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Chapter 14: Problem 27

A \(0.150 \mathrm{~kg}\) toy is undergoing \(\mathrm{SHM}\) on the end of a horizontal spring with force constant \(k=300 \mathrm{~N} / \mathrm{m}\). When the toy is \(0.0120 \mathrm{~m}\) from its equilibrium position, it is observed to have a speed of \(0.400 \mathrm{~m} / \mathrm{s} .\) What are the toy's (a) total energy at any point of its motion; (b) amplitude of motion; (c) maximum speed during its motion?

Short Answer

Expert verified
The total energy at any point of its motion is contained within the energy equation in SHM, which combines kinetic and potential energy. The amplitude of motion is obtained from potential energy at the point of maximum displacement. The maximum speed is found from the kinetic energy at equilibrium position, where potential energy is zero.

Step by step solution

01

Calculate total energy

The total energy in a simple harmonic motion is said to be constant and it equals to kinetic energy plus potential energy. At the given position, where the toy is observed, we have these two forms of energy. Kinetic energy is \(1/2 m v^2\), where \(m = 0.150 ~kg\) is the mass of the toy, and \(v = 0.400 ~m/s\) its speed. The potential energy of the spring at this position can be calculated by \(1/2 k x^2\), where \(k=300 ~N/m\) is the spring constant, and \(x = 0.0120 ~m\) is the displacement from the equilibrium position. The total energy \(E\) is then given by: \(E = 1/2 m v^2 + 1/2 k x^2\).
02

Compute the amplitude of motion

The amplitude of motion is the maximum displacement from the equilibrium position. This is reached when the potential energy is at a maximum which equals the total energy, because at this point the kinetic energy is zero (the object momentarily stops before reversing its motion). We express the potential energy at the maximum displacement \(A\) giving \(1/2 k A^2 = E\). We can solve for \(A\) by rearranging this equation: \(A = sqrt(2E/k)\).
03

Find the maximum speed

The maximum speed is attained at the equilibrium position where the total energy is fully in the form of kinetic energy as potential energy is zero. Given the total energy \(E\) and the mass \(m\), we can express kinetic energy at the maximum speed \(V\), as \(1/2 m V^2 = E\). Rearranging this equation gives: \(V = sqrt(2E/m)\).

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Key Concepts

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

Total Energy in SHM
When exploring the concept of Simple Harmonic Motion (SHM), the idea of total energy provides a constant measure throughout the motion. Total energy in SHM is the sum of kinetic and potential energy that an object has as it moves back and forth.

In SHM, energy shifts back and forth between kinetic and potential forms, but their sum remains constant. The kinetic energy, given by the formula \( \frac{1}{2} mv^2 \) where \( m \) is the object's mass and \( v \) its velocity, is highest when the object passes through the equilibrium point. On the other hand, potential energy, which for a spring system is expressed as \( \frac{1}{2} kx^2 \) where \( k \) is the spring constant and \( x \) the displacement from equilibrium, peaks when the object is at its furthest points from equilibrium.

The conservation of energy principle ensures that the sum of these energies remains constant, even though each energy form individually varies throughout the cycle of SHM. This total is an important aspect when analyzing motion properties, predicting movement, and determining other characteristics of SHM like amplitude and maximum speed.
Amplitude of SHM
A fundamental parameter in Simple Harmonic Motion is the amplitude, often denoted by \( A \). The amplitude represents the maximum displacement of the object from its equilibrium position and is a key indicator of the motion's extent.

It is critical to note that amplitude does not depend on mass or speed; rather, it is purely a function of the energy within the system and the spring constant (for spring-mass systems). The relationship \( A = \sqrt{\frac{2E}{k}} \) connects the amplitude to the total energy \( E \) and the spring constant \( k \) of the system. This equation reveals that a higher total energy or a softer spring (with a lower \( k \) value) leads to a greater amplitude.

Amplitude serves as a crucial concept in understanding the limits within which the object will operate during its motion and is integral for calculating various other characteristics of SHM, including the maximum potential energy and the point at which kinetic energy is zero.
Maximum Speed in SHM
The maximum speed in SHM occurs as the object passes through the equilibrium point. At this juncture, all the total energy is converted into kinetic energy since the potential energy is zero. Understanding the maximum speed is vital as it reflects the most dynamic part of the object's motion.

Calculated using the formula \( V = \sqrt{\frac{2E}{m}} \), where \( V \) is the maximum speed, \( E \) the total energy, and \( m \) the object's mass, this velocity is a direct indicator of the system鈥檚 energy. When the mass is constant, an increase in total energy directly translates into a higher maximum speed.

Maximum speed in SHM not only helps in predicting the fastest point of an object鈥檚 oscillatory motion but is also essential when analyzing forces and accelerations within the system, given that these values are also at their highest when the object is at equilibrium.

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Most popular questions from this chapter

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A\). (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium.) (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?

In a physics lab, you attach a \(0.200 \mathrm{~kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is \(2.60 \mathrm{~s}\). Find the spring's force constant.

Don't Miss the Boat. While on a visit to Minnesota ("Land of 10,000 Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the \(1500 \mathrm{~kg}\) boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude \(20 \mathrm{~cm}\) The boat takes \(3.5 \mathrm{~s}\) to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass \(60 \mathrm{~kg}\) ) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within \(10 \mathrm{~cm}\) of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?

A uniform, solid metal disk of mass \(6.50 \mathrm{~kg}\) and diameter \(24.0 \mathrm{~cm}\) hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of \(4.23 \mathrm{~N}\) tangent to the rim of the disk to turn it by \(3.34^{\circ},\) thus twisting the wire. You now remove this force and release the disk from rest. (a) What is the torsion constant for the metal wire? (b) What are the frequency and period of the torsional oscillations of the disk? (c) Write the equation of motion for \(\theta(t)\) for the disk.

BIO Weighing a Virus. In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just \(30 \mathrm{nm}\) long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{V}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by \(f_{\mathrm{S}+\mathrm{V}} / f_{\mathrm{S}}=1 / \sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)},\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that it is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{~g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{~Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{~Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?
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