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shusha [124]
3 years ago
15

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?
Physics
1 answer:
zlopas [31]3 years ago
8 0

Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

\rho = Density of medium

A = Area

\nu = Frequency

\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

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The period is how long each one takes, or seconds per swing.
It's exactly the flip of frequency.
So we could just take the frequency, flip it, and find   1 / 0.2 ,
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<h3><u>Given</u><u>:</u><u>-</u></h3>

Acceleration,a = 3 m/s²

Initial velocity,u = 0 m/s

Final velocity,v = 12 m/s

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

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