Question

What is the ratio of the displacement amplitudes of two sound waves given that they are both5.0 kHz but have a 3.0 dB intensity level difference?

Answer 1

The ratio of the displacement amplitudes of two sound waves is 1.16.

Step-by-step 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.

`I\propto(1)/(r^2)`

The sound intensity for first wave is

`\beta_(1)=10\log(I_(1))/(I_(0))`...(I)

The sound intensity for second wave is

`\beta_(2)=10\log(I_(2))/(I_(0))`...(II)

We need to calculate the ratio of intensity

From equation (I) and (II)

`\beta_(2)-\beta_(1)=10\log(I_(2))/(I_(0))-10\log(I_(1))/(I_(0))`

`\Delta \beta=10\log((I_(2))/(I_(1)))`

Put the value into the formula

`3.0=10\log((I_(2))/(I_(1)))`

`(I_(2))/(I_(1))=e^{(3.0)/(10)}`

`(I_(2))/(I_(1))=1.34`

We need to calculate the ratio of the displacement

Using formula of displacement

`(r_(1))/(r_(2))=\sqrt{(I_(2))/(I_(1))}`

Put the value into the formula

`(r_(1))/(r_(2))=√(1.34)`

`(r_(1))/(r_(2))=1.16`

Hence, The ratio of the displacement amplitudes of two sound waves is 1.16.