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Chapter 14: Problem 27

A 0.500 -kg glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m},\) undergoes \(\mathrm{SHM}\) 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 \(x=-0.015 \mathrm{m} ;\) (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x=-0.015 \mathrm{m} ;\) (e) the total mechanical energy of the glider at any point in its motion.

Short Answer

Expert verified
(a) 1.2 m/s; (b) 1.1115 m/s; (c) 36 m/s虏; (d) 13.5 m/s虏; (e) 0.36 J.

Step by step solution

01

Identify Maximum Speed Formula

The maximum speed of an object in simple harmonic motion (SHM) is given by the formula \(v_{max} = A \omega\), where \(A\) is the amplitude and \(\omega\) is the angular frequency. The angular frequency \(\omega\) is calculated using the formula \(\omega = \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant and \(m\) is the mass of the glider.
02

Calculate Angular Frequency

Substitute the given values into the angular frequency formula: \(\omega = \sqrt{\frac{450}{0.500}} = \sqrt{900} = 30 \, \text{rad/s}\). This means the angular frequency of the glider is \(30 \, \text{rad/s}\).
03

Calculate Maximum Speed

Substitute the amplitude and angular frequency into the maximum speed formula: \(v_{max} = 0.040 \, \text{m} \times 30 \, \text{rad/s} = 1.2 \, \text{m/s}\). Thus, the maximum speed of the glider is \(1.2 \, \text{m/s}\).
04

Determine Speed at x = -0.015 m

The speed at a position \(x\) can be determined using the formula \(v = \omega \sqrt{A^2 - x^2}\). Substitute the values: \(v = 30 \, \text{rad/s} \times \sqrt{0.040^2 - (-0.015)^2}\). Simplifying: \(v = 30 \times \sqrt{0.0016 - 0.000225} = 30 \times \sqrt{0.001375}\). So \(v \approx 30 \times 0.03705 = 1.1115 \, \text{m/s}\).
05

Calculate Maximum Acceleration

The maximum acceleration in SHM is given by \(a_{max} = A \omega^2\). Substitute the values: \(a_{max} = 0.040 \, \text{m} \times (30 \, \text{rad/s})^2 = 0.040 \times 900 = 36 \, \text{m/s}^2\). Hence, the maximum acceleration is \(36 \, \text{m/s}^2\).
06

Find Acceleration at x = -0.015 m

The acceleration at any point \(x\) is given by \(a = -\omega^2 x\). Substitute the values: \(a = -(30^2) \times (-0.015) = -900 \times (-0.015) = 13.5 \, \text{m/s}^2\). Thus, the acceleration at \(x = -0.015 \, \text{m}\) is \(13.5 \, \text{m/s}^2\).
07

Calculate Total Mechanical Energy

The total mechanical energy in a spring-mass system is given by \(E = \frac{1}{2} k A^2\). Substitute the given values to calculate: \(E = \frac{1}{2} \times 450 \, \text{N/m} \times (0.040 \, \text{m})^2 = 0.36 \, \text{J}\). Therefore, the total mechanical energy is \(0.36 \, \text{J}\).

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Key Concepts

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

Maximum Speed in SHM
In Simple Harmonic Motion (SHM), an object's speed constantly changes. However, there's a point at which it reaches maximum speed. This happens when the object passes through the equilibrium point.
To find this maximum speed, we use the formula:\( v_{\text{max}} = A \omega \)where:
  • \( A \) is the amplitude, or the maximum distance from the equilibrium.
  • \( \omega \) is the angular frequency, which we will discuss further in another section.
For the glider in question, with an amplitude of 0.040 m and an angular frequency of 30 rad/s, the calculation is straightforward:\[ v_{\text{max}} = 0.040 \times 30 = 1.2 \text{ m/s} \]This means the glider moves fastest, at 1.2 m/s, as it passes through the center of its motion.
Angular Frequency Calculation
Angular frequency is a measure of how quickly the object oscillates back and forth. It's essential in determining various properties of SHM, including speed and energy.
The formula to calculate angular frequency \( \omega \) in a spring-mass system is:\( \omega = \sqrt{\frac{k}{m}} \)where:
  • \( k \) is the spring constant, indicating the stiffness of the spring.
  • \( m \) is the mass of the object.
For our glider:\( \omega = \sqrt{\frac{450}{0.500}} = \sqrt{900} = 30 \text{ rad/s} \)Thus, the angular frequency of the glider is 30 radians per second. This value helps in calculating the maximum speed and acceleration, as well as other characteristics of motion.
Mechanical Energy in Spring-Mass System
In any simple harmonic motion system, the sum of kinetic and potential energy remains constant. This total energy, known as mechanical energy, plays a crucial role in understanding the system's dynamics.
For a spring-mass system, the total mechanical energy \( E \) is calculated using the equation:\( E = \frac{1}{2} k A^2 \)where:
  • \( k \) is the spring constant.
  • \( A \) is the amplitude of motion.
In this scenario with a spring constant of 450 N/m and amplitude 0.040 m:\( E = \frac{1}{2} \times 450 \times (0.040)^2 \)Simplifying, we get:\( E = 0.36 \text{ J} \)
This energy indicates that despite the changes in speed and position of the glider, the total energy remains constant at 0.36 Joules throughout its motion.

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Most popular questions from this chapter

CP A holiday ornament in the shape of a hollow sphere with mass \(M=0.015 \mathrm{kg}\) and radius \(R=0.050 \mathrm{m}\) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (Hint: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

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?

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 \(\mathrm{kg}\) , and a 175 -kg sack of gravel sits on the middle of it. The beam is oscillating in \(\mathrm{SHM}\) , with an amplitude of 40.0 \(\mathrm{cm}\) and a frequency of 0.600 cycle/s. (a) The sack of gravel falls off the beam when the beam has its maximum upward displacement. What are the frequency and amplitude of the subsequent SHM of the beam? (b) If the gravel instead falls off when the beam has its maximum speed, what are the frequency and amplitude of the subsequent SHM of the beam?

A 0.150 -kg toy is undergoing SHM on the end of a horizontal spring with force constant \(k=300 \mathrm{N} / \mathrm{m} .\) When the object is 0.0120 \(\mathrm{m}\) from its equilibrium position, it is observed to have a speed of 0.300 \(\mathrm{m} / \mathrm{s} .\) What are (a) the total energy of the object at any point of its motion; (b) the amplitude of the motion; \((\mathrm{c})\) the maximum speed attained by the object during its motion?

A 1.50 -kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 \(\mathrm{N} / \mathrm{m}\) and a \(275-\mathrm{g}\) metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point \(A\) , which is 15.0 \(\mathrm{cm}\) below the equilibrium point, and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?
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