Question

Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the intensity of wave 1 to the intensity of wave 2?

Answer 1

`(I_1)/(I_2)=(4)/(9)`

Step-by-step explanation:

c = Speed of wave

`\rho` = Density of medium

A = Area

`\\u` = Frequency

`\\u_1=(2)/(3)\\u_2`

Intensity of sound is given by

`I=(1)/(2)\rho c(A\omega)^2\\\Rightarrow I=(1)/(2)\rho c(A2\pi \\u)^2`

So,

`I\propto \\u^2`

We get

`(I_1)/(I_2)=(\\u_1^2)/(\\u_2^2)\\\Rightarrow (I_1)/(I_2)=((2)/(3)^2\\u_2^2)/(\\u_2^2)\\\Rightarrow (I_1)/(I_2)=(4)/(9)`

The ratio is
`(I_1)/(I_2)=(4)/(9)`