Question

These exercises deal with undamped vibrations of a spring-mass system, $$ m y^{\prime \prime}+k y=0, \quad y(0)=y_{0}, \quad y^{\prime}(0)=y_{0}^{\prime} . $$ Use a value of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) or \(32 \mathrm{ft} / \mathrm{sec}^{2}\) for the acceleration due to gravity. A 10-kg mass, when attached to the end of a spring hanging vertically, stretches the spring \(30 \mathrm{~mm}\). Assume the mass is then pulled down another \(70 \mathrm{~mm}\) and released (with no initial velocity). (a) Determine the spring constant \(k\). (b) State the initial value problem (giving numerical values for all constants) for \(y(t)\), where \(y(t)\) denotes the displacement (in meters) of the mass from its equilibrium rest position. Assume that \(y\) is measured positive in the downward direction. (c) Solve the initial value problem formulated in part (b).

Short answer

Answer: The displacement of the mass from its equilibrium rest position as a function of time is given by the function y(t) = 0.07cos(√326.667 * t).

Step-by-step solution

Determine the spring constant k

Begin by using Hooke's Law, which states that the force exerted by a spring is proportional to its displacement from its rest position: $$ F = -ky $$ In this case, the force is the weight of the mass, which is given by mg (where m is the mass and g is the acceleration due to gravity). The displacement y is the initial elongation of the spring due to the weight, which is 30 mm or 0.03 m. Plug in these values: $$ mg = -k(0.03) $$ We are given the mass m (10 kg) and can use the given value for the acceleration due to gravity g (9.8 m/s^2). Solve for k: $$ k = -\frac{(10)(9.8)}{0.03} $$ $$ k = 3266.67 \, N/m $$ The spring constant k is approximately 3266.67 N/m.

State the initial value problem

Given the equation $$ m y^{\prime \prime}+k y=0 $$ we need to state the initial value problem for the displacement y(t) and its derivative. The initial conditions are given as follows: 1. The initial displacement y(0) is the amount the mass is pulled down after the initial elongation, which is 70 mm or 0.07 m. 2. The initial velocity y'(0) is given as having no initial velocity, or y'(0) = 0. With these initial conditions, we have the following initial value problem: $$ 10 y^{\prime \prime} + 3266.67 y = 0, \quad y(0) = 0.07, \quad y^{\prime}(0) = 0 $$

Solve the initial value problem

In order to solve the initial value problem, first, divide the equation by the mass m=10: $$ y^{\prime \prime} + 326.667 y = 0 $$ Notice that this is a simple harmonic oscillator equation, which can be solved by assuming a solution of the form $$y(t) = A\cos(\omega t) + B\sin(\omega t)$$, where A and B are unknown constants, and $$\omega = \sqrt{326.667}$$. We can now make use of the initial conditions given. First, we know that y(0) = 0.07, thus $$ A\cos(0) + B\sin(0) = 0.07 \Rightarrow A = 0.07 $$ Next, we have to find the initial velocity function, which is the derivative of the displacement function: $$ y'(t) = -A\omega\sin(\omega t) + B\omega\cos(\omega t) $$ We are given that y'(0) = 0, thus $$ -A\omega\sin(0) + B\omega\cos(0) = 0 \Rightarrow B = 0 $$ With A=0.07 and B=0, the solution we obtain for y(t) is $$ y(t) = 0.07\cos(\sqrt{326.667} \cdot t) $$ The displacement of the mass from its equilibrium rest position as a function of time y(t) is given by: $$ y(t) = 0.07\cos(\sqrt{326.667} \cdot t) $$

Key concepts

Undamped Vibrations

When a mass is connected to a spring, its motion typically features oscillations, or vibrations. Undamped vibrations refer to oscillations where the energy is conserved over time. In these ideal systems, no energy is lost due to friction, air resistance, or other damping forces. In a real-world scenario, some damping usually occurs, but assuming an undamped system can often be a good approximation for short time periods or systems with minimal friction.

Characterizing Undamped Motion

In the given exercise, the spring-mass system is described by a differential equation that does not account for any damping forces. Mathematically, this scenario is represented by the second-order linear differential equation
\( m y'' + k y = 0 \) where \( m \) is the mass and \( k \) is the spring constant. The solution to this type of equation models the motion of the mass-spring system through time, illustrating how the mass moves back and forth from its rest position in perpetual motion if undamped.

Initial Value Problem

To predict the future behavior of a physical system, we often set up an initial value problem. This involves differential equations coupled with specific starting conditions—initial values. The initial values in the context of the spring-mass system are the initial displacement and velocity of the mass.

Setting Up the Problem

The goal is to solve for the displacement <\( y(t) \)> as a function of time, knowing the spring constant <\( k \)>, the mass <\( m \)>, and the initial displacement and velocity. The exercise details these conditions precisely, giving us the necessary variables to establish the initial value problem for <\( y(t) \)>. These initial conditions form the foundation for solving the differential equation and understanding the resulting motion.

Simple Harmonic Motion

In physics, simple harmonic motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. This restoring force is what causes the mass attached to the spring to oscillate.

The Nature of SHM

The classic example of SHM is a perfect mass-spring system, free of damping. The oscillations occur at a constant frequency and amplitude; this is because the total energy in the system remains constant over time. In the context of the problem we're discussing, after applying the initial conditions to the general solution of the differential equation, the function <\( y(t) \)> represents the simple harmonic motion of the mass.

The spring-mass system's periodic motion is characterized by an angular frequency <\( \omega \)>, derived from the spring constant <\( k \)> and the mass <\( m \)>. The motion is sinusoidal, with the displacement <\( y(t) \)> described by a cosine function, showcasing the beautiful predictability and symmetry of SHM.