College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 12 - Problems - Page 463: 26


(a) The intensity increases by 125%. (b) The intensity level increases by $3.52~dB$

Work Step by Step

(a) If the original amplitude was $A$, then the new amplitude is $1.5~A$. Since the intensity is proportional to the square of the amplitude, the new intensity is proportional to $(1.5~A)^2 = 2.25~A^2$, while the original amplitude was proportional to $A^2$. Let $I_1$ be the original intensity and let $I_2$ be the new intensity. Note that $I_2 = 2.25~I_1$. We can see that the intensity increases by 125%. (b) We can find the new intensity level $\beta_2$: $\beta_2-\beta_1 = 10~log~\frac{I_2}{I_0}-10~log~\frac{I_1}{I_0}$ $\beta_2-\beta_1 = 10~log~\frac{I_2}{I_1}$ $\beta_2-\beta_1 = 10~log~\frac{2.25~I_1}{I_1}$ $\beta_2-\beta_1 = 10~log~2.25$ $\beta_2 = \beta_1 + 10~log~2.25$ $\beta_2 = \beta_1+3.52~dB$ The intensity level increases by $3.52~dB$
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