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Halliday 2 Cap 18 - P18 026

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    December 2018

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26. We use ∆β12 = β1 − β2 = (10 dB) log(I1 /I2 ). (a) Since ∆β12 = (10 dB) log(I1 /I2 ) = 37 dB, we get I1 /I2 = 1037 dB/10 dB = 103.7 = 5.0 × 103 .  √ √ (b) Since ∆pm ∝ sm ∝ I, we have ∆pm 1 /∆pm 2 = I1 /I2 = 5.0 × 103 = 71.  (c) The displacement amplitude ratio is sm 1 /sm 2 = I1 /I2 = 71.