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Chapter 13: Problem 31

A 175 -g glider ona horizontal, frictionless air trackis 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
a) Amplitude is 0.0404 m, b) max speed is 1.2 m/s, c) angular frequency is 29.75 rad/s.

Step by step solution

01

Understanding Given Data

We have a 175 g glider on a frictionless track connected to a spring with a constant of 155 N/m. It's moving at 0.815 m/s and is 0.03 m from equilibrium. We will use these to find the amplitude, maximum speed, and angular frequency.
02

Converting Mass to Kg

Convert the mass of the glider from grams to kilograms: \[ 175 \text{ g} = 0.175 \text{ kg} \]
03

Using Energy Conservation for Amplitude

Energy conservation gives us: \[ \frac{1}{2} m v^2 + \frac{1}{2} k x^2 = \frac{1}{2} k A^2 \]Substitute the known values: \[ \frac{1}{2} (0.175) (0.815)^2 + \frac{1}{2} (155) (0.03)^2 = \frac{1}{2} (155) A^2 \]Solve for A.
04

Calculating Amplitude

Simplify and solve:\[ 0.05707 + 0.06975 = 77.5 A^2 \]\[ A^2 = \frac{0.05707 + 0.06975}{77.5} \]\[ A = \sqrt{0.0016335} \approx 0.0404 \text{ m} \]
05

Finding Maximum Speed

The maximum kinetic energy is equal to the total energy. Thus, \[ \frac{1}{2} k A^2 = \frac{1}{2} m v_{max}^2 \]Substitute A and solve for \(v_{max}\):\[ \frac{1}{2} (155) (0.0404)^2 = \frac{1}{2} (0.175) v_{max}^2 \]Calculate \(v_{max}\).
06

Calculating Maximum Speed

Simplify and solve: \[ 0.125999 = 0.0875 v_{max}^2 \]\[ v_{max}^2 = \frac{0.125999}{0.0875} \]\[ v_{max} = \sqrt{1.44} \approx 1.2 \text{ m/s} \]
07

Finding Angular Frequency

The angular frequency \( \omega \) is calculated as: \[ \omega = \sqrt{\frac{k}{m}} \]Substitute the values:\[ \omega = \sqrt{\frac{155}{0.175}} \]Compute \( \omega \).
08

Calculating Angular Frequency

Simplify and solve:\[ \omega = \sqrt{885.71} \approx 29.75 \text{ rad/s} \]

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Key Concepts

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

Amplitude of Harmonic Motion
Amplitude in harmonic motion represents the maximum displacement from the equilibrium position. For a glider attached to an ideal spring, the energy conservation principle helps us understand the interplay between kinetic and potential energy. At any point during the glider's oscillation, the total mechanical energy remains constant.

This total energy consists of kinetic energy when the glider is moving and potential energy when it is displaced by the spring. The formula for energy conservation is given as:
  • \[ \frac{1}{2} mv^2 + \frac{1}{2} kx^2 = \frac{1}{2} kA^2 \]
These terms respectively represent kinetic energy, potential energy, and the total energy represented by the maximum amplitude squared. By substituting the known values, one can solve for amplitude \(A\), which in this exercise is found to be approximately 0.0404 meters. Understanding amplitude allows us to determine the extent to which the glider can move from the equilibrium position during its harmonic motion.
Maximum Speed in Harmonic Oscillations
The maximum speed of an oscillating object such as a glider on a spring is a crucial part of understanding its motion dynamics. It occurs when the object passes through the equilibrium position, where the potential energy is zero, and all the energy is kinetic. This implies that the maximum kinetic energy is equal to the total mechanical energy in the system.

The formula used to calculate the maximum speed \(v_{max}\) is:
  • \[ \frac{1}{2} kA^2 = \frac{1}{2} mv_{max}^2 \]
Rearranging, we solve for \(v_{max}\) using the previously determined amplitude \(A\). In this particular exercise, the maximum speed of the glider is found to be approximately 1.2 meters per second. This speed reflects the energetic transfer from potential energy to kinetic energy as the glider moves through its equilibrium point.
Angular Frequency Calculation
Angular frequency \(\omega\) is a fundamental concept in harmonic motion, measuring how fast the object oscillates. It provides insight into the dynamic behavior of the system, including how quickly the oscillation cycle repeats.

The formula to compute angular frequency is:
  • \[ \omega = \sqrt{\frac{k}{m}} \]
This equation directly ties angular frequency to both the spring constant \(k\) and the mass \(m\) of the object. For a given spring constant and mass, substituting into the formula gives us the angular frequency of the system. In the example provided, the angular frequency is calculated to be approximately 29.75 radians per second.

Understanding \(\omega\) helps anticipate the time behavior of the oscillating system and is a key element in predicting the position and velocity of the system as a function of time.

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Most popular questions from this chapter

A tuning fork labeled 392 Hz has the tip of each of its two prongs vibrating with an amplitude of 0.600 \(\mathrm{mm}\) . (a) What is the maximum speed of the tip of a prong? (b) A housefly (Musca domestica) with mass 0.0270 \(\mathrm{g}\) is holding on to the tip of one of the prongs. As the prong vibrates, what is the fly's maximum kinetic energy? Assume that the fly's mass has a negligible effect on the frequency of oscillation.

A 0.500 \(\mathrm{kg}\) glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m}\) , undergoes 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.

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 reased, 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.)

When a 0.750 -kg mass oscillates on an ideal spring, the frequency is 1.33 Hz. What will the frequency be if 0.220 kg are added to the original mass, and (b) subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

A \(0.400-\mathrm{kg}\) object undergoing \(\mathrm{SHM}\) has \(a_{x}=-2.70 \mathrm{m} / \mathrm{s}^{2}\) when \(x=0.300 \mathrm{m}\) . What is the time for one oscillation?
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