Chapter 13, Problem 83QAP

To determine

(a)

The factor by which a point source moves away from the screen in order to lower the intensity by factor of 10.

Answer to Problem 83QAP

The factor by which a point source moves away from the screen in order to lower the intensity by factor of 10 is 3.16.

Explanation of Solution

Given:

The intensity of the new wave is,
  $I_2=\frac{I_1}{10}$

Formula used:

The intensity of the sound wave is given by,
  $I=\frac{P}{4\pi r^2}$

Where,
  $I$ =Intensity
  $P$ =Power
  $r$ =Distance

Calculation:

From inverse square law of sound waves, we get
  $\begin{array}{l}I=\frac{P}{4\pi r^2}\\ I_1\propto \frac{1}{r_1^2}\\ I_2\propto \frac{1}{r_2^2}\end{array}$

  $\begin{array}{l}\frac{\raisebox{1ex}{$I_1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}{I_1}=\frac{r_1^2}{r_2^2}\\ \frac{1}{10}=\frac{r_1^2}{r_2^2}\end{array}$

  $r_2^2=10r_1^2$

  $\begin{array}{l}{r}_2=r_1\sqrt{10}\\ r_2=3.16r_1\end{array}$

Conclusion:

By factor 3.16 the point source of sound wave moves away from the screen.

To determine

(b)

The factor by which a point source moves away from the screen in order to lower the intensity by factor of 3.

Answer to Problem 83QAP

The factor by which a point source moves away from the screen in order to lower the intensity by factor of 3 is 1.73.

Explanation of Solution

Given:

The intensity of the new wave is,
  $I_2=\frac{I_1}{3}$

Formula used:

The intensity of the sound wave is given by,
  $I=\frac{P}{4\pi r^2}$

Where,
  $I$ =Intensity
  $P$ =Power
  $r$ =Distance

Calculation:

From inverse square law of sound waves, we get
  $\begin{array}{l}I=\frac{P}{4\pi r^2}\\ I_1\propto \frac{1}{r_1^2}\\ I_2\propto \frac{1}{r_2^2}\end{array}$

  $\begin{array}{l}\frac{\raisebox{1ex}{$I_1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{I_1}=\frac{r_1^2}{r_2^2}\\ \frac{1}{3}=\frac{r_1^2}{r_2^2}\end{array}$

  $r_2^2=3r_1^2$

  $\begin{array}{l}{r}_2=r_1\sqrt{3}\\ r_2=1.73r_1\end{array}$

Conclusion:

By factor 1.73 the point source of sound wave moves away from the screen.

To determine

(c)

The factor by which a point source moves away from the screen in order to lower the intensity by factor of 2.

Answer to Problem 83QAP

The factor by which a point source moves away from the screen in order to lower the intensity by factor of 2 is 1.41.

Explanation of Solution

Given:

The intensity of the new wave is,
  $I_2=\frac{I_1}{2}$

Formula used:

The intensity of the sound wave is given by,
  $I=\frac{P}{4\pi r^2}$

Where,
  $I$ =Intensity
  $P$ =Power
  $r$ =Distance

Calculation:

From inverse square law of sound waves, we get
  $\begin{array}{l}I=\frac{P}{4\pi r^2}\\ I_1\propto \frac{1}{r_1^2}\\ I_2\propto \frac{1}{r_2^2}\end{array}$

  $\begin{array}{l}\frac{\raisebox{1ex}{$I_1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{I_1}=\frac{r_1^2}{r_2^2}\\ \frac{1}{2}=\frac{r_1^2}{r_2^2}\end{array}$

  $r_2^2=2r_1^2$

  $\begin{array}{l}{r}_2=r_1\sqrt{2}\\ r_2=1.41r_1\end{array}$

Conclusion:

By factor 1.41 the point source of sound wave moves away from the screen.

To determine

(d)

The changes observed when the point source is replaced with hemispherical instead of spherical.

Explanation of Solution

Introduction:

The intensity of sound wave at a distance due to a spherical point source is given by inverse square law of sound.

  $I=\frac{P}{4\pi r^2}$

From inverse square law of sound waves, we get
  $I=\frac{P}{4\pi r^2}$

For hemisphere the surface area is given by,
$\text{Surface}\text{area=3}\pi r^2$

Thus, intensity is given by
  $I=\frac{P}{3\pi r^2}$

When the shape of point source is changed from spherical to hemispherical the intensity of it is increased and becomes $\frac{P}{3\pi r^2}$.

Conclusion:

By changing the point source from spherical to hemispherical the intensity is $I=\frac{P}{3\pi r^2}$