The acoustic intensity, I, of a sound wave is proportional to the square of the pressure amplitude, P, and inversely proportional to the velocity, v, of the wave.
a) what is the proportionality relationship?
b) what happens to the intensity if both the pressure and velocity are doubled?
thanks alot :)
a. I = P^2/V
b. I = (2P)^2/2V = 4P^2/2V = 2P^2/V.
Therefore, I is doubled.
a) The proportionality relationship between the acoustic intensity (I), pressure amplitude (P), and velocity (v) of a sound wave can be expressed as:
I ∝ P^2/v
This means that the intensity is directly proportional to the square of the pressure amplitude and inversely proportional to the velocity of the wave.
b) If both the pressure amplitude (P) and velocity (v) of the sound wave are doubled, we can assess the effect on the intensity (I) using the proportionality relationship mentioned above.
Let's assume the initial values of P and v are P1 and v1, and the initial intensity is I1.
When both P and v are doubled, their new values become 2P1 and 2v1, respectively.
Plugging in these new values into the proportionality relationship, we have:
I2 ∝ (2P1)^2 / (2v1)
= 4P1^2 / 2v1
= 2P1^2 / v1
So, the new intensity (I2) is equal to 2 times the initial intensity (I1).
In simpler terms, if both the pressure amplitude and velocity of a sound wave are doubled, the intensity of the wave will increase by a factor of 2.