Question

A mass of one \(\mathrm{kg}\) is attached to a spring with constant \(k=4 \mathrm{~N} / \mathrm{m}\). An external force \(F(t)=\) \(-\cos \omega t-2 \sin \omega t \mathrm{n}\) is applied to the mass. Find the displacement \(y\) for \(t>0\) if \(\omega\) equals the natural frequency of the spring-mass system. Assume that the mass is initially displaced \(3 \mathrm{~m}\) above equilibrium and given an upward velocity of \(450 \mathrm{~cm} / \mathrm{s}\).

Short answer

Answer: The final expression for the displacement y(t) of the mass is \(y(t) = \frac{8}{3} \cos(2t) + \frac{7}{4} \sin(2t) + \frac{1}{3} \cos(2t) + \frac{1}{2} \sin(2t)\).

Step-by-step solution

Write the equation of motion

Using the Hooke's Law and Newton's second law, we can write the equation of motion for a spring-mass system subjected to an external force F(t) as: \(m \frac{d^2 y}{d t^2} + k y = F(t)\), where m is the mass, k is the spring constant, and y(t) is the displacement at time t. Given that the mass is 1kg, the spring constant k is 4 N/m, and the external force is F(t) = -cos(ωt) - 2 sin(ωt) N, we can write the equation of motion as: \(\frac{d^2 y}{d t^2} + 4 y = -\cos (\omega t) - 2 \sin (\omega t)\).

Recognize the differential equation's form

The given equation is a second-order non-homogeneous differential equation: \(\frac{d^2 y}{d t^2} + 4 y = -\cos (\omega t) - 2 \sin (\omega t)\). Given that ω is the natural frequency of the system (ω = √(4)): \(\frac{d^2 y}{d t^2} + 4 y = -\cos (2 t) - 2 \sin (2 t)\).

Solve the differential equation

In order to solve a second-order non-homogeneous differential equation of the form: \(\frac{d^2 y}{d t^2} + 4 y = g(t)\) We need to find the particular solution, \(y_p(t)\), and the complementary solution, \(y_c(t)\). So the general solution will be: \(y(t) = y_c(t) + y_p(t)\). Now, let's find the complementary solution which corresponds to the homogeneous equation (without the external force): \(\frac{d^2 y}{d t^2} + 4 y = 0\). The characteristic equation is: \(r^2 + 4 = 0\), which gives us the roots: r = ±2i. Thus, the complementary solution will be: \(y_c(t) = C_1 \cos(2t) + C_2 \sin(2t)\), where \(C_1\) and \(C_2\) are constants. In order to find the particular solution, we need to guess a function for \(y_p(t)\), then differentiate it and plug it back into the non-homogeneous differential equation. Since the external force is composed of cosine and sine functions, we can use the following guess for the particular solution: \(y_p(t) = A \cos(2t) + B \sin(2t)\). Let's differentiate \(y_p(t)\) twice: \(\frac{dy_p(t)}{dt} = -2 A \sin(2t) + 2 B \cos(2t)\), \(\frac{d^2 y_p(t)}{dt^2} = -4 A \cos(2t) - 4 B \sin(2t)\). Now, plug \(y_p(t)\) and its derivatives back into the non-homogeneous equation: \(-4 A \cos(2t) - 4 B \sin(2t) + 4 (A \cos(2t) + B \sin(2t))= -\cos(2t) - 2 (\sin (2 t))\). Comparing the coefficients, we find that A = 1/3, B = 1/2. Thus, the particular solution is: \(y_p(t) = \frac{1}{3} \cos(2t) + \frac{1}{2} \sin(2t)\). So, the general solution is: \(y(t) = y_c(t) + y_p(t) = C_1 \cos(2t) + C_2 \sin(2t) + \frac{1}{3} \cos(2t) + \frac{1}{2} \sin(2t)\).

Apply the initial conditions

Now, let's apply the initial conditions to find the constants C1 and C2. We are given that at t=0, the displacement is 3m and the velocity is 450 cm/s, which is 4.5 m/s: - For t = 0, y(0) = 3: \(C_1 + \frac{1}{3} = 3 \implies C_1 = \frac{8}{3}\). - For the initial velocity, let's find y'(t), then plug t=0: \(y'(t)= -2 C_1 \sin(2t) + 2 C_2 \cos(2t) - \frac{2}{3} \sin(2t) + \cos(2t)\), \(y'(0) = 4.5 \implies 2 C_2 + 1 = 4.5 \implies C_2 = \frac{7}{4}\).

Write the final equation for displacement y(t)

So, the displacement of the mass for t > 0 is given by: \(y(t) = \frac{8}{3} \cos(2t) + \frac{7}{4} \sin(2t) + \frac{1}{3} \cos(2t) + \frac{1}{2} \sin(2t)\). This is the final expression for the displacement y(t) of the mass attached to the spring under the given conditions.

Key concepts

Understanding the Spring-Mass System

A spring-mass system is a classic model in physics used to describe a variety of oscillatory motions. It consists of a mass attached to a spring that can stretch or compress, obeying Hooke's Law. When the mass is pulled or pushed, it creates a restoring force trying to bring the system back to equilibrium.

The dynamics of the spring-mass system are governed by Newton's second law, which relates the net force acting on the mass to the acceleration it experiences. Since the only force considered (when there's no damping or external force) is the spring's restoring force, the resulting motion is harmonic, and it can be described by a second-order differential equation.

This equation reflects the relationship between displacement, mass, spring constant, and any external force. It's essential to understand the nature of this system for analyzing the behavior of oscillating systems in mechanics, electrical engineering (as in LC circuits), and beyond.

Grasping Hooke's Law

Hooke's Law is a principle of physics that relates the force necessary to extend or compress a spring to the distance it is stretched or compressed. Mathematically, Hooke's Law is represented as F = -kx, where F stands for the force applied to the spring, k is the spring constant, indicating the spring's stiffness, and x is the displacement of the spring from its equilibrium position.

The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement, which explains why it's considered a restoring force. In the context of second-order non-homogeneous differential equations, Hooke's Law plays a crucial role in defining the properties of the system's natural response.

Importance of Initial Conditions

Initial conditions in a differential equation problem specify the state of the system at a starting time, usually t = 0. For a second-order differential equation, they usually involve the initial position and initial velocity of the mass in a spring-mass system.

These conditions are vital because they allow us to solve for the constants in the general solution of a differential equation, making a particular solution that corresponds to the physical situation. Without initial conditions, we might end up with an infinite number of solutions as there would be unknown constants. Initial conditions ensure that we find a solution that matches the reality of the scenario being modeled.

Characteristic Equation in Action

The characteristic equation arises when solving the homogeneous part of a second-order differential equation. It is a tool that helps determine the form of the complementary solution (or general solution to the homogeneous equation).

For the equation \( m \frac{d^2 y}{d t^2} + k y = 0 \), the characteristic equation would be r² + \(\frac{k}{m}\) = 0. The roots of this equation, which can be real or complex, give insight into the behavior of the system—whether it will oscillate, grow exponentially, or decay. In the case of a spring-mass system, complex roots lead to oscillatory motion, which we often see as sinusoidal functions in the complementary solution.

Understanding the characteristic equation is crucial for interpreting the system's natural frequencies and periods, and it underlies the behavior of many physical systems beyond just spring and mass configurations.