Question

Sound is detected when a sound wave causes the tympanic membrane (the eardrum) to vibrate (see the figure ). Typically, the diameter of this membrane is about 8.40 {\rm mm} in humans.

a.)how much energy is delivered to the eardrum each second when someone whispers (20.0 {\rm 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 2.00 {\rm mg} mosquito would have to fly (in {\rm mm/s}) to have this amount of kinetic energy.

Answer 1

Answer:

a) Energy delivered per second to the tympanic membrane = 5.54 × 10⁻¹⁵ J/s

b) velocity of mosquito that will generate that amount of energy, v = 0.0000744 m/s = 0.0744 mm/s.

Explanation:

a) [D] = 10 log (I/I₀)

I₀ = 10⁻¹² W/m²

Given the sound intensity level in decibels, we need to obtain the corresponding sound intensity.

20 = 10 log (I/(10⁻¹²))

2 = log (I/(10⁻¹²))

100 = (I/(10⁻¹²))

I = 10⁻¹⁰ W/m²

Power experienced by the tympanic membrane of the ear due to the sound intensity = Intensity × Area of the membrane

Area of the membrane = πD²/4 = π(8.4 × 10⁻³)²/4 = 5.54 × 10⁻⁵ m²

Power = 10⁻¹⁰ × 5.54 × 10⁻⁵ = 5.54 × 10⁻¹⁵ W

Energy delivered per second to the tympanic membrane = 5.54 × 10⁻¹⁵ J/s

b) Kinetic energy = mv²/2

5.54 × 10⁻¹⁵ = (2 × 10⁻⁶)v²/2

v² = (2 × 5.54 × 10⁻¹⁵)/(2 × 10⁻⁶)

v = 0.0000744 m/s = 0.0744 mm/s.

Answer 2

The definition of decibels and the relationship between work and kinetic energy allows us to find the results for the questions about the system:

     a) The energy supplied to the eardrum is I = 5.67 10⁻¹⁵ W.

     b) The speed of the mosquito with this energy is: v= 7.53 10⁻² m/s

Given parameters.

• Eardrum diameter d= 8.40 mm = 8.40 10-3 m.

• Mosquito mass m= 2.00 mg = 2.00 10-6 kg

• Whisper sound intensity. Beta = 20dB

To find.

    a) The energy per second.

    b) The kinetic energy of a mass mosquito

Decibels definition.

The intensity of sound is in a very wide range of magnitudes, to simplify its use, the decibel is defined as a logarithmic unit.

                  $\beta = 10 \ log (\frac{I}{I_o})$  

where β are the decibels, I the intensity and I₀ the reference intensity. In the case of humans, the sensitivity threshold is of the order of 10⁻¹² W/m²

The intensity of the expression is:

              $\frac{I}{I_o} = 10^{\beta/10 }$  

              $I = I_o\ 10^{\beta/10}$

Let's calculate

               I = $10^{-12} \ 10^{20/10}$  

               I = 10⁻¹⁰ W/m²

The intensity is defined by the energy deposited per unit of time and area.

              I =  $\frac{P}{A}$  

              P = I A

Let's calculate the area of ​​the eardrum.

            A = π r² = $\pi \ \frac{d^2}{4}$  

Let's calculate.

            A = $\frac{\pi}{4} (8.4 \ 10^{-3} )^2$  

            A = 5.67 10⁻⁵ m²

Let's calculate the power.

            P = 10⁻¹⁰  5.67 10⁻⁵

            P = 5.67 10⁻¹⁵5W

b) Power is work per unit of time.

            P = $\frac{W}{t}$

            W= P t

The work is equal to the change in kinetic energy, if we assume that the mosquito starts from rest.

            W = ΔK = $K_f - K_o$

            W = ½ mv²

           v² = $\frac{2 K}{m}$  

Let's calculate

          v² =   = 5.67  10⁻⁹

          v= 7.53 10⁻⁵ m/s

         v= 7.53 10⁻² mm/s

In conclusion using the definition of decibels and the relationship of work and kinetic energy we can find the results for the questions about the system are:

      a) The energy supplied to the eardrum is I = 5.67 10⁻¹⁵  W.

       b) The speed of the mosquito with this energy is: v= 7.53 10⁻² mm/s

Learn more about sound intensity here: brainly.com/question/19920717