Chapter 15, Problem 67A

a.

To determine

To find: the force on the eardrum.

Answer to Problem 67A

The force on ear drum is $1.0\times {10}^{-6}\text{N}$ .

Explanation of Solution

Given data:

Amplitude of a pressure wave is $0.020\text{ Pa}$ .

Area of the eardrum is $0.52{\text{ cm}}^2$ .

Formula used:

  $F=PA$ where F is the force, P is the pressure and A is the area.

Calculation:

  $\begin{array}{l}F=PA\\ P=\left(0.020{\text{ N/m}}^2\right)\left(0.52\times {10}^{-4}{\text{ m}}^2\right)\\ =1.0\times {10}^{-6}\text{ N}\end{array}$

Conclusion:

The force on the eardrum is $1.0\times {10}^{-6}\text{ N}$ .

b.

To determine

To find:the force that the bones exert on the oval window.

Answer to Problem 67A

The force exerted on the oval window is $1.5\times {10}^{-6}\text{N}$

Explanation of Solution

Given data:

Force on eardrum is ${\text{(F}}_e)=1.0\times {10}^{-6}\text{N}$

The mechanical advantage of the three bones in the middle ear is $1.5$ .

Formula used:

As $\text{MA=}\frac{{\text{F}}_r}{{\text{F}}_e}$ , so the formula used here is ${\text{F}}_r=\left(\text{MA}\right)\left({\text{F}}_e\right)$ .

Calculation:

Substituting the values, we get

  $\begin{array}{l}{\text{F}}_r=\left(1.5\right)\left(1.0\times {10}^{-6}\text{ N}\right)\\ =1.5\times {10}^{-6}\text{ N}\end{array}$

Conclusion:

The force exerted on the oval window is $1.5\times {10}^{-6}\text{N}$

c.

To determine

To find: the pressure increase that is transmitted to the liquid in the cochlea.

Answer to Problem 67A

The pressure increase is $0.58\text{}\text{ Pa}$ .

Explanation of Solution

Given data:

Area of the oval window is $0.026{\text{ cm}}^2$ .

Force exerted on the oval window is $1.5\times {10}^{-6}\text{N}$

Formula used:

  $\text{P}=\frac{\text{F}}{\text{A}}$ where F is the force, P is the pressure and A is the area.

Calculation:

  $\begin{array}{l}\text{P}=\frac{1.5\times {10}^{-6}\text{N}}{0.026\times {10}^{-4}{\text{ m}}^2}\\ =0.58\text{ Pa}\end{array}$

Conclusion:

The pressure increase that is transmitted to the liquid in the cochlea is $0.58\text{}\text{ Pa}$ .