Question

exam lesson name: light and sound
exam number: 700699rr
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question 11 of 20 :
select the best answer for the question.

1. a surface receiving sound is moved from its original position to a position three times farther away from the source of the sound. the intensity of the received sound thus becomes

○ a. three times higher.
○ b. nine times higher.
○ c. nine times lower.
○ d. three times lower.

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Explanation:

Step1: Recall the inverse square law for sound intensity

The intensity \( I \) of sound (or any spherical wave) follows the inverse square law, which is given by \( I \propto \frac{1}{r^2} \), where \( r \) is the distance from the source.

Step2: Let the initial distance be \( r_1 \) and the final distance be \( r_2 \)

Given that \( r_2 = 3r_1 \). Let the initial intensity be \( I_1 \) and the final intensity be \( I_2 \). From the inverse square law, \( I_1 \propto \frac{1}{r_1^2} \) and \( I_2 \propto \frac{1}{r_2^2} \).

Step3: Find the ratio of the intensities

Divide the equation for \( I_2 \) by the equation for \( I_1 \): \(\frac{I_2}{I_1}=\frac{\frac{1}{r_2^2}}{\frac{1}{r_1^2}}=\frac{r_1^2}{r_2^2}\). Substitute \( r_2 = 3r_1 \) into the formula: \(\frac{I_2}{I_1}=\frac{r_1^2}{(3r_1)^2}=\frac{r_1^2}{9r_1^2}=\frac{1}{9}\). So \( I_2=\frac{1}{9}I_1 \), which means the intensity becomes nine times lower.

Answer:

C. nine times lower.