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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 0.60-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 5.70 to 4.80 m/s. What is the spring constant of the spring?

To measure the static friction coefficient between a 1.6-kg block and a vertical wall, the setup shown in the drawing is used. A spring (spring constant = 510 N/m) is attached to the block. Someone pushes on the end of the spring in a direction perpendicular to the wall until the block does not slip downward. The spring is compressed by 0.039 m. What is the coefficient of static friction?

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of \(7.0 \mathrm{rad} / \mathrm{s}\). The drawing indicates the position of the block when the spring is unstrained. This position is labeled " \(x=0 \mathrm{m}\)." The drawing also shows a small bottle located \(0.080 \mathrm{m}\) to the right of this position. The block is pulled to the right, stretching the spring by \(0.050 \mathrm{m},\) and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding \(v_{0} .\) Ignoring the width of the block, find \(v_{0}\).

A 1.1-kg object is suspended from a vertical spring whose spring constant is \(120 \mathrm{N} / \mathrm{m}\). (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of \(0.20 \mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 59 N/m and 33 N/m. Find the angle that the rod makes with the horizontal.
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