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Sound pressure level and power level calculations

Short answer

(b) SPL = 20 log(0.2/20×10⁻⁶) = 80 dB. (c) PWL = 10 log(0.01/10⁻¹²) = 100 dB. For hemispherical spreading, SPL at 10 m = 100 – 20 log(10) – 8 = 72 dB. Results: 80 dB SPL, 100 dB PWL, 72 dB at 10 m.

(a) Define sound pressure level (SPL) and sound power level (PWL). (b) A sound has an rms pressure of 0.2 Pa. Find its SPL. (c) A machine radiates 0.01 W of sound power. Find its PWL, and its SPL at 10 m on open flat ground (hemispherical radiation). Reference values: p_ref = 20 micro-Pa and W_ref = 10^-12 W.

(a) Definitions

  • SPL: Lp=10log⁡10(p2pref2)=20log⁡10(ppref)L_p = 10\log_{10}\left(\dfrac{p^2}{p_{ref}^2}\right) = 20\log_{10}\left(\dfrac{p}{p_{ref}}\right) dB, with pref=20 μp_{ref} = 20\ \muPa (threshold of hearing at 1 kHz). It depends on the distance from the source.
  • PWL: Lw=10log⁡10(WWref)L_w = 10\log_{10}\left(\dfrac{W}{W_{ref}}\right) dB, with Wref=10−12W_{ref} = 10^{-12} W. It is the property of the source and is independent of distance and surroundings.

(b) SPL

Lp=20log⁡10(0.220×10−6)=20log⁡10(104)=80 dBL_p = 20\log_{10}\left(\frac{0.2}{20\times10^{-6}}\right) = 20\log_{10}(10^4) = 80\ \text{dB}

(c) PWL and SPL at 10 m

Lw=10log⁡10(0.0110−12)=10log⁡10(1010)=100 dBL_w = 10\log_{10}\left(\frac{0.01}{10^{-12}}\right) = 10\log_{10}(10^{10}) = 100\ \text{dB}

For hemispherical spreading, intensity I=W/(2πr2)I = W/(2\pi r^2), so

Lp=Lw−20log⁡10r−8=100−20log⁡10(10)−8=100−20−8=72 dBL_p = L_w - 20\log_{10} r - 8 = 100 - 20\log_{10}(10) - 8 = 100 - 20 - 8 = 72\ \text{dB}

(The constant is 10log⁡102π=810\log_{10}2\pi = 8.)

Answer: (b) SPL = 80 dB; (c) PWL = 100 dB; SPL at 10 m = 72 dB.

✓ Answer checked by Claude Opus 5.5 (high effort)

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