Question

part a
two sound waves have equal displacement amplitudes, but one has 24 times the frequency of the other. what is the ratio of their intensities? express your answer using two significant figures.

$\frac{i_{high - frequency}}{i_{low - frequency}}=$

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Explanation:

Step1: Recall intensity formula for sound waves

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. Let the frequency of the first wave be $f_1$ and the second wave be $f_2 = 24f_1$. The displacement amplitudes are equal, i.e., $A_1=A_2 = A$, and $
ho$ and $v$ are the same for both waves in the same medium.

Step2: Write the intensity ratio

The intensity of the first wave $I_1=\frac{1}{2}
ho v(2\pi f_1)^{2}A^{2}$ and the intensity of the second wave $I_2=\frac{1}{2}
ho v(2\pi f_2)^{2}A^{2}$. Then the ratio $\frac{I_2}{I_1}=\frac{(2\pi f_2)^{2}}{(2\pi f_1)^{2}}$.

Step3: Substitute $f_2 = 24f_1$ into the ratio

Substitute $f_2 = 24f_1$ into $\frac{I_2}{I_1}=\frac{(2\pi f_2)^{2}}{(2\pi f_1)^{2}}$, we get $\frac{I_2}{I_1}=\frac{(24f_1)^{2}}{f_1^{2}}=\frac{24^{2}f_1^{2}}{f_1^{2}} = 576$.

Answer:

$5.8\times10^{2}$ (rounded to two significant - figures)