Question

A 10-kg object suspended from the end of a vertically hanging spring stretches the spring \(9.8 \mathrm{~cm}\). At time \(t=0\), the resulting spring-mass system is disturbed from its rest state by the given applied force, \(F(t)\). The force \(F(t)\) is expressed in newtons and is positive in the downward direction; time is measured in seconds. (a) Determine the spring constant, \(k\). (b) Formulate and solve the initial value problem for \(y(t)\), where \(y(t)\) is the displacement of the object from its equilibrium rest state, measured positive in the downward direction. (c) Plot the solution and determine the maximum excursion from equilibrium made by the object on the \(t\)-interval \(0 \leq t<\infty\) or state that there is no such maximum. \(F(t)=20 \cos 10 t\)

Short answer

The governing differential equation for the displacement of the mass-spring system with an applied force is: $(10)y''(t) = -101.96y(t) + 20\cos{(10t)}$.

Step-by-step solution

Determine the Spring Constant

Hooke's law states that the force exerted by a spring is proportional to its displacement from its equilibrium position, with the spring constant \(k\) being the proportionality constant. Mathematically, this can be expressed as \(F_s=-ky\). In this case, since the spring is vertically hanging, the weight of the mass (10 kg) causes the initial extension in the spring of \(9.8 \mathrm{~cm}\). At equilibrium, the spring force equals the object's weight, so \(F_s=-F_w\). The weight of the object is given by \(F_w=mg\) where \(m\) is the mass and \(g\) is the gravitational acceleration (approx. \(9.8 \mathrm{~m/s^2}\)). Hence, we have: \[-k(0.098)=-(10)(9.8)\] We can solve this equation for the spring constant, \(k\).

Set up the Differential Equation for Displacement

The net force on the object, which is the sum of the spring force, \(F_s=-ky\), and the applied force, \(F(t)=20\cos{10t}\), must be equal to the product of the object's mass, \(m\), and its acceleration, which is the second derivative of the displacement with respect to time, \(y''(t)\). Mathematically, this can be expressed as: \[my''(t)=-ky(t)+F(t)\] Substitute the values for \(m\), \(k\), and \(F(t)\) to form a second-order nonhomogeneous linear differential equation: \[(10)y''(t)=-101.96y(t)+20\cos{10t}\]

Solve the Differential Equation

In order to solve the differential equation, we first need to find the general solution of its homogeneous equation \(\displaystyle (10)y''(t) = -101.96y(t)\), which can be solved using the characteristic equation \(\displaystyle \rho (\rho) = -10.196\). The characteristic equation has complex roots, which means that the general solution of the homogeneous part will be of the form \(\displaystyle y(t) = Ae^{(\alpha t)}\cos{(\beta t)} + Be^{(\alpha t)}\sin{(\beta t)}\), where \(\displaystyle \alpha\) and \(\displaystyle \beta\) are constants. Now, we need to find a particular solution to the nonhomogeneous part of the differential equation. Since the forcing function is of the form \(\displaystyle 20\cos{10t}\), we can assume a particular solution of the form \(\displaystyle y_p(t) = C\cos{(10t)}+D\sin{(10t)}\), where \(\displaystyle C\) and \(\displaystyle D\) are constants to be determined. Substitute \(\displaystyle y_p(t)\) into the original nonhomogeneous equation and solve for \(\displaystyle C\) and \(\displaystyle D\). Finally, the general solution of the nonhomogeneous part will be given by the sum of the general solutions of the homogeneous and the particular parts.

Apply Initial Conditions and Find the Displacement Function

At time \(\displaystyle t=0\), the given force \(\displaystyle F(t)\) disturbs the system from its rest state; thus, the initial displacement and initial velocity are both zero. Mathematically, this can be expressed as \(\displaystyle y(0) = 0\) and \(\displaystyle y'(0) = 0\). Plugging these initial conditions into the general solution obtained in step 3 will help us find the exact values of the constants involved in the displacement function, \(\displaystyle y(t)\).

Plot the Displacement Function and Find the Maximum Excursion

We can plot the displacement function, \(\displaystyle y(t)\), that we derived in step 4 using a suitable plotting software or by hand. Knowing the behavior of the function and by using calculus (finding the critical points where \(\displaystyle y'(t) = 0\)), we can determine the maximum excursion from equilibrium, if it exists, on the interval \(\displaystyle 0 \leq t < \infty\).

Key concepts

Hooke's Law

Hooke's Law is a principle of physics that relates the force exerted by a spring to the distance it is stretched or compressed. This law is essential for understanding how spring-mass systems behave.

According to Hooke's Law, the force \( F \) that a spring exerts is directly proportional to the displacement \( y \) from its natural length, which is mathematically represented as \( F_s = -ky \) where \( k \) is the spring constant, showing the stiffness of the spring. The negative sign indicates that the force the spring exerts is in the opposite direction to the displacement.

This law is crucial for solving many mechanical problems involving springs and can be applied to calculate the spring constant as shown in the exercise: by equating the spring force due to the weight of a suspended object to the force required to stretch or compress the spring itself. This is how \( k \) was determined using the weight of the 10-kg object and the stretch of the spring.

Differential Equations

Differential equations play a pivotal role in modeling the behavior of various physical systems. They relate a function with its derivatives, representing rates of change.

In the context of a spring-mass system, we particularly deal with second-order linear differential equations to describe the motion of the mass attached to the spring. The general form of such an equation is \( m y''(t) + ky(t) = F(t) \) where \( m \) is the mass, \( y''(t) \) is the acceleration, \( k \) is the spring constant, and \( F(t) \) represents external forces applied to the system.

To solve these equations, as seen in the exercise, we need to break them down into their homogeneous and particular parts, find solutions for each, and then add them together to get the general solution. The solution method requires both mathematical techniques and an understanding of how physical forces influence the system's dynamics.

Initial Value Problem

In the context of differential equations, an initial value problem involves finding a specific solution that satisfies both the differential equation and the initial conditions provided.

To solve initial value problems, we first find the general solution to the differential equation and then apply the initial conditions to determine the constants within that general solution. In the case of the spring-mass system, the initial conditions specify the state of the system at the start of the observation. For instance, the exercise revealed that at \( t=0 \) the system is disturbed, hence we had initial conditions for the displacement \( y(0) \) and the velocity \( y'(0) \) to be zero. Utilizing these initial values allows us to pinpoint the exact motion of the system over time.

Harmonic Motion

Harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction. This is typical of systems like the spring-mass system.

When dealing with harmonic motion, we expect the mass to oscillate about an equilibrium position, usually resulting in a sine or cosine wave pattern over time. The motion of a weight attached to a spring is a classic example of simple harmonic motion when it is undisturbed by other forces. In our exercise, the applied force \( F(t) = 20 \( \cos{10t} \) \) also has a sinusoidal form, hence the resulting motion is a type of forced harmonic motion.

The importance of harmonic motion lies in its predictability and the widespread occurrence in physical systems, from microscopic atomic lattices to large-scale structures like bridges. Identifying this type of motion helps in predicting and analyzing system responses to various stimuli.