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Chapter 14: Problem 13

Some passengers on an ocean cruise may suffer from motion sickness as the ship rocks back and forth on the waves. At one position on the ship, passengers experience a vertical motion of amplitude \(1 \mathrm{m}\) with a period of \(15 \mathrm{s}\). a. To one significant figure, what is the maximum acceleration of the passengers during this motion? b. What fraction is this of \(g ?\)

Short Answer

Expert verified
The maximum acceleration of passengers during this motion to one significant figure is 1.7 m/s^2 and its fraction compared with g is approximately 0.17

Step by step solution

01

Calculate Angular Frequency

Firstly, derive the angular frequency \(\omega\) from the period \(T\) of the motion using the formula \(\omega = \frac{2\pi}{T}\). Substituting given \(T = 15s\) into the formula, we obtain \(\omega = \frac{2\pi}{15}\)
02

Calculate Maximum Acceleration

The maximum acceleration \(a_{max}\) in harmonic motion is given by the formula \(a_{max} = \omega^2 \times \text{Amplitude}\). Substituting the Amplitude = 1m and the derived value of \(\omega\), we obtain \(a_{max} = \left(\frac{2\pi}{15}\right)^2 \times 1\). Calculate the value for \(a_{max}\) and round it to one significant figure
03

Calculate Fraction of g

To find the fraction of the maximum acceleration of the motion with respect to g, divide the maximum acceleration by \(g = 9.8 m/s^2\). Hence, fraction = \(\frac{\text{max acceleration}} {g}\)

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Key Concepts

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

Angular Frequency
In the context of simple harmonic motion, angular frequency ( \( \omega \) ) is a crucial concept that represents how fast the system oscillates. It is particularly helpful when analyzing the motion's dynamics in a circular framework. Angular frequency is related to the period of motion ( \( T \) ), which is the time taken for one complete oscillation. The relationship between them is given by the formula:
  • \( \omega = \frac{2\pi}{T} \)
In our exercise, the ship's motion has a period of 15 seconds. By substituting this value into the formula, we determine that \( \omega = \frac{2\pi}{15} \), which describes how quickly the passengers experience one complete cycle of vertical motion.
Understanding \( \omega \) helps in predicting other characteristics of the motion, such as velocity and acceleration.
Amplitude
Amplitude is a term that describes the maximum displacement from the equilibrium position in oscillatory motion. It signifies the peak level of disturbance caused by the motion. In simple harmonic motion, amplitude is constant and defines the bounds of the oscillation.
  • In this exercise, the amplitude is given as 1 meter. This means that as the ship rocks, passengers can experience a vertical displacement up to 1 meter above or below the equilibrium level.

  • Amplitude affects the energy of the motion:
    • A larger amplitude indicates more potential energy at the peaks of the motion.
The amplitude is integral for determining the forces acting on the passengers and helps in assessing comfort levels during the cruise.
Maximum Acceleration
The maximum acceleration experienced during simple harmonic motion is derived from both angular frequency and amplitude. It defines the greatest change in velocity that an object undergoes during a cycle of motion. The formula to calculate maximum acceleration \( a_{max} \) is:
  • \( a_{max} = \omega^2 \times \text{Amplitude} \)
Using the values from the exercise, where \( \omega = \frac{2\pi}{15} \) and amplitude = 1 meter, we find:
  • \( a_{max} = \left(\frac{2\pi}{15}\right)^2 \times 1 \)
Calculating this gives us approximately 0.17 m/s2 as the maximum acceleration. This value tells us how much force change the passengers experience at the peak of each oscillation before direction reversal.
Period of Motion
The period of motion, associated with the time for one complete cycle of oscillation, is a fundamental property in simple harmonic motion.
  • In this exercise, the period \( T \) is specified as 15 seconds.
This implies:
  • It takes 15 seconds for the motion to complete one sequence of upward and downward movement.
  • The period is inversely related to frequency: a longer period hints at slower oscillation.
Knowing the period aids in understanding how frequently events in the motion recur, and aids designers in predicting motion impacts on passengers.

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Most popular questions from this chapter

The New England Merchants Bank Building in Boston is \(152 \mathrm{m}\) high. On windy days it sways with a frequency of \(0.17 \mathrm{Hz},\) and the acceleration of the top of the building can reach \(2.0 \%\) of the free-fall acceleration, enough to cause discomfort for occupants. What is the total distance, side to side, that the top of the building moves during such an oscillation?

A \(200 \mathrm{g}\) mass attached to a horizontal spring oscillates at a frequency of \(2.0 \mathrm{Hz}\). At one instant, the mass is at \(x=5.0 \mathrm{cm}\) and has \(v_{x}=-30 \mathrm{cm} / \mathrm{s} .\) Determine: a. The period. b. The amplitude. c. The maximum speed. d. The total energy.

\(\mathrm{A} 500 \mathrm{g}\) air-track glider attached to a spring with spring constant \(10 \mathrm{N} / \mathrm{m}\) is sitting at rest on a frictionless air track. A 250 g glider is pushed toward it from the far end of the track at a speed of \(120 \mathrm{cm} / \mathrm{s}\). It collides with and sticks to the \(500 \mathrm{g}\) glider. What are the amplitude and period of the subsequent oscillations?

A 200 g ball is tied to a string. It is pulled to an angle of \(8.00^{\circ}\) and released to swing as a pendulum. A student with a stopwatch finds that 10 oscillations take 12.0 s. How long is the string?

A mass on a string of unknown length oscillates as a pendulum with a period of 4.00 s. What is the period if a. The mass is doubled? b. The string length is doubled? c. The string length is halved? d. The amplitude is halved? Parts a to d are independent questions, each referring to the initial situation.
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