. Hearing Loss from MP3 Players Recent research has shownthat the use of earbud-style headphones packaged with MP3players can cause permanent hearing loss.(a) The intensity of the sound from the speakers of acertain MP3 player (without earbuds) is measured at3.1 x 10- W/m. Find the decibel level.(b) If earbuds are used with the MP3 player in part (a), thedecibel level is 95 dB. Find the intensity.(c) Find the ratio of the intensity of the sound from the MP3player with earbuds to that of the sound without earbuds.

(a) It is given that: Intensity of sound from speaker=I=3.1×10-5 W/m2 According to the definition,…

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Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

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Question

. Hearing Loss from MP3 Players Recent research has shown
that the use of earbud-style headphones packaged with MP3
players can cause permanent hearing loss.
(a) The intensity of the sound from the speakers of a
certain MP3 player (without earbuds) is measured at
3.1 x 10- W/m. Find the decibel level.
(b) If earbuds are used with the MP3 player in part (a), the
decibel level is 95 dB. Find the intensity.
(c) Find the ratio of the intensity of the sound from the MP3
player with earbuds to that of the sound without earbuds.

Expert Solution

Step 1

(a)

It is given that:

$Intensity of sound from speaker = I = 3.1 \times 10^{- 5} W / m^{2}$

According to the definition, the decibel level $B$ of sound can be calculated as:

$B = 10 \log (\frac{I}{I_{o}}) . . . . . . . . (i) Here, I_{o} = 10^{- 12} W / m^{2} Substituting all the known values in equation (i) : B = 10 \log (\frac{3.1 \times 10^{- 5}}{10^{- 12}}) B = 10 [\log (3.1 \times 10^{- 5}) - \log (10^{- 12})] \begin{array}{rcl} B \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} [\log (3.1 \times 10^{- 5}) - (- 12) \log (10)] \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} B \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} [(\log 3.1 \times \log 10^{- 5}) + 12] \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} B \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} [(12.49136 \times \log 10^{- 5})] \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} B \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 74 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 9136 \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} dB \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} ( \end{array} \begin{array}{rcl} Ans \end{array} \begin{array}{rcl} ) \end{array} \begin{array}{rcl} \end{array}$

B=10logII0.

Step 2

(b)

It is given that the decibel level of MP3 player = 95 dB.

By the definition, decibel level B of sound is given by:

$B = 10 \log (\frac{I}{I_{0}}) Substituting all the known values : 95 = 10 \log \frac{I}{10^{- 12}} 95 = 10 [\log (I) - \log 10^{- 12}] \begin{array}{rcl} 95 \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} [\log (I) - (- 12) \log 10] \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} 95 \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} [\log (I) + 12] \end{array} \begin{array}{rcl} \frac{95}{10} \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \frac{10 [\log (I) + 12]}{10} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} (Dividing by 10 on both the sides :) \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} 9 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 5 \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \log \end{array} \begin{array}{rcl} (I) \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} 12 \end{array} \begin{array}{rcl} \end{array}$

Then,

Adding (-12) on both the sides:

$9.5 + (- 12) = \log (I) + 12 + (- 12) - 2.5 = \log (I)$

Using:

$\log_{a} (x) = y a^{y} = x$

So,

$10^{- 2.5} = I I = 10^{- 2.5} I \approx 10^{- 3} W / m^{2}$

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