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A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by \(0.065 \mathrm{m},\) released from rest, and subsequently oscillate back and forth with an angular frequency of 11.3 \(\mathrm{rad} / \mathrm{s}\) . What is the speed of the object at the instant when the spring is stretched by 0.048 \(\mathrm{m}\) relative to its unstrained length?

Short Answer

Expert verified
Speed of the object is approximately 0.624 m/s.

Step by step solution

01

Understand the Problem

We have a spring system where the spring compresses and then stretches, oscillating with a given angular frequency. We're tasked with finding the speed of the object when the spring is stretched from its equilibrium position.
02

Identify Given Values

Initial compression of the spring, \( x_0 = 0.065 \mathrm{m} \); angular frequency, \( \omega = 11.3 \ \mathrm{rad/s} \); displacement from equilibrium at the desired speed, \( x = 0.048 \mathrm{m} \).
03

Apply Conservation of Energy

Using energy conservation: Initial total energy (at compression) = Total energy at the new stretch position. The total mechanical energy in oscillation is given by:\[ E = \frac{1}{2} k x_0^2 = \frac{1}{2} k x^2 + \frac{1}{2} m v^2 \]where \(k\) is the spring constant, and \(v\) is the speed of the object at \(x\).
04

Calculate Spring Constant using Angular Frequency

The angular frequency \( \omega \) is related to the spring constant \( k \) and mass \( m \) by:\[ \omega = \sqrt{\frac{k}{m}} \Rightarrow k = m \omega^2 \]
05

Rearrange Energy Equation for Speed

Using the relation for \(k\) from Step 4 in the energy conservation equation, we get:\[ \frac{1}{2} m \omega^2 x_0^2 = \frac{1}{2} m \omega^2 x^2 + \frac{1}{2} m v^2 \]Cancel the mass and rearrange for \(v\):\[ v^2 = \omega^2 (x_0^2 - x^2) \]
06

Calculate the Speed of the Object

Substitute \( \omega = 11.3 \ \, \mathrm{rad/s} \), \( x_0 = 0.065 \mathrm{m} \), and \( x = 0.048 \mathrm{m} \) in:\[ v = \sqrt{(11.3)^2 ((0.065)^2 - (0.048)^2)} \]Calculate the numerical value to find:\[ v \approx 0.624 \, \mathrm{m/s} \]
07

Finalize the Solution

The speed of the object when the spring is stretched to \(0.048 \, \mathrm{m}\) is approximately \(0.624 \, \mathrm{m/s}\).

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Key Concepts

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

Conservation of Energy
The conservation of energy is a fundamental concept in physics important for understanding oscillating systems. In the context of a spring oscillation, energy is constantly transformed between kinetic energy (the energy of motion) and potential energy (stored energy in the spring).
When the spring is compressed or stretched, it stores potential energy. As the spring returns to its equilibrium position, the potential energy is converted into kinetic energy. The principle of conservation of energy states that the total mechanical energy of the system remains constant if there are no external forces, like friction.
This can be represented by the equation:
  • \[ E = rac{1}{2} k x_0^2 = rac{1}{2} k x^2 + rac{1}{2} m v^2 \]
where:
  • \(E\) is the total mechanical energy,
  • \(k\) is the spring constant,
  • \(x_0\) is the initial compression,
  • \(x\) is the displacement from equilibrium at any point,
  • \(v\) is the speed of the object.
At any instance of the spring's oscillation, energy transformations comply with this equation, which ensures the overall energy is conserved.
Angular Frequency
Angular frequency is a measure of how quickly an object moves back and forth in oscillation, measured in radians per second. It is related to the oscillation's frequency and the properties of the system involved. For a spring-mass system, the angular frequency \( \omega \) is tied to the spring constant \( k \) and the mass \( m \) of the oscillating object.The formula connecting angular frequency to these parameters is:
  • \[ \omega = \sqrt{\frac{k}{m}} \]
Here:
  • \(\omega\) is the angular frequency,
  • \(k\) is the spring constant,
  • \(m\) is the mass of the object.
This equation shows that the angular frequency increases with a stronger spring (higher \( k \)) and decreases with a heavier mass (higher \( m \)).
This is because a stiffer spring or a lighter mass enables the system to oscillate faster. Understanding this relationship is crucial for analyzing the speed and behavior of oscillating systems.
Mechanical Energy
Mechanical energy in a spring system encompasses both kinetic and potential energy. It determines how energy is distributed within the system over its oscillation path. In a spring system, this energy storage and transfer allows the object connected to the spring to continue moving between compression and elongation.
When the spring is at the maximum compression or elongation, the energy is at its potential maximum. When it passes through the equilibrium position, energy is at its kinetic maximum, as the object is moving fastest.
The expression for mechanical energy is:
  • \[ E = rac{1}{2} k x^2 + rac{1}{2} m v^2 \]
Where:
  • \(k\) is the spring constant,
  • \(x\) is the current displacement,
  • \(v\) is the current speed of the object,
  • \(m\) is the mass of the object.
By setting up the energy equation with both kinetic and potential energy, we can solve for unknowns like the velocity of the object at different points in its motion. Recognizing the role of mechanical energy helps in understanding the interaction and balance within oscillating systems.
Spring Constant
The spring constant \(k\) is a measure of a spring's stiffness or resistance to compression and elongation. It defines how much force is needed to deform the spring by a given amount. A higher spring constant indicates a stiffer spring, requiring more force for the same amount of stretch or compression.
The spring constant is critical for determining the spring's behavior and is used in several equations related to spring mechanics.
For an oscillating spring system, the spring constant is connected to the angular frequency and mass as shown:
  • \[ k = m \omega^2 \]
This formula implies:
  • \(k\) can be directly found if \(m\) (mass) and \(\omega\) (angular frequency) are known,
  • the stiffer the spring (higher \(k\)), the higher the angular frequency \(\omega\).
Understanding the spring constant helps in analyzing the dynamical properties of the spring, predicting how it will behave under certain conditions, and solving for variables like force and energy within the spring system. Knowledge of the spring constant allows for precise calculations in various applications, including engineering and acoustics.

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Most popular questions from this chapter

A 15.0 -kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.00 \(\mathrm{m} / \mathrm{s}\) in 0.500 \(\mathrm{s}\) . In the process, the spring is stretched by 0.200 \(\mathrm{m}\) . The block is then pulled at a constant speed of 5.00 \(\mathrm{m} / \mathrm{s}\) , during which time the spring is stretched by only 0.0500 \(\mathrm{m} .\) Find \((\mathrm{a})\) the spring constant of the spring and \((\mathrm{b})\) the coefficient of kinetic friction between the block and the table.

A die is designed to punch holes with a radius of \(1.00 \times 10^{-2} \mathrm{m}\) in a metal sheet that is \(3.0 \times 10^{-3} \mathrm{m}\) thick, as the drawing illustrates. To punch through the sheet, the die must exert a shearing stress of \(3.5 \times 10^{8} \mathrm{Pa}\) . What force \(\overrightarrow{\mathrm{F}}\) must be applied to the die?

Depending on how you fall, you can break a bone easily. The severity of the break depends on how much energy the bone absorbs in the accident, and to evaluate this let us treat the bone as an ideal spring. The maximum applied force of compression that one man's thighbone can endure without breaking is \(7.0 \times 10^{4} \mathrm{N}\) . The minimum effective cross-sectional area of the bone is \(4.0 \times 10^{-4} \mathrm{m}^{2}\) , its length is \(0.55 \mathrm{m},\) and Young's modulus is \(Y=9.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}\) . The mass of the man is 65 \(\mathrm{kg}\) . He falls straight down without rotating, strikes the ground stiff-legged on one foot, and comes to a halt without rotating. To see that it is easy to break a thighbone when falling in this fashion, find the maximum distance through which his center of gravity can fall without his breaking a bone.

Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{N} / \mathrm{m}\) . Four identical passengers it down inside the car, and it is set into a vertical oscillation that has a period of 0.370 s. If the mass of the empty car is 1560 \(\mathrm{kg}\) , determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.

A 1.1\(\cdot \mathrm{kg}\) object is suspended from a vertical spring whose spring constant is 120 \(\mathrm{N} / \mathrm{m}\) . (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of 0.20 \(\mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

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