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Chapter 10: Problem 15

When responding to sound, the human eardrum vibrates about its equilibrium position. Suppose an eardrum is vibrating with an amplitude of \(6.3 \times 10^{-7}\) m and a maximum speed of \(2.9 \times 10^{-3} \mathrm{m} / \mathrm{s}\) (a) What is the frequency (in Hz) of the eardrum鈥檚 vibration? (b) What is the maximum acceleration of the eardrum?

Short Answer

Expert verified
Frequency is approximately 732.7 Hz, and maximum acceleration is about 13.4 m/s虏.

Step by step solution

01

Understand the relationship between amplitude, speed, and frequency

In harmonic motion, the maximum speed (\(v_{max}\)) of an object is related to its amplitude (\(A\)) and angular frequency (\(\omega\)) by the equation \(v_{max} = A \times \omega\). We need to find \(\omega\) first to calculate the frequency.
02

Calculate angular frequency (蠅)

Given the maximum speed \(v_{max} = 2.9 \times 10^{-3} \ \mathrm{m/s}\) and amplitude \(A = 6.3 \times 10^{-7} \ \mathrm{m}\), we can rearrange the formula to find \(\omega\):\[ \omega = \frac{v_{max}}{A} = \frac{2.9 \times 10^{-3}}{6.3 \times 10^{-7}} \approx 4603.17 \ \mathrm{rad/s} \]
03

Find the frequency from angular frequency

The relationship between frequency \(f\) and angular frequency \(\omega\) is given by \(\omega = 2\pi f\). To find \(f\), use:\[ f = \frac{\omega}{2\pi} = \frac{4603.17}{2\pi} \approx 732.7 \ \mathrm{Hz} \]
04

Compute maximum acceleration

The maximum acceleration \(a_{max}\) in harmonic motion is given by \(a_{max} = A \times \omega^2\). Substitute the given values:\[ a_{max} = 6.3 \times 10^{-7} \times (4603.17)^2 \approx 13.4 \ \mathrm{m/s^2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Frequency
Angular frequency, denoted by \(\omega\), is a fundamental concept in harmonic motion. It represents how quickly an object oscillates in radians per second. Angular frequency links closely to both maximum speed and amplitude. When the eardrum vibrates, its angular frequency describes the rapidity of these oscillations.
Angular frequency is determined through the formula \(\omega = \frac{v_{max}}{A}\), where \(v_{max}\) is the maximum speed, and \(A\) is the amplitude. This formula allows us to see how both speed and the extent of motion contribute to the frequency of oscillation.
The angular frequency helps determine how vibrations translate into sound frequencies, which are critical for hearing.
Maximum Speed
Maximum speed in harmonic motion refers to the highest velocity an oscillating object reaches. For an eardrum responding to sound, this is the point at which it moves the fastest between oscillations. In this context, the maximum speed is given as \(2.9 \times 10^{-3} \ \mathrm{m/s}\).
This speed can be found using the formula \(v_{max} = A \times \omega\). It provides a connection between amplitude and angular frequency.
  • Higher amplitude or angular frequency results in greater maximum speed.
  • Maximum speed indicates how vigorously an object like the eardrum is responding to external forces, such as sound waves.
Understanding this concept is crucial in acoustics, as it affects how sound is perceived.
Amplitude
Amplitude is a measure of how far an object moves from its equilibrium position during oscillation. For the eardrum, the amplitude is \(6.3 \times 10^{-7} \ \mathrm{m}\).
Amplitude is a pivotal factor in determining the intensity of vibrations. It describes the extent of motion, affecting the energy transported by a wave.
  • Larger amplitudes often correspond to louder sounds, as the eardrum moves further back and forth.
  • In the context of maximum speed, amplitude is used to calculate the highest velocity the object achieves during vibration.
In acoustics, the amplitude helps us understand how changes in motion intensity influence what we hear.
Frequency
Frequency, measured in hertz (Hz), defines the number of complete cycles an oscillation undergoes per second. In the eardrum's vibration, the calculated frequency is approximately \(732.7\ \mathrm{Hz}\).
Frequency is directly related to angular frequency through the relationship \(f = \frac{\omega}{2\pi}\), where \(\omega\) is the angular frequency.
  • Higher frequencies result in higher-pitched sounds.
  • Frequency reveals information about the tone and pitch of the sound produced by vibrating objects.
Understanding frequency enables us to explore and interpret different sounds and musical notes our ears perceive.

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Most popular questions from this chapter

A vertical spring (spring constant 5 112 N/m) is mounted on the fl oor. A 0.400-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

A spring is resting vertically on a table. A small box is dropped onto the top of the spring and compresses it. Suppose the spring has a spring constant of 450 N/m and the box has a mass of 1.5 kg. The speed of the box just before it makes contact with the spring is 0.49 m/s. (a) Determine the magnitude of the spring鈥檚 displacement at an instant when the acceleration of the box is zero. (b) What is the magnitude of the spring鈥檚 displacement when the spring is fully compressed?

Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man's thighhone can endure without hreaking is \(7.0 \times 10^{4} \mathrm{N}\). The minimum effective cross-sectional area of the bone is \(4.0 \times 10^{-4} \mathrm{m}^{2},\) its length is \(0.55 \mathrm{m},\) and Young's modulus is \(Y=9.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} .\) The mass of the man is 65 kg. He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his center of gravity can fall without his breaking a bone.

"93 -" ssm A 75-kg diver is standing at the end of a diving board while it is vibrating up and down in simple harmonic motion, as indicated in the figure. The diving board has an effective spring constant of \(k=4100 \mathrm{N} / \mathrm{m}\), and the vertical distance between the highest and lowest points in the motion is \(0.30 \mathrm{m}\). Concepts: (i) How is the amplitude \(A\) related to the vertical distance between the highest and lowest points of the diver's motion? (ii) Starting from the top, where is the diver located one- quarter of a period later, and what can be said about his speed at this point? (iii) If the amplitude were to double, would the period also double? Explain. Calculations: (a) What is the amplitude of the motion? (b) Starting when the diver is at the highest point, what is his speed one-quarter of a period later? (c) If the vertical distance between his highest and lowest points were changed to \(0.10 \mathrm{m},\) what would be the time required for the diver to make one complete motional cycle?

A copper rod (length \(=2.0 \mathrm{m},\) radius \(\left.=3.0 \times 10^{-3} \mathrm{m}\right)\) hangs down from the ceiling.A 9.0 -kg object is attached to the lower end of the rod. The rod acts as a "spring," and the object oscillates vertically with a small amplitude. Ignoring the rod's mass, find the frequency \(f\) of the simple harmonic motion.
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