Question

two sound waves have equal displacement amplitudes, but one has 2.4 times the frequency of the other. what is the ratio of their intensities? express your answer using two significant figures.

Explanation:

Step1: Recall intensity formula

The intensity formula for a sound - wave is $I = \frac{1}{2}
ho v\omega^{2}A^{2}$, where $
ho$ is the density of the medium, $v$ is the speed of sound in the medium, $\omega = 2\pi f$ is the angular frequency, and $A$ is the displacement amplitude. Since the displacement amplitudes are equal and the medium is the same (so $
ho$ and $v$ are constant), we have $I\propto\omega^{2}\propto f^{2}$.

Step2: Set up the ratio of intensities

Let $f_1$ be the frequency of the low - frequency wave and $f_2 = 2.4f_1$ be the frequency of the high - frequency wave. Then $\frac{I_2}{I_1}=\frac{f_2^{2}}{f_1^{2}}$.

Step3: Calculate the ratio

Substitute $f_2 = 2.4f_1$ into the ratio: $\frac{I_2}{I_1}=\frac{(2.4f_1)^{2}}{f_1^{2}}=\frac{2.4^{2}f_1^{2}}{f_1^{2}} = 2.4^{2}=5.76$.

Answer:

$5.8$