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\(\bullet\) A 0.500 kg glider on an air track is attached to the end of an ideal spring with force constant \(450 \mathrm{N} / \mathrm{m} ;\) it undergoes simple harmonic motion with an amplitude of 0.040 \(\mathrm{m} .\) Compute (a) the maximum speed of the glider, (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m},(\mathrm{c})\) the magnitude of the maximum acceleration of the glider, (d) the acceleration of the glider at \(x=-0.015 \mathrm{m},\) and \((\mathrm{e})\) the total mechanical energy of the glider at any point in its motion.

Short Answer

Expert verified
(a) 1.2 m/s; (b) 1.107 m/s; (c) 36 m/s²; (d) 13.5 m/s²; (e) 0.36 J.

Step by step solution

01

Determine the natural frequency

First, we need to find the natural frequency \(\omega\) of the simple harmonic oscillator. The formula for angular frequency \(\omega\) is\[\omega = \sqrt{\frac{k}{m}}\]where \(k = 450 \, \text{N/m}\) and \(m = 0.500 \, \text{kg}\). Substituting these values gives:\[\omega = \sqrt{\frac{450}{0.500}} = 30 \, \text{rad/s} \].
02

Calculate the maximum speed

The maximum speed \(v_{max}\) of the glider is given by the equation:\[v_{max} = \omega A\]where \(A = 0.040 \, \text{m}\) is the amplitude. Using \(\omega = 30 \, \text{rad/s}\), we find:\[v_{max} = 30 \times 0.040 = 1.2 \, \text{m/s}.\]
03

Calculate the speed at x = -0.015 m

The speed \(v\) at any position \(x\) for a simple harmonic oscillator is given by:\[v = \omega \sqrt{A^2 - x^2}\]Substituting \(\omega = 30 \, \text{rad/s}\), \(A = 0.040 \, \text{m}\), and \(x = -0.015 \, \text{m}\), we get:\[v = 30 \sqrt{0.040^2 - (-0.015)^2} = 30 \sqrt{0.0016 - 0.000225}\]\[v = 30 \sqrt{0.001375} \approx 1.107 \text{ m/s}\].
04

Find the maximum acceleration

The maximum acceleration \(a_{max}\) is given by:\[a_{max} = \omega^2 A\]Using \(\omega = 30 \, \text{rad/s}\) and \(A = 0.040 \, \text{m}\), we have:\[a_{max} = 30^2 \times 0.040 = 36 \, \text{m/s}^2.\]
05

Calculate the acceleration at x = -0.015 m

The acceleration \(a\) at any position \(x\) is given by:\[a = -\omega^2 x\]Substituting \(\omega = 30 \, \text{rad/s}\) and \(x = -0.015 \, \text{m}\), we get:\[a = -(30)^2 \times (-0.015) = 13.5 \, \text{m/s}^2.\]
06

Determine the total mechanical energy

The total mechanical energy \(E\) of the system at any point is constant and given by:\[E = \frac{1}{2}kA^2\]Substituting \(k = 450 \, \text{N/m}\) and \(A = 0.040 \, \text{m}\), we find:\[E = \frac{1}{2} \times 450 \times 0.040^2 = 0.36 \, \text{J}.\]

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Key Concepts

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

Natural Frequency
In simple harmonic motion, the natural frequency represents how fast an object oscillates without external forces. It's the constant rate at which the system moves back and forth. For a system with a mass \(m\) and a spring constant \(k\), the natural frequency, often expressed as angular frequency \(\omega\), is determined by the equation:\[ \omega = \sqrt{\frac{k}{m}} \]For our given values, where the mass \(m\) of the glider is 0.500 kg, and the spring constant \(k\) is 450 N/m, we find:\[ \omega = \sqrt{\frac{450}{0.500}} = 30 \, \text{rad/s} \]This calculated natural frequency, 30 rad/s, defines the rate at which the glider will naturally oscillate on the air track when disturbed by an external force. Understanding the natural frequency helps explain the dynamic behavior of the system and its stability under different conditions.
Maximum Speed
The maximum speed in simple harmonic motion occurs when the oscillating object passes through the equilibrium position, where the energy is entirely kinetic. This speed is determined by the product of the natural frequency and the amplitude of motion. The formula to find the maximum speed \(v_{max}\) of the glider is:\[ v_{max} = \omega A \]Here, \(A\) is 0.040 m (the amplitude), and we've calculated \(\omega\) as 30 rad/s. Substituting these values gives:\[ v_{max} = 30 \times 0.040 = 1.2 \, \text{m/s} \]This result shows the fastest speed the glider achieves as it moves back and forth. In practical terms, this speed indicates how quickly the system can respond to disturbances.
Mechanical Energy
Mechanical energy in simple harmonic motion comprises two parts: potential energy stored in the spring and kinetic energy of the moving object. The total mechanical energy \(E\) is conserved and constant throughout the motion. It is calculated using:\[ E = \frac{1}{2}kA^2 \]With \(k\) being the spring constant (450 N/m) and \(A\) the amplitude (0.040 m), we find:\[ E = \frac{1}{2} \times 450 \times 0.040^2 = 0.36 \, \text{J} \]This constant energy signifies that the sum of kinetic and potential energies remains unchanged throughout. At the maximum amplitude, energy is entirely potential; at equilibrium, it's entirely kinetic. This energy conservation ensures that the system continues its oscillation perpetually (assuming no dissipative forces).
Maximum Acceleration
During oscillations, acceleration is maximized when the restoring force pulls or pushes the object to its peak displacement. The object's maximum acceleration \(a_{max}\) can be calculated using:\[ a_{max} = \omega^2 A \]Substituting \(\omega = 30 \, \text{rad/s}\) and \(A = 0.040 \, \text{m}\), we obtain:\[ a_{max} = 30^2 \times 0.040 = 36 \, \text{m/s}^2 \]This represents the highest acceleration experienced by the glider as it is furthest from its equilibrium position. Understanding this concept is crucial for analyzing how quickly the object can change its velocity.

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Most popular questions from this chapter

\(\bullet\) A 2.50 kg rock is attached at the end of a thin, very light rope 1.45 \(\mathrm{m}\) long and is started swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after 10\(\frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."

\(\bullet\) \(\bullet\) A steel cable with cross-sectional area of 3.00 \(\mathrm{cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8}\) Pa. Find the maximum upward acceleration that can be given to a 1200 kg elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

\(\bullet\) \(\bullet\) Stress on the shinbone. The compressive strength of our bones is important in everyday life. Young's modulus for bone is approximately 14 GPa. Bone can take only about a 1.0\(\%\) change in its length before fracturing. If Hooke's law were to hold up to fracture: (a) What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 3.0 \(\mathrm{cm}^{2} .\) . This is approximately the cross-sectional area of a tibia, or shinbone, at its narrowest point.) (b) Estimate the maximum height from which a 70 \(\mathrm{kg}\) man can jump and not fracture the tibia. Take the time between when he first touches the floor and when he has stopped to be \(0.030 \mathrm{s},\) and assume that the stress is distributed equally between his legs.

\(\bullet\) In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 \(\mathrm{km}\) and the pressure is \(1.16 \times 10^{8}\) Pa (about 1150 atmospheres). (a) If a cubic meter of water is taken to this depth from the surface (where the normal atmospheric pressure is about \(1.0 \times 10^{3} \mathrm{Pa}\) , what is the change in its volume? Assume that the bulk modulus for seawater is the same as for freshwater \(\left(2.2 \times 10^{9} \mathrm{Pa}\right) .\) (b) At the surface, seawater has a density of \(1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) What is the density of sea-water at the depth of the Challenger Deep?

\(\bullet\) \(\bullet\) 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 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 $$\frac{f_{\mathrm{S}+\mathrm{V}}}{f_{\mathrm{S}}}=\frac{1}{\sqrt{1+\frac{m_{\mathrm{v}}}{m_{\mathrm{S}}}}},$$ 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 femtograms?

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