Respuesta :
Answer:
The ratio of the displacement amplitudes of two sound waves is 1.16.
Explanation:
Given that,
Frequency = 5.0 kHz
Intensity level difference = 3.0 dB
We know that,
The sound intensity is inversely proportional to the square of distance.
[tex]I\propto\dfrac{1}{r^2}[/tex]
The sound intensity for first wave is
[tex]\beta_{1}=10\log\dfrac{I_{1}}{I_{0}}[/tex]...(I)
The sound intensity for second wave is
[tex]\beta_{2}=10\log\dfrac{I_{2}}{I_{0}}[/tex]...(II)
We need to calculate the ratio of intensity
From equation (I) and (II)
[tex]\beta_{2}-\beta_{1}=10\log\dfrac{I_{2}}{I_{0}}-10\log\dfrac{I_{1}}{I_{0}}[/tex]
[tex]\Delta \beta=10\log(\dfrac{I_{2}}{I_{1}})[/tex]
Put the value into the formula
[tex]3.0=10\log(\dfrac{I_{2}}{I_{1}})[/tex]
[tex]\dfrac{I_{2}}{I_{1}}=e^{\dfrac{3.0}{10}}[/tex]
[tex]\dfrac{I_{2}}{I_{1}}=1.34[/tex]
We need to calculate the ratio of the displacement
Using formula of displacement
[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{\dfrac{I_{2}}{I_{1}}}[/tex]
Put the value into the formula
[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{1.34}[/tex]
[tex]\dfrac{r_{1}}{r_{2}}=1.16[/tex]
Hence, The ratio of the displacement amplitudes of two sound waves is 1.16.
