Key to "Unless indicated otherwise, assume the speed of sound..."

High School
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Salvatore Boone

Answered question

2022-01-10

Unless indicated otherwise, assume the speed of sound in air to be

v = 344 \frac{m}{s}

. Sound is detected when a sound wave causes the tympanic membrane (the eardrum) to vibrate. Typically, the diameter of this membrane is about 8.4 mm in humans.
(a) How much energy is delivered to the eardrum each second when someone whispers (20 dB) a secret in your ear?
(b) To comprehend how sensitive the ear is to very small amounts of energy, calculate how fast a typical 20-mg mosquito would have to fly (in

\frac{m m}{s}

) to have this amount of kinetic energy.

Answer & Explanation

Navreaiw

Beginner2022-01-11Added 34 answers

a) Our goal is to find the intensity of the sound wave, then to make use of the value of the intensity to determine the energy being delivered to the eardrum each second.
We can calculate the intensity of the sound wave using the relation that describes the intensity level of a sound wave, which is

β = 10 \log (\frac{I}{I_{0}})

(1)
Where

β = 20 d B, I_{0} = 10^{- 12} \frac{W}{m^{2}}

and I is the intensity of the sound wave that we would like to determine.
Hence

20 d B = 10 \log (\frac{I}{10^{- 12} \frac{W}{m^{2}}})

2 = \log (\frac{I}{10^{- 12} \frac{W}{m^{2}}})

Remember that

10^{\log (x)} = x

, hence

10^{2} = 10^{\log (\frac{I}{10^{- 12} \frac{W}{m^{2}}})}

\frac{I}{10^{- 12} \frac{W}{m^{2}}} = 10^{2}

I = 10^{- 10} \frac{W}{m^{2}}

But we know that intensity is energy per unit time per unit area (Power per unit area), which could be represented mathematically as follows

E = I A △

(2)
Where A is the area of the tympanic membrane and it can be calculated as follows

A = π r^{2} = π \times {(4.2 \times 10^{- 3} m)}^{2} = 5.54 \times 10^{- 5} m^{2}

Now, substitute into (2) with

5.54 \times 10^{- 5} m^{2}

for

A, 10^{- 10} \frac{W}{m^{2}}

for I and 1 s for

△

Notice that

△ t = 1 s

since we want to calculate the energy is delivered each second

E = (10^{- 10} \frac{W}{m^{2}}) \times (5.54 \times 10^{- 5} m^{2}) \times (1 s)

E = 5.54 \times 10^{- 15} J

l \in e (1, 0) {370}

Deufemiak7

Beginner2022-01-12Added 34 answers

b) The velocity of a mosquito with a mass of 2.0 mg and has a kinetic energy of

5.54 \times 10^{- 15} J

can be calculated as follows

K E = \frac{1}{2} m v^{2}

v = \sqrt{\frac{2 \times K E}{m}}

To get v, substitute with

5.54 \times 10^{- 15} J

tor KE and

2 \times 10^{- 6} k g

v = \sqrt{\frac{2 \times (5.54 \times 10^{- 15} J)}{2 \times 10^{- 6} k g}}

v = 0.074 \times 10^{- 3} \frac{m}{s}

v = 0.074 \frac{m m}{s}

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