Chapter 12, Problem 18P

(a)

To determine

To Calculate: The increasing factor of the intensity.

Answer to Problem 18P

The intensity is increased by a factor of 9 if the amplitude of a sound wave is tripled.

Explanation of Solution

Given info:

The amplitude of a sound wave is triple.

Formula used:

The intensity is given as,

  $I=2\rho v{\pi }^2{f}^2{A}^2$

Here, $A$ is the amplitude of the wave,

  $\rho$ is the density of the medium,

  $f$ is the frequency of the sound wave,

  $v$ is the speed of the wave.

Calculation:

According to the given formula, the intensity is directly proportional to the square of the amplitude of the wave. Therefore, when the amplitude of the sound wave is tripled, then intensity becomes,

  $\begin{array}{l}I={\left(3A\right)}^2\\ =9A^2\end{array}$

Conclusion:

Hence, the intensity is increased by a factor of 9.

(b)

To determine

To Calculate: The change in the sound level in dB.

Answer to Problem 18P

The sound level is increased by a factor of $9.5\text{}\text{dB}$ .

Explanation of Solution

Given:

The intensity is increased by a factor of 9 if the amplitude of a sound wave is tripled.

Formula used:

The intensity is given as,

  $\beta \left(\text{in}\text{dB}\right)=10\log \frac{I}{I_0}$

Here,

  $I$ is intensity of given sound,

  $I_0$ is the hearing intensity.

Calculation:

Now, substitute the value of first intensity and second intensity which is the 9 times of the first one.

The sound level is,

  $\begin{array}{l}\beta \left(\text{in}\text{dB}\right)=10\log \frac{I}{I_0}\\ =10\log \frac{9I_0}{I_0}\\ =10\log 9\\ \beta \left(\text{in}\text{dB}\right)=9.5\text{dB}\end{array}$

Conclusion:

The change in sound level is $9.5\text{dB}$ .