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Chapter 17: Problem 8

A 29.4-N weight stretches a spring 1.225 m. The spring-mass system resides in a medium offering a resistance to the motion equal to 18 times the instantaneous velocity. If the weight is released at a position \(0.6 \mathrm{m}\) above its equilibrium position with a downward velocity of \(0.9 \mathrm{m} / \mathrm{s},\) find its position relative to the equilibrium position 2 s later.

Short Answer

Expert verified
The position is -0.01375 meters relative to equilibrium after 2 seconds.

Step by step solution

01

Find Spring Constant

The force exerted by the spring is given by Hooke's Law: \( F = kx \), where \( F \) is the force in Newtons, \( k \) is the spring constant, and \( x \) is the displacement in meters. We know:\[F = 29.4 \, \text{N} \x = 1.225 \, \text{m}\]So, solving for \( k \) gives:\[k = \frac{F}{x} = \frac{29.4}{1.225} \approx 24 \, \text{N/m}\]
02

Setup Differential Equation

The motion of the system can be modeled by a second-order linear differential equation considering damping:\[m\frac{d^2y}{dt^2} + c\frac{dy}{dt} + ky = 0\]where \( m \) is the mass, \( c \) is the damping coefficient, and \( k \) is the spring constant. The weight \( W = mg = 29.4 \, \text{N} \) implies \( m = 3 \, \text{kg} \) (since \( g \approx 9.8 \, \text{m/s}^2 \)). The damping coefficient is given by 18 times the velocity; thus, \( c = 18 \). Substituting, the equation becomes:\[3\frac{d^2y}{dt^2} + 18\frac{dy}{dt} + 24y = 0\]
03

Solve the Differential Equation

We can simplify the equation by dividing through by 3:\[\frac{d^2y}{dt^2} + 6\frac{dy}{dt} + 8y = 0\]This is a standard form of a second-order linear differential equation. Its characteristic equation is:\[r^2 + 6r + 8 = 0\]Solving gives:\[r = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \implies r = -2, r = -4\]The solution is:\[y(t) = C_1e^{-2t} + C_2e^{-4t}\]
04

Determine Constants using Initial Conditions

The initial conditions are: when \( t = 0 \), \( y = -0.6 \) m (since release was with a 0.6 m drop from equilibrium), and \( \frac{dy}{dt} = 0.9 \) m/s. Using these, we solve:\[-0.6 = C_1 \cdot 1 + C_2 \cdot 1 \, 0.9 = -2C_1 \cdot 1 - 4C_2 \cdot 1\]Solving these equations simultaneously:\[C_1 + C_2 = -0.6 \-2C_1 - 4C_2 = 0.9\]From the first equation, \( C_1 = -0.6 - C_2 \). Substitute in the second:\[-2(-0.6 - C_2) - 4C_2 = 0.9 \ 1.2 + 2C_2 - 4C_2 = 0.9\]\[1.2 - 2C_2 = 0.9 \ C_2 = 0.15 ,\; C_1 = -0.75\]
05

Calculate Position at t=2s

Using the solution with constants:\[y(t) = -0.75e^{-2t} + 0.15e^{-4t}\]Substitute \( t = 2 \):\[y(2) = -0.75e^{-4} + 0.15e^{-8}\]Computing the exponential terms using a calculator,\[e^{-4} \approx 0.0183 \quad \text{and} \quad e^{-8} \approx 0.0003\]Thus,\[y(2) = -0.75 \times 0.0183 + 0.15 \times 0.0003 \approx -0.01375 \, \text{m}\]
06

Conclusion: Interpret Result

The position of the mass relative to the equilibrium position after 2 seconds is -0.01375 meters. The negative sign indicates it is slightly below the equilibrium point at this time.

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Key Concepts

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

Spring Constant
The spring constant, often denoted as \( k \), is a fundamental property of a spring that describes its stiffness. According to Hooke's Law, the force \( F \) exerted by a spring is proportional to the displacement \( x \) from its equilibrium position. This relationship is mathematically expressed as \( F = kx \). A stiffer spring will have a larger spring constant, indicating more force is needed to produce a certain amount of displacement.
  • A known force of 29.4 N stretches the spring by 1.225 m.
  • Using \( k = \frac{F}{x} \), the spring constant computes to approximately 24 N/m.
  • The spring constant helps determine how the spring will react in various scenarios, especially when placed in mechanical systems.
Understanding the spring constant is key to predicting the behavior of spring-mass systems under various forces and conditions.
Damping Coefficient
The damping coefficient, represented by \( c \), describes the resistance a medium offers to the motion of an object within it. In our problem, the resistance is proportional to the velocity, specifically 18 times the instantaneous velocity. The damping coefficient affects how quickly the oscillations in a system decrease over time, leading to a gradual reduction in motion, known as damping.
  • The system in the exercise has a damping coefficient of 18, indicating significant resistance to motion.
  • This factor, combined with mass and spring constant, shapes the differential equation governing the system's motion.
  • High damping coefficients lead to faster decay of oscillations, possibly preventing oscillations entirely in critical cases.
The damping coefficient is crucial in designing systems that require specific oscillation control, such as shock absorbers in vehicles or sound-dampening frameworks.
Initial Conditions
Initial conditions are the specific values at the start of observation, crucial in solving differential equations for dynamic systems. They provide the necessary constants to define a particular solution from a family of solutions. In this scenario, the initial conditions are:
  • The weight is released at 0.6 m above its equilibrium position.
  • It also has a downward velocity of 0.9 m/s at the release time.
These initial conditions allow us to solve the differential equation by plugging these values into the solutions, determining the exact constants \( C_1 \) and \( C_2 \). Thus, precise initial conditions define not only the starting state but also the entire motion of the system, influencing how solutions evolve over time.

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Most popular questions from this chapter

A 39.2 -N weight stretches a spring \(1.225 \mathrm{m}\). The spring-mass system resides in a medium offering a resistance to the motion equal to 24 times the instantaneous velocity, and an external force given by \(f(t)=28.8+19.2 e^{-t}\) (in newtons) is being applied. If the weight is released at a position \(0.6 \mathrm{m}\) above its equilibrium position with downward velocity of \(0.9 \mathrm{m} / \mathrm{s},\) find its position relative to the equilibrium after 2 s have elapsed.

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