{2}\end{array}$

Thus the value of $I_{\text{a}}$ is $\frac{I_0}{2}$.

For Case (b);

The section is again moved out by $0.1\text{m}$, then the path length $r_2$ is increased by $0.2\text{m}$

The path length $r_2$ is increased so, the path difference is,

$\begin{array}{c}\left|r_2-r_1\right|=1\text{m}+\text{0}\text{.2}\text{m}\\ =\text{1}\text{.2}\text{m}\end{array}$

Thus, the value of $\left|r_2-r_1\right|$ is $1.2\text{m}$.

Write the expression for phase difference,

$\varphi =\left(\frac{2\pi }{\lambda }\right)\left(r_2-r_1\right)$

Substitute $1.2\text{m}$ for $\left(r_2-r_1\right)$ and $0.8\text{m}$ for $\lambda$ in above equation.

$\begin{array}{c}\varphi =\left(\frac{2\pi }{0.8\text{m}}\right)\left(\text{1}\text{.2}\text{m}\right)\\ =3\pi \end{array}$

Thus, the value of $\varphi$ is $3\pi$.

Substitute $3\pi$ for $\varphi$ and $I_{\text{b}}$ for $I$ in equation (1)

$\begin{array}{c}{I}_{\text{b}}=I_0{\cos }^2\frac{3\pi }{2}\\ =0\end{array}$

Thus the value of $I_{\text{b}}$ is $0$.

For Case (c);

The section is again moved out by $0.1\text{m}$, then the path length $r_2$ is increased by $0.2\text{m}$

The path length $r_2$ is increased so, the path difference is,

$\begin{array}{c}\left|r_2-r_1\right|=1.2\text{m}+\text{0}\text{.2}\text{m}\\ \left(r_2-r_1\right)=\text{1}\text{.4}\text{m}\end{array}$

Thus, the value of $\left|r_2-r_1\right|$ is $\text{1}\text{.4}\text{m}$.

Write the expression for phase difference,

$\varphi =\left(\frac{2\pi }{\lambda }\right)\left(r_2-r_1\right)$

Substitute $\text{1}\text{.4}\text{m}$ for $\left(r_2-r_1\right)$ and $0.8\text{m}$ for $\lambda$ in above equation.

$\begin{array}{c}\varphi =\left(\frac{2\pi }{0.8\text{m}}\right)\left(\text{1}\text{.4}\text{m}\right)\\ =3.5\pi \end{array}$

Thus, the value of $\varphi$ is $3.5\pi$.

Substitute $3.5\pi$ for $\varphi$ and $I_{\text{c}}$ for $I$ in equation (1)

$\begin{array}{c}{I}_{\text{c}}=I_0{\cos }^2\frac{3.5\pi }{2}\\ =\frac{I_0}{2}\end{array}$

Thus, the value of $I_{\text{c}}$ is $\frac{I_0}{2}$.

For Case (d);

The section is again moved out by $0.1\text{m}$, then the path length $r_2$ is increased by $0.2\text{m}$

The path length $r_2$ is increased so, the path difference is,

$\begin{array}{c}\left|r_2-r_1\right|=1.4\text{m}+\text{0}\text{.2}\text{m}\\ =\text{1}\text{.6}\text{m}\end{array}$

Thus, the value of $\left(r_2-r_1\right)$ is $\text{1}\text{.6}\text{m}$.

Write the expression for phase difference,

$\varphi =\left(\frac{2\pi }{\lambda }\right)\left(r_2-r_1\right)$

Substitute $\text{1}\text{.6}\text{m}$ for $\left(r_2-r_1\right)$ and $0.8\text{m}$ for $\lambda$ in above equation.

$\begin{array}{c}\varphi =\left(\frac{2\pi }{0.8\text{m}}\right)\left(\text{1}\text{.6}\text{m}\right)\\ =4\pi \end{array}$

Thus, the value of $\varphi$ is $4\pi$.

Substitute $4\pi$ for $\varphi$ and $I_{\text{d}}$ for $I$ in equation (1)

$\begin{array}{c}{I}_{\text{d}}=I_0{\cos }^2\frac{4\pi }{2}\\ =I_0\end{array}$

Thus, the value of $I_{\text{d}}$ is $I_0$.

The ranking of the intensity of each case is,

$\begin{array}{l}{I}_0>\frac{I_0}{2}=\frac{I_0}{2}>0\\ I_{\text{d}}>I_{\text{a}}=I_{\text{c}}>I_{\text{b}}\\ d>a=c>b\end{array}$

Conclusion:

Therefore, the ranking of the situations according to the intensity of sound at the receiver in descending order is $d>a=c>b$.