Question

A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance \(x\) (in inches) of the mass from its equilibrium position after \(t\) seconds is given by the function \(x(t)=10 \sin t-10 \cos t,\) where \(x\) is positive when the mass is above the equilibrium position. a. Graph and interpret this function. b. Find \(\frac{d x}{d t}\) and interpret the meaning of this derivative. c. At what times is the velocity of the mass zero? d. The function given here is a model for the motion of an object on a spring. In what ways is this model unrealistic?

Short answer

Answer: The times when the velocity of the mass is zero are given by: \(t = \pm \frac{2\pi}{3} + 2k\pi\), where \(k\) is an integer.

Step-by-step solution

Plot the function

To graph the function \(x(t)=10 \sin t - 10 \cos t\), it is recommended to use any available graphing tool such as a graphing calculator, a computer software, or an online graphing utility. The function represents a combination of sinusoidal functions. The graph will show a periodic behavior as expected for the oscillatory motion of a spring-mass system. b) Find \(\frac{dx}{dt}\) and interpret the meaning of this derivative.

Differentiate x(t)

To find the derivative \(\frac{dx}{dt}\), we have to differentiate the function with respect to time \(t\). By applying the differentiation rules, we obtain: \(\frac{dx}{dt}=10\cos t + 10\sin t\)

Interpret the derivative

The derivative \(\frac{dx}{dt}\) represents the rate of change of displacement with respect to time, which is the velocity of the mass. Thus, the function \(\frac{dx}{dt}=10\cos t + 10\sin t\) gives the velocity of the mass at any given time. c) At what times is the velocity of the mass zero?

Find when the derivative is zero

To find the times when the velocity is zero, we need to solve the equation \(\frac{dx}{dt}=0\). \(10\cos t + 10\sin t = 0\) To solve this equation, we can divide both sides by 10: \(\cos t + \sin t = 0\) Now, let's use the substitution \(u = \cos t\): \(u + \sqrt{1-u^2} = 0\) Square both sides: \(u^2 + 2u + 1 - u^2 = 0\) \(2u + 1 = 0\) Solve for \(u\): \(u = -\frac{1}{2}\) Then, we need to find \(t\): \(\cos t = -\frac{1}{2}\) \(t = \pm \frac{2\pi}{3} + 2k\pi\), where \(k\) is an integer.

Answer for times when velocity is zero

The times when the velocity of the mass is zero are given by: \(t = \pm \frac{2\pi}{3} + 2k\pi\), where \(k\) is an integer. d) The function given here is a model for the motion of an object on a spring. In what ways is this model unrealistic?

Limitations of the model

The function \(x(t)=10\sin t - 10\cos t\) provides an idealized model of a spring-mass system. Some of the limitations and unrealistic assumptions of this model include: 1. No friction or air resistance: The model assumes that there is no friction or air resistance acting on the mass. In reality, these forces would cause the oscillations to eventually dampen and come to a halt. 2. No amplitude decay: The model assumes constant amplitude of oscillation over time, which is not true for real-world systems as energy is lost due to various factors such as internal friction in the spring, damping, and air resistance. 3. Simple harmonic motion: The model assumes that the motion of the mass follows simple harmonic motion, which is not always true for real-life objects that may not behave in such a predictable and linear way. 4. Ideal spring: The model assumes an ideal spring with a linear force-displacement relationship (Hooke's law), which may not hold true for real springs, especially when the spring is stretched or compressed beyond its elastic limit.

Key concepts

Spring-Mass System

A spring-mass system consists of a mass attached to a spring, which can move vertically or horizontally. When the mass is pulled away from its equilibrium position and released, it starts oscillating back and forth. This motion is influenced by the spring's stiffness and the mass of the object. The system can be analyzed using different equations of motion, capturing its characteristics.

• Equilibrium Position: This is the point where the spring is neither stretched nor compressed, and the force on the mass is zero.
• Simple Harmonic Motion: Often, the spring-mass system exhibits simple harmonic motion, where the restoring force is proportional to the displacement from equilibrium.
• Hooke's Law: This principle describes the force exerted by a spring: \(F = -kx\), where \(k\) is the spring constant and \(x\) is the displacement.

By understanding these key ideas, we get a comprehensive picture of how the spring-mass system behaves under different conditions.

Oscillations

Oscillations refer to the repetitive variations in motion of an object around an equilibrium point. In the case of a spring-mass system, oscillations can be described using sinusoidal functions, capturing the system's periodic nature.

• Amplitude: This is the maximum displacement from the equilibrium position, reflecting the energy in the system.
• Period: The time it takes to complete one full cycle of oscillation is called the period and is calculated as \(T = rac{2\pi}{ ext{angular frequency}}\).
• Frequency: The number of oscillations per unit time, inversely related to the period, indicates how quickly the system oscillates.

These parameters help us describe the oscillatory motion, revealing insights into the dynamics and energy distribution in the system.

Differential Calculus

Differential calculus is vital when analyzing the motion of a spring-mass system. It allows us to determine how quantities change over time, providing deeper understanding of motion.

• Velocity: By differentiating the displacement function \(x(t)\), we can find the velocity \(\frac{dx}{dt}\) of the mass, which indicates how fast it is moving and in what direction.
• Acceleration: Further differentiation gives acceleration \(\frac{d^2x}{dt^2}\), showing how quickly the velocity changes with time.
• Rate of change: Calculus helps track changes in motion parameters, essential in understanding the forces in play and predicting future behavior.

By mastering these differential concepts, students can better comprehend the dynamic behavior of oscillating systems like a spring-mass setup.

Limitations of Mathematical Models

Mathematical models provide simplifications of real-world behaviors and, thus, have inherent limitations. They are crucial for understanding dynamics yet can overlook certain practical aspects.

• Idealized Conditions: Models often assume conditions like no friction or air resistance, leading to perpetual motion. In reality, these forces slow motion over time.
• No Energy Loss: Real systems lose energy due to damping effects, but simple models usually ignore this, predicting constant oscillation amplitudes.
• Linearity Assumption: Many models assume linear relationships (e.g., Hookean behavior), which may not hold if the system is stretched beyond normal limits.
• Complex Forces: External or nonlinear forces can affect motion unpredictably, often disregarded in basic models.

Understanding these limitations helps us place models in context, recognizing when they might need adjustments or refinements.