/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 Although California is known for... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 64

Although California is known for earthquakes, it has large regions dotted with precariously balanced rocks that would be easily toppled by even a mild earthquake. Apparently no major earthquakes have occurred in those regions. If an earthquake were to put such a rock into sinusoidal oscillation (parallel to the ground) with a frequency of \(2.2 \mathrm{~Hz}\), an oscillation amplitude of \(1.0\) \(\mathrm{cm}\) would cause the rock to topple. What would be the magnitude of the maximum acceleration of the oscillation, in terms of \(g\) ?

Short Answer

Expert verified
The maximum acceleration is approximately 0.195 times the acceleration due to gravity, \( g \).

Step by step solution

01

Understand the Relationship Between Oscillation and Acceleration

To find the maximum acceleration of an object in sinusoidal motion, we use the formula for the maximum acceleration of simple harmonic motion: \[ a_{max} = (2\pi f)^2 \times A \] where \( f \) is the frequency and \( A \) is the amplitude of the motion.
02

Identify Given Values

We know from the problem that the frequency \( f = 2.2 \) Hz and the amplitude \( A = 1.0 \) cm (which is equal to 0.01 meters).
03

Plug in Values and Solve for Maximum Acceleration

Using the formula from Step 1, substitute the given values into the equation:\[ a_{max} = (2\pi \times 2.2)^2 \times 0.01 \]Calculate \((2\pi \times 2.2)^2\):- First calculate \(2\pi \times 2.2 = 4.4\pi \)- Then calculate \((4.4\pi)^2\): \[ 4.4^2 \approx 19.36 \] and \( \pi^2 \approx 9.87 \) Therefore, \( (4.4\pi)^2 \approx 19.36 \times 9.87 \approx 191.23 \)- Now calculate \[ a_{max} = 191.23 \times 0.01 = 1.9123 \text{ m/s}^2 \]
04

Convert Acceleration to Terms of \( g \)

The acceleration due to gravity \( g \) is approximately \( 9.81 \text{ m/s}^2 \). To express \( a_{max} \) in terms of \( g \), divide the calculated maximum acceleration by \( g \):\[ \frac{1.9123}{9.81} \approx 0.195 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Frequency
Frequency describes how many times a motion or cycle repeats itself in one second. It is measured in Hertz (Hz). In the context of Simple Harmonic Motion (SHM), frequency tells us how often the oscillating body (like the precariously balanced rock) completes one full oscillation within a second.

Understanding frequency is important because it helps in determining other properties of SHM, such as period and energy. The formula related to frequency is:
  • Frequency (\(f\)) = 1/Period (\(T\))
This relationship tells us that higher frequency means a shorter period, indicating quicker oscillations.

In our exercise, the frequency is given as 2.2 Hz, meaning the motion repeats itself 2.2 times every second. This information is crucial as it is part of the formula used to calculate maximum acceleration in SHM.
Amplitude
Amplitude in Simple Harmonic Motion (SHM) refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium. In simpler terms, it is the maximum distance that the object moves from its central position.

The amplitude is crucial because it indicates how far an oscillating object travels during each cycle. It provides an idea of the energy involved in the motion鈥攍arger amplitudes generally mean higher energy. The formula for amplitude is simple:
  • Amplitude (\(A\)) is usually given in units like meters, centimeters, etc.
In our specific problem, the amplitude is 1.0 cm, or 0.01 meters. This means the rock would move 0.01 meters from its resting position during each oscillation cycle. Amplitude is essential when calculating the maximum acceleration because it directly influences how an object speeds up or slows down during its oscillation.
Acceleration
In Simple Harmonic Motion (SHM), acceleration is not constant; it varies as the object moves back and forth. The acceleration is highest at the turning points of the motion and zero at the central equilibrium position.

The formula for maximum acceleration in SHM is:
  • Maximum Acceleration (\(a_{max}\)) = \((2\pi f)^2 \times A\)
This equation shows that maximum acceleration depends on both the frequency and the amplitude of the motion. In the given problem, we use the known values of frequency and amplitude to calculate this maximum acceleration.

After performing the calculations, the maximum acceleration is found to be approximately 1.9123 m/s虏. To express this in terms of gravity (\(g\)), we divide it by the standard acceleration due to gravity, 9.81 m/s虏. Thus, the result is about 0.195 \(g\), indicating that any acceleration causing oscillation greater than this value may result in the rock toppling over.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A (hypothetical) large slingshot is stretched \(2.30 \mathrm{~m}\) to launch a \(170 \mathrm{~g}\) projectile with speed sufficient to escape from Earth (11.2 \(\mathrm{km} / \mathrm{s}\) ). Assume the elastic bands of the slingshot obey Hooke's law. (a) What is the spring constant of the device if all the elastic potential energy is converted to kinetic energy? (b) Assume that an average person can exert a force of \(490 \mathrm{~N}\). How many people are required to stretch the elastic bands?

A pendulum is formed by pivoting a long thin rod about a point on the rod. In a series of experiments, the period is measured as a function of the distance \(x\) between the pivot point and the rod's center. (a) If the rod's length is \(L=2.20 \mathrm{~m}\) and its mass is \(m=22.1 \mathrm{~g}\), what is the minimum period? (b) If \(x\) is cho- sen to minimize the period and then \(L\) is increased, does the period increase, decrease, or remain the same? (c) If, instead, \(m\) is increased without \(L\) increasing, does the period increase, decrease, or remain the same?

An oscillating block-spring system has a mechanical energy of 1.00 J, an amplitude of \(10.0 \mathrm{~cm}\), and a maximum speed of \(1.20 \mathrm{~m} / \mathrm{s}\). Find (a) the spring constant, (b) the mass of the block, and (c) the frequency of oscillation.

A grandfather clock has a pendulum that consists of a thin brass disk of radius \(r=15.00 \mathrm{~cm}\) and mass \(1.000\) kg that is attached to a long thin rod of negligible mass. The pendulum swings freely about an axis perpendicular to the rod and through the end of the rod opposite the disk, as shown in Fig. \(15.56 .\) If the pendulum is to have a period of \(2.000 \mathrm{~s}\) for small oscillations at a place where \(g=9.800 \mathrm{~m} / \mathrm{s}^{2}\), what must be the rod length \(L\) to the nearest tenth of a millimeter?

A particle executes linear SHM with frequency \(0.25 \mathrm{~Hz}\) about the point \(x=0\). At \(t=0\), it has displacement \(x=0.37 \mathrm{~cm}\) and zero velocity. For the motion, determine the (a) period, (b) angular frequency, (c) amplitude, (d) displacement \(x(t),(\mathrm{e})\) velocity \(v(t)\), (f) maximum speed, (g) magnitude of the maximum acceleration, (h) displacement at \(t=3.0 \mathrm{~s}\), and (i) speed at \(t=3.0 \mathrm{~s}\)
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Physics of Motion

Read Explanation

Rotational Dynamics

Read Explanation

Mechanics and Materials

Read Explanation

Dynamics

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.