practice problem 1
- on the surface of the Earth
- in the ocean, near the equator, 1000 m below seal level (in the equatorial SOFAR channel)
Sound is a variation in pressure that ranges from…
|Pmin = P0 − ∆P||to||Pmax = P0 + ∆P|
|P =||instantaneous pressure|
|P0 =||static pressure|
|∆P =||maximum dynamic pressure|
No sound wave may vary in pressure more than the static pressure of the medium itself. If it did, the minimum pressure would be less than zero, which is impossible. Sound intensity levels are computed relative to an agreed upon standard reference pressure. This number is different for gases and liquids.
- For this part, we'll use the standard value of atmospheric pressure (101,325 Pa) and the reference pressure in air (20 μPa).
LP = 20 log ⎛
P0 LP = 20 log ⎛
101,325 Pa ⎞
20 × 10−6 Pa LP = 194 db
- For this part, we need to first compute the pressure at depth.
P = P0 + ρgh P = (101,325 Pa) + (1,025 kg/m3)(9.8 m/s2)(1,000 m) P = 10,146,325 Pa LP = 20 log ⎛
P0 LP = 20 log ⎛
10,146,325 Pa ⎞
1 × 10−6 Pa LP = 260 db
practice problem 2
- pressure amplitude
- density amplitude
- displacement amplitude
- velocity amplitude
- acceleration amplitude
- For pressure amplitude, compare 60 dB to the reference intensity of 20 μPa at 0 dB. Start with the equation for pressure level in decibels.
I = ∆P2max 2ρv
Solve for ∆Pmax. (I'm going to drop the subscript P from the level symbol L as well as the unit reminder [dB].)
L 20 = log ∆Pmax ∆P0 10L/20 = ∆Pmax ∆P0 ∆Pmax = ∆P010L/20
Substitute and compute (a.k.a. plug and chug).
∆Pmax = (20 µPa)1060 dB/20 ∆Pmax = (20 µPa)103 ∆Pmax = 20 mPa
- We have an equation that relates density amplitude to pressure amplitude.
∆ρ = ∆P v2
Let's assume the conversation is taking place in air at room temperature (20 ℃) with a speed of sound of 343 m/s. Divide and conquor!
∆ρ = (0.020 Pa) (343 m/s)2 ∆ρ = 2.5 × 10−7 kg/m3
- Before we can compute the displacement amplitude, we'll need the intensity. The density of air at room temperature (20 ℃) is 1.207 kg/m3.
I = ∆P2max 2ρv I = (0.020 Pa)2 2(343 m/s)(1.207 kg/m3) I = 4.8 × 10−7 W/m2
Intensity allows us to find displacement amplitude.
I = 2π2ρƒ2vΔx2max Δxmax = √ I 2π2ρƒ2v
Human speech doesn't occur at just one frequency. Early Twentieth Century telephone engineers determined that most of the power in human speech fit between 300 and 3400 Hz. Let's take the geometric mean of this range…
√(300 Hz × 3400 Hz) ≈ 1000 Hz
Δxmax = √ 4.8 × 10−7 W/m2 2π2(1.207 kg/m3)(1000 Hz)2(343 m/s) Δxmax = 7.7 × 10−9m = 7.7 nm
- Velocity amplitude is most easily computed from displacement amplitude.
∆vmax = 2πƒ ∆xmax ∆vmax = 2π(1000 Hz)(7.7 × 10−9m) ∆vmax = 4.8 × 10−5 m/s
- For acceleration amplitude.
∆amax = 2πƒ ∆vmax ∆amax = 2π(1000 Hz)(4.8 × 10−5m/s) ∆amax = 0.30 m/s2
practice problem 3
practice problem 4