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A \(175-\) glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 155 \(\mathrm{N} / \mathrm{m}\) . At the instant you make measurements on the glider, it is moving at 0.815 \(\mathrm{m} / \mathrm{s}\) and is 3.00 \(\mathrm{cm}\) from its equilibrium point. Use energy conservation to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

Short Answer

Expert verified
The amplitude is 3.14 cm, the maximum speed is 0.970 m/s, and the angular frequency is 29.8 rad/s.

Step by step solution

01

Understand the Problem

We need to find the amplitude of the motion, the maximum speed of the glider, and the angular frequency of the oscillations. The glider's mass is moving at a certain speed and is at a certain displacement from the equilibrium point. We will use the concepts of energy conservation and simple harmonic motion (SHM).
02

Identify Given Information

The mass of the glider, \(m\), is 175 g or 0.175 kg (since we need it in kg). The spring constant, \(k\), is 155 N/m. The initial speed, \(v\), of the glider is 0.815 m/s, and the displacement from the equilibrium point, \(x\), is 3.00 cm or 0.03 m.
03

Apply Conservation of Energy for Amplitude

Using the conservation of energy principle, the total mechanical energy is constant. It includes kinetic energy (\(KE\)) and potential energy (\(PE\)). At an arbitrary point, we have \(KE + PE = \frac{1}{2} mv^2 + \frac{1}{2} kx^2 = \frac{1}{2} kA^2\), where \(A\) is the amplitude.Calculate: \[\frac{1}{2} (0.175)(0.815)^2 + \frac{1}{2} (155)(0.03)^2 = \frac{1}{2} (155)A^2\]
04

Solve for Amplitude

Rearrange the equation to solve for \(A\):\[A^2 = \frac{0.5 \times 0.175 \times 0.815^2 + 0.5 \times 155 \times 0.03^2}{0.5 \times 155} \]Calculate \(A\):\[A \approx 0.0314 \text{ m or } 3.14 \text{ cm}\]
05

Calculate Maximum Speed

The maximum speed occurs when all the energy is kinetic, i.e., at the equilibrium point where potential energy is zero:\[\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2\]Solve for \(v_{max}\):\[v_{max}^2 = \frac{kA^2}{m}\]\[v_{max} = \sqrt{\frac{155 \times 0.0314^2}{0.175}}\]\[v_{max} \approx 0.970 \text{ m/s}\]
06

Determine Angular Frequency

The angular frequency \(\omega\) of an oscillating spring-mass system is given by:\[\omega = \sqrt{\frac{k}{m}}\]Calculate \(\omega\):\[\omega = \sqrt{\frac{155}{0.175}}\]\[\omega \approx 29.8 \, \text{rad/s}\]

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Key Concepts

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

Energy Conservation
In physics, energy conservation is a crucial concept, especially in harmonic motion. It implies that within a closed system, the total mechanical energy remains unchanged over time. This energy is comprised of two components: kinetic energy (energy due to motion) and potential energy (energy stored in the spring). In the context of harmonic motion with a spring and a mass, the total energy can be expressed as:
  • Kinetic Energy, \( KE = \frac{1}{2}mv^2 \)
  • Potential Energy, \( PE = \frac{1}{2}kx^2 \)

For a glider attached to a spring, energy conservation means that the sum of these energies at any point in time is equal to the maximum potential energy stored when the spring is at its maximum compression or elongation. Hence, the energy equation for this system is given by:
\[ KE + PE = \frac{1}{2} kA^2 \]Where:
  • \(A\) is the amplitude of the oscillation
  • \(k\) is the spring constant

  • By using this equation, you can find the amplitude when given the mass, velocity, and displacement of the glider.
Spring Constant
The spring constant, denoted by \(k\), is a measure of the stiffness of a spring. It indicates how much force is needed to extend or compress the spring by a unit length. The spring constant is expressed in units of \( \, \mathrm{N}/\mathrm{m} \).
A larger spring constant means a stiffer spring, which requires more force for the same amount of stretch compared to a spring with a smaller constant. In our problem, the spring constant is 155 \(\mathrm{N}/\mathrm{m}\). This means:
  • For every meter the spring is stretched (or compressed), it needs a force of 155 newtons.

  • The spring constant is crucial for calculating the potential energy stored in a spring and for determining the angular frequency in simple harmonic motion. It directly affects the amplitude of the oscillations and the maximum speed of the object attached.
Oscillations
Oscillations refer to the repeated back and forth motion of an object. In this context, it is the motion of the glider as it moves around the equilibrium point when attached to a spring. Such movement is periodic and can be described through simple harmonic motion (SHM). In SHM:
  • The amplitude is the maximum displacement from the equilibrium position.
  • The motion is sinusoidal, meaning it can be described by sine and cosine functions.

  • Each complete cycle of oscillation involves movement from one position, through the equilibrium, to the opposite extreme, and back again. This type of motion is key in calculating elements like the glider's maximum speed and amplitude based on conservation of energy. Understanding oscillations allows us to predict the motion behavior of objects in systems like pendulums or springs.
Angular Frequency
Angular frequency, denoted by \(\omega\), is an important parameter in oscillatory motion, measured in radians per second (rad/s). It represents the rate of rotation. In the context of spring-mass systems, angular frequency relates to how quickly an object oscillates back and forth.
The formula for angular frequency in a spring-mass system is given by:\[ \omega = \sqrt{\frac{k}{m}} \]Where:
  • \(k\) is the spring constant
  • \(m\) is the mass of the object

With a higher angular frequency, the object completes more oscillations in a given time. In this problem, the calculated angular frequency tells us how fast the glider oscillates along the track. Understanding \(\omega\) is essential for grasping the dynamics of oscillatory systems and is fundamental in designing systems where timing of oscillations matters, like in clocks or measuring devices.

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Most popular questions from this chapter

A 40.0 -N force stretches a vertical spring 0.250 \(\mathrm{m}\) . (a) What mass must be suspended from the spring so that the system will oscillare with a period of 1.00 \(\mathrm{s} ?\) (b) If the amplitude of the motion is 0.050 \(\mathrm{m}\) and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 s after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 m below the equilibrium position, moving upward?

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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 the formula \(\frac{f_{\mathrm{S}+\mathrm{V}}}{f_{\mathrm{S}}}=\frac{1}{\sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)}}\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that 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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