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Chapter 14: Problem 27

(I) A 1.15 -kg mass oscillates according to the equation \(x=0.650 \cos 7.40 t\) where \(x\) is in meters and \(t\) in seconds. Determine \((a)\) the amplitude, \((b)\) the frequency, \((c)\) the total energy, and \((d)\) the kinetic energy and potential energy when \(x=0.260 \mathrm{m} .\)

Short Answer

Expert verified
(a) Amplitude = 0.650 m, (b) Frequency = 1.18 Hz, (c) Total energy = 20.34 J, (d) Kinetic energy = 15.70 J, Potential energy = 4.64 J

Step by step solution

01

Determine the Amplitude

The amplitude of the oscillating mass can be determined directly from the equation of motion. In the equation \(x = 0.650 \cos(7.40t)\), the amplitude is the coefficient of the cosine function. Thus, the amplitude \(A\) is \(0.650\) meters.
02

Calculate the Angular Frequency and Frequency

In the equation \(x = 0.650 \cos(7.40t)\), the term \(7.40\) represents the angular frequency (\(\omega\)) in radians per second. To find the frequency (\(f\)), use the relation \(f = \frac{\omega}{2\pi}\). Thus, \(f = \frac{7.40}{2\pi} \approx 1.18\) Hz.
03

Compute the Total Energy of the System

The total mechanical energy \(E\) in simple harmonic motion is given by \(E = \frac{1}{2} m \omega^2 A^2\), where \(m\) is the mass, \(\omega\) is the angular frequency, and \(A\) is the amplitude. Substituting the values: \(E = \frac{1}{2} \times 1.15 \times 7.40^2 \times 0.650^2 \approx 20.34\) Joules.
04

Determine the Potential Energy when \(x = 0.260\ \mathrm{m}\)

The potential energy \(U\) at displacement \(x\) is given by \(U = \frac{1}{2} m \omega^2 x^2\). Substituting for \(x = 0.260\ \mathrm{m}\), \(U = \frac{1}{2} \times 1.15 \times 7.40^2 \times 0.260^2 \approx 4.64\) Joules.
05

Calculate the Kinetic Energy when \(x = 0.260\ \mathrm{m}\)

The total energy is the sum of kinetic energy (\(K\)) and potential energy (\(U\)). Thus, \(K = E - U = 20.34 - 4.64 = 15.70\) Joules.

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Key Concepts

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

Amplitude
In simple harmonic motion, the amplitude is a crucial concept. It represents the maximum extent of oscillation from the equilibrium position. You can think of it as the "height" of the wave formed by the motion. In this exercise, the amplitude is given by the coefficient of the cosine term in the equation of motion.
This is why, in the equation, \(x = 0.650 \cos(7.40t)\), the amplitude is directly identified as \(0.650\) meters.
Understanding amplitude helps you visualize just how far the object moves from the center point in either direction during its motion.
Frequency
Frequency is another fundamental aspect of simple harmonic motion. It tells you how many times the oscillation repeats in one second and is measured in Hertz (Hz).
From the equation \(x = 0.650 \cos(7.40t)\), the term \(7.40\) is the angular frequency, represented by \(\omega\).
To find the frequency \(f\), use the formula:
  • \( f = \frac{\omega}{2\pi} \)
  • where \( \omega = 7.40 \) radians per second
So, the frequency is calculated as approximately \(1.18\) Hz. This means the cycle completely repeats a little over once every second.
Kinetic Energy
Kinetic energy in simple harmonic motion illustrates the energy associated with the motion of the mass. As the mass moves fastest at the equilibrium point, its kinetic energy is also maximum there.
The total mechanical energy of a system is a combination of both kinetic and potential energy. At any point during the oscillation, you can find the kinetic energy \(K\) by subtracting the potential energy \(U\) from the total energy \(E\).
In the exercise, when \(x = 0.260\;\mathrm{m}\), the kinetic energy becomes:
  • \( K = E - U = 20.34 - 4.64 = 15.70 \) Joules
Understanding this is vital for visualizing how energy changes form but not in total amount, in a closed system.
Potential Energy
Potential energy is the energy stored within a system due to its position or arrangement. In simple harmonic motion, potential energy is stored when the mass is displaced from the equilibrium position.
The formula to calculate potential energy when the mass is at a position \(x\) is given by:
  • \(U = \frac{1}{2} m \omega^2 x^2\)
  • where \(m\) is mass, \(\omega\) is angular frequency, and \(x\) is the displacement
For the mass at \(x = 0.260\;\mathrm{m}\), we find the potential energy:
  • \(U = \frac{1}{2} \times 1.15 \times 7.40^2 \times 0.260^2 \approx 4.64 \) Joules
This concept helps in understanding how the position affects the energy dynamics of the system.

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Most popular questions from this chapter

(II) An 1150 kg automobile has springs with \(k=16,000 \mathrm{N} / \mathrm{m}\) . One of the tires is not properly balanced; it has a little extra mass on one side compared to the other, causing the car to shake at certain speeds. If the tire radius is \(42 \mathrm{cm},\) at what speed will the wheel shake most?

(a) Show that the total mechanical energy, \(E=\frac{1}{2} m v^{2}+\frac{1}{2} k x^{2},\) as a function of time for a lightly damped harmonic oscillator is $$ E=\frac{1}{2} k A^{2} e^{-(b / m) t}=E_{0} e^{-(b / m) t} $$ where \(E_{0}\) is the total mechanical energy at \(t=0 .\) (Assume \(\omega^{\prime} \gg b / 2 m .\) ) (b) Show that the fractional energy lost per period is $$ \frac{\Delta E}{E}=\frac{2 \pi b}{m \omega_{0}}=\frac{2 \pi}{Q} $$ where \(\omega_{0}=\sqrt{k / m}\) and \(Q=m \omega_{0} / b\) is called the quality factor or \(Q\) value of the system. A larger \(Q\) value means the system can undergo oscillations for a longer time.

A vertical spring with spring stiffness constant \(305 \mathrm{~N} / \mathrm{m}\) oscillates with an amplitude of \(28.0 \mathrm{~cm}\) when \(0.260 \mathrm{~kg}\) hangs from it. The mass passes through the equilibrium point \((y=0)\) with positive velocity at \(t=0 .\) (a) What equation describes this motion as a function of time? (b) At what times will the spring be longest and shortest?

(II) A simple pendulum is 0.30 \(\mathrm{m}\) long. At \(t=0\) it is released from rest starting at an angle of \(13^{\circ} .\) Ignoring friction, what will be angular position of the pendulum at (a) \(t=0.35 \mathrm{s},\) (b) \(t=3.45 \mathrm{s},\) and \((c) t=6.00 \mathrm{s?}\)

(II) At \(t=0,\) a \(785-\mathrm{g}\) mass at rest on the end of a horizontal spring \((k=184 \mathrm{N} / \mathrm{m})\) is struck by a hammer which gives it an initial speed of 2.26 \(\mathrm{m} / \mathrm{s} .\) Determine \((a)\) the period and frequency of the motion, \((b)\) the amplitude, (c) the maximum acceleration, (d) the position as a function of time, \((e)\) the total energy, and \((f)\) the kinetic energy when \(x=0.40 A\) where \(A\) is the amplitude.
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