Physics
Physics
5th Edition
ISBN: 9781260487008
Author: GIAMBATTISTA, Alan
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 10, Problem 40P

(a)

To determine

Whether the maximum velocity and the greatest acceleration of the eardrum are for high-frequency sounds or low-frequency sounds for a given vibration amplitude.

(a)

Expert Solution
Check Mark

Answer to Problem 40P

The maximum velocity and the greatest acceleration of the eardrum are for high-frequency sounds for a given vibration amplitude.

Explanation of Solution

Write the expression for the maximum velocity.

vm=ωA (I)

Here, vm is the maximum velocity, ω is the angular frequency and A is the vibration amplitude

Write the expression for ω.

ω=2πf (II)

Here, f is the frequency

Put equation (II) in equation (I).

vm=2πfA (III)

Equation (III) implies that vmf.

Write the expression for the maximum acceleration.

am=ω2A (IV)

Here, am is the maximum acceleration

Put equation (II) in equation (IV).

am=(2πf)2A=4π2f2A (V)

Equation (V) implies that amf2.

Conclusion:

Since maximum velocity is proportional to frequency and the maximum acceleration is proportional to square of the frequency, maximum velocity and acceleration will occur for high frequency sounds.

Therefore, the maximum velocity and the greatest acceleration of the eardrum are for high-frequency sounds for a given vibration amplitude.

(b)

To determine

The maximum velocity and acceleration of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 Hz.

(b)

Expert Solution
Check Mark

Answer to Problem 40P

The maximum velocity of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 Hz is 1.3×106 m/s and the maximum acceleration is 1.6×104 m/s2.

Explanation of Solution

Equation (III) can be used to find the maximum velocity and equation (V) can be used to find the maximum acceleration.

Conclusion:

Substitute 20.0 Hz for f and 1.0×108 m for A in equation (III) to find vm.

vm=2π(20.0 Hz)(1.0×108 m)=1.3×106 m/s

Substitute 20.0 Hz for f and 1.0×108 m for A in equation (V) to find am.

am=4π2(20.0 Hz)2(1.0×108 m)=1.6×104 m/s2

Therefore, the maximum velocity of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 Hz is 1.3×106 m/s and the maximum acceleration is 1.6×104 m/s2.

(c)

To determine

The maximum velocity and acceleration of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 kHz.

(c)

Expert Solution
Check Mark

Answer to Problem 40P

The maximum velocity of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 kHz is 0.0013 m/s and the maximum acceleration is 160 m/s2.

Explanation of Solution

Equation (III) can be used to find the maximum velocity and equation (V) can be used to find the maximum acceleration.

Conclusion:

Substitute 20.0 kHz for f and 1.0×108 m for A in equation (III) to find vm.

vm=2π(20.0 kHz)(1.0×108 m)=2π(20.0×103 Hz)(1.0×108 m)=0.0013 m/s

Substitute 20.0 kHz for f and 1.0×108 m for A in equation (V) to find am.

am=4π2(20.0 kHz)2(1.0×108 m)=4π2(20.0×103 Hz)2(1.0×108 m)=160 m/s2

Therefore, the maximum velocity of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 kHz is 0.0013 m/s and the maximum acceleration is 160 m/s2.

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