Question

part a
two sound waves have equal displacement amplitudes, but one has 2.9 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}}=

Explanation:

Step1: Recall intensity - frequency relation

The intensity $I$ of a sound wave is proportional to the square of the frequency $f$ when the displacement amplitudes are equal, i.e., $I\propto f^{2}$. Let $f_1$ be the lower - frequency and $f_2 = 2.9f_1$ be the higher - frequency.

Step2: Calculate the ratio of intensities

The ratio of the intensities $\frac{I_2}{I_1}=\frac{f_2^{2}}{f_1^{2}}$. Substitute $f_2 = 2.9f_1$ into the formula: $\frac{I_2}{I_1}=\frac{(2.9f_1)^{2}}{f_1^{2}}=\frac{2.9^{2}f_1^{2}}{f_1^{2}} = 2.9^{2}$.

Step3: Evaluate the result

$2.9^{2}=8.41$. Rounding to two significant figures, we get 8.4.

Answer:

8.4