Sound with an intensity exceeding 120 dB appears painful to a person. A small speaker delivers 2.0 W of audio output. Determine how close the person can get to the speaker without hurting his ears.

Correct answer: C. 40 cm The intensity level β in decibels is given by β = 10 _ 10 ( (I)/(I_0) ), where I_0 = 10^ -12 W/m^2 . Substituting β = 120 dB: 120 = 10 _ 10 ( I 10^ -12 ) _ 10 ( I 10^ -12 ) = 12 I = 10^ 12 × 10^ -12 = 1 W/m^2 The intensity I at a distance r from a point source of power P is given by I = (P)/(4 π r^2) . Substituting I = 1 W/m^2 and P = 2.0 W : 1 = (2.0)/(4 π r^2) r^2 = (1)/(2 π) r = 1 2 π m r (1)/(2.506) m 0.399 m r 40 cm

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Sound Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Sound Waves

Sound with an intensity exceeding 120 dB appears painful to a...

Sound with an intensity exceeding

120 \text{ dB}

appears painful to a person. A small speaker delivers

2.0 \text{ W}

of audio output. Determine how close the person can get to the speaker without hurting his ears.

Options

A $20 \text{ cm}$
B $71 \text{ cm}$
C $40 \text{ cm}$
D $80 \text{ cm}$

Correct answer

C. $40 \text{ cm}$

Step-by-step solution

The intensity level

\beta

in decibels is given by

\beta = 10 \log_{10} \left( \dfrac{I}{I_0} \right), where

I_0 = 10^{-12}

\text{ W/m}^2

.

Substituting \beta =

120

\text{ dB}:

120 = 10 \log_{10} \left( \dfrac{I}{10^{-12}} \right)

\log_{10} \left( \dfrac{I}{10^{-12}} \right) = 12

I = 10^{12} \times 10^{-12} = 1 \text{ W/m}^2

The intensity

I

at a distance

r

from a point source of power

P

is given by

I = \dfrac{P}{4 \pi r^2}

.

Substituting

I = 1 \text{ W/m}^2

and

P = 2.0 \text{ W}

1 = \dfrac{2.0}{4 \pi r^2}

r^2 = \dfrac{1}{2 \pi}

r = \dfrac{1}{\sqrt{2 \pi}} \text{ m}

r \approx \dfrac{1}{2.506} \text{ m} \approx 0.399 \text{ m}

r \approx 40 \text{ cm}

Practice Sound Waves on Quantrex Academy →

More from Sound Waves