Power and Pressure of a whisper and shout

Question

A quiet whisper is measured at 30 dB and a loud shout at 110 dB. (A_{ear canal} = 1.54 x 10^-4 m², Z_air = 413 kg m^-2 s^-1).

a) What power, in watts, is delivered to the opening of the ear canal during the whisper?

b) How many times larger is the power (in watts) delivered to the ear canal by the shout?

c) What us the pressure variation during the whisper?

d) How many times larger is the pressure variation during the shout?

Answer 1

Hi Stephany B.,

As you would find out if you poked around a bit on Google, 0 dB is defined as 10^(-12) W. And the scaling is logarithmic, every +10 dB = x 10X on power.

So, 30 dB = 10^(-9) W, 110 dB = 10^(-1) W.

A Watt is a Joule/second hence kg m^2 s(-2) / s == kg m^2 s(-3) . Reality check: yes, this is an acceleration of a mass (kg m s^-2) operating for 1 m, so as to constitute an energy (kg m^2 s^-2) , delivered per second, so as to constitute power (kg m^2 s^-3).

Power scales as (pressure)^2 , where pressure refers to instantaneous air pressure in the wave.

In part (C), we want pressure == pressure variation, in units of N m^-2 or (kg m s^-2) x m^-2 = kg m^-1 s^-2 .

So dimensionally, we need to convert kg m^2 s^-3 to kg m^-1 s^-2 by multiplying by m^-3 s . I'l confess I don't recognize Z(air) (acoustic impedance) as useful here.

Do let me know what you come up with (make a comment, I'm following your question!)