Displacement Of A Damped Spring-Mass System: A Detailed Solution
Hey guys! Ever wondered how to calculate the displacement of an object attached to a spring and a dashpot when it's set in motion? It sounds like a mouthful, but it's a classic problem in physics and engineering! We're going to break down a problem involving a 16 kg object connected to a spring and a dashpot, and we'll walk through the steps to find its displacement. So, grab your thinking caps, and let's dive in!
Understanding the Problem
In this problem, we have a 16 kg object connected to a spring and a dashpot. The spring constant (k) is 768 kg/s^2, which tells us how stiff the spring is. The damping constant (c) is 256 N-sec/m, which represents the resistance to motion provided by the dashpot (think of it like a shock absorber). The object is initially at equilibrium, but it's given an upward push with a velocity of 1 m/s. Our mission, should we choose to accept it, is to find the object's displacement as a function of time. This means we want to know where the object is at any given moment after it's pushed. To achieve this, we will solve a second-order differential equation that describes the motion of the system. Let's explore this further, shall we?
This problem combines concepts from mechanics, differential equations, and a bit of physics intuition. Before we jump into the calculations, it's crucial to understand the physical principles at play. The spring exerts a restoring force proportional to the displacement from equilibrium (Hooke's Law), pulling the object back towards its resting position. The dashpot, on the other hand, provides a damping force proportional to the object's velocity, resisting its motion. These forces, along with the object's mass and initial conditions (initial position and velocity), determine the object's displacement over time. Visualizing this system can be incredibly helpful. Imagine a weight hanging from a spring, submerged in a viscous fluid (the dashpot). When you push the weight, it oscillates up and down, but the oscillations gradually decrease due to the damping effect of the fluid. This is precisely the behavior we're trying to model mathematically.
The problem also introduces the concept of equilibrium. Equilibrium, in this context, is the position where the object would naturally rest if left undisturbed. It's the point where the forces acting on the object (gravity, spring force) are balanced. Pushing the object upwards from equilibrium disrupts this balance, causing the spring and dashpot to react and influence the object's motion. Understanding equilibrium is crucial for setting up the initial conditions for our differential equation, which we'll see in the next steps. So, keep this mental picture in mind as we move forward – a weight, a spring, a dashpot, and a push – that's the essence of the problem we're about to solve!
Setting up the Differential Equation
The motion of the object can be described by a second-order linear homogeneous differential equation. This might sound intimidating, but don't worry, we'll break it down. The general form of this equation is:
m * x''(t) + c * x'(t) + k * x(t) = 0
Where:
- m is the mass of the object (16 kg).
- x''(t) is the acceleration of the object (the second derivative of displacement with respect to time).
- c is the damping constant (256 N-sec/m).
- x'(t) is the velocity of the object (the first derivative of displacement with respect to time).
- k is the spring constant (768 kg/s^2).
- x(t) is the displacement of the object from equilibrium at time t.
Plugging in the given values, we get:
16 * x''(t) + 256 * x'(t) + 768 * x(t) = 0
Now, to make things a bit simpler, we can divide the entire equation by 16:
x''(t) + 16 * x'(t) + 48 * x(t) = 0
This is our differential equation! It's a mathematical representation of the forces acting on the object and how they influence its motion. Let's think about why this equation works. The term m * x''(t) represents the force due to the object's inertia (resistance to acceleration). The term c * x'(t) represents the damping force, which opposes the motion and is proportional to the velocity. The term k * x(t) represents the spring force, which pulls the object back towards equilibrium and is proportional to the displacement. The sum of these forces equals zero because we're considering a homogeneous system, meaning there's no external force acting on the object after the initial push. So, the equation essentially says that the forces of inertia, damping, and the spring balance each other out.
But how do we actually solve this equation? That's where the characteristic equation comes in, which we'll discuss in the next section. But before we move on, let's highlight the importance of this differential equation. It's not just a random jumble of symbols; it's a powerful tool that allows us to predict the behavior of a wide range of physical systems, from oscillating springs to electrical circuits. By understanding the underlying principles and how to set up these equations, we can gain valuable insights into the world around us. So, give yourself a pat on the back for making it this far – you're one step closer to understanding the motion of a damped spring-mass system!
Solving the Characteristic Equation
To solve the differential equation, we assume a solution of the form:
x(t) = e^(rt)
Where r is a constant we need to determine. To do this, we substitute this assumed solution into our differential equation:
x''(t) + 16 * x'(t) + 48 * x(t) = 0
First, we need to find the derivatives of x(t):
- x'(t) = r * e^(rt)
- x''(t) = r^2 * e^(rt)
Now, substitute these into the differential equation:
r^2 * e^(rt) + 16 * r * e^(rt) + 48 * e^(rt) = 0
We can factor out e^(rt), which is never zero:
e^(rt) * (r^2 + 16r + 48) = 0
This leaves us with the characteristic equation:
r^2 + 16r + 48 = 0
Now, we need to solve this quadratic equation for r. We can do this by factoring, using the quadratic formula, or completing the square. In this case, factoring is the easiest approach:
(r + 4)(r + 12) = 0
This gives us two roots:
- r1 = -4
- r2 = -12
These roots are crucial! They tell us about the nature of the solution. Since we have two distinct real roots, the system is overdamped. This means that the object will return to equilibrium slowly without oscillating. If the roots were complex, the system would be underdamped, and the object would oscillate before settling down. If the roots were real and repeated, the system would be critically damped, representing the fastest return to equilibrium without oscillation. The roots of the characteristic equation are like a fingerprint of the system's behavior, revealing its fundamental response to disturbances.
The characteristic equation itself is a direct result of our initial assumption about the form of the solution, x(t) = e^(rt). This exponential form is a common choice for linear differential equations because its derivatives are simply multiples of itself, making the substitution and simplification process much easier. The roots, then, are the specific values of 'r' that make this exponential function a valid solution to the equation. So, by solving the characteristic equation, we're essentially finding the fundamental building blocks of the system's response.
Now that we have the roots, we can construct the general solution to the differential equation. This involves combining the exponential solutions corresponding to each root, with arbitrary constants to account for the initial conditions. It's like having the ingredients to a recipe – the roots – and now we're going to put them together to bake the cake – the general solution. So, let's move on to the next section and see how these roots translate into a meaningful description of the object's displacement over time!
Finding the General Solution
Since we have two distinct real roots (r1 = -4 and r2 = -12), the general solution to the differential equation is:
x(t) = A * e^(-4t) + B * e^(-12t)
Where A and B are arbitrary constants. These constants are determined by the initial conditions of the problem. This equation represents the superposition of two exponential decays. Each term represents a mode of decay, with the rate of decay determined by the corresponding root. The constants A and B determine the amplitude of each mode, essentially how much each exponential contributes to the overall displacement. Think of it like mixing two different colors of paint – the roots determine the colors, and A and B determine how much of each color you use to get the final shade.
But why this form of the solution? It stems from the fact that the exponential functions e^(-4t) and e^(-12t) are linearly independent solutions to the differential equation. This means that neither function can be expressed as a constant multiple of the other. In the context of differential equations, linearly independent solutions form a basis for the solution space, meaning that any solution to the equation can be written as a linear combination of these basis functions. This is a powerful concept that underlies many techniques for solving linear differential equations.
The constants A and B are like the finishing touches on our masterpiece. They personalize the solution to fit the specific circumstances of the problem, in this case, the initial position and velocity of the object. Without these constants, we would have a family of solutions, but only one of them will accurately describe the object's displacement. Finding A and B involves using the initial conditions to create a system of equations, which we'll tackle in the next section. So, we've got the general shape of the solution – a combination of exponential decays – and now we're ready to fine-tune it by finding the values of A and B. Stay tuned, the final answer is within reach!
Applying Initial Conditions
We are given the following initial conditions:
- x(0) = 0 (The object starts at equilibrium).
- x'(0) = 1 m/s (The object is pushed upwards with a velocity of 1 m/s).
We need to use these conditions to find the values of A and B in our general solution:
x(t) = A * e^(-4t) + B * e^(-12t)
First, let's apply the condition x(0) = 0:
0 = A * e^(0) + B * e^(0)
0 = A + B
This gives us our first equation: A + B = 0. Now, we need to use the second initial condition, x'(0) = 1 m/s. To do this, we first need to find the derivative of x(t):
x'(t) = -4A * e^(-4t) - 12B * e^(-12t)
Now, apply the condition x'(0) = 1:
1 = -4A * e^(0) - 12B * e^(0)
1 = -4A - 12B
This is our second equation: -4A - 12B = 1. We now have a system of two equations with two unknowns:
- A + B = 0
- -4A - 12B = 1
We can solve this system using substitution or elimination. Let's use substitution. From equation 1, we have A = -B. Substitute this into equation 2:
-4(-B) - 12B = 1
4B - 12B = 1
-8B = 1
B = -1/8
Now, substitute B back into A = -B:
A = -(-1/8)
A = 1/8
So, we have found the values of A and B: A = 1/8 and B = -1/8. These values are the key to unlocking the specific solution that describes our object's motion. They tell us the precise contribution of each exponential decay mode to the overall displacement. Without the initial conditions, we would have an infinite number of possible solutions, each corresponding to different values of A and B. The initial conditions act as constraints, narrowing down the possibilities until we arrive at the unique solution that matches the physical situation described in the problem.
Think of it like tuning a radio – the general solution is like the radio itself, capable of receiving many different frequencies, but the initial conditions are like the tuning knob, allowing us to select the specific frequency (solution) that we want. The process of applying initial conditions is a fundamental step in solving many types of differential equations, and it highlights the importance of incorporating specific information about the system being modeled. We're almost there, guys! We've found A and B, so now we can plug them back into the general solution and get the final answer!
The Final Solution
Now that we have A = 1/8 and B = -1/8, we can plug these values back into the general solution:
x(t) = (1/8) * e^(-4t) + (-1/8) * e^(-12t)
Or, we can write it as:
x(t) = (1/8) * (e^(-4t) - e^(-12t))
This is the displacement of the object as a function of time! It tells us exactly where the object is at any given moment after it's pushed upwards. Let's break down what this equation means. It's a combination of two exponential decay terms, e^(-4t) and e^(-12t). The term e^(-4t) decays more slowly than e^(-12t), meaning it has a longer-lasting effect on the object's displacement. The coefficients (1/8) and (-1/8) determine the amplitude and direction of each term's contribution. The positive (1/8) * e^(-4t) term represents a positive displacement (upwards), while the negative (-1/8) * e^(-12t) term represents a negative displacement (downwards). The difference between these two terms gives us the net displacement at any time t.
This solution is characteristic of an overdamped system. As we discussed earlier, the overdamped nature is reflected in the distinct real roots of the characteristic equation. The object doesn't oscillate; instead, it gradually returns to equilibrium. The e^(-12t) term decays quickly, so initially, it contributes to a rapid movement towards equilibrium. Then, the e^(-4t) term dominates, causing a slower, more gradual return. This behavior is typical of systems with strong damping forces, like a door closer or a shock absorber in a car.
We've successfully navigated the world of differential equations, sprung into action with spring constants and damping coefficients, and finally, we've landed on the solution! Give yourselves a round of applause, guys, because you've just witnessed the power of mathematics to describe the motion of a physical system. This equation isn't just a bunch of symbols; it's a story – the story of how a 16 kg object, connected to a spring and a dashpot, responds to an initial push. And the best part is, the same principles we've used here can be applied to a wide range of other problems, from analyzing electrical circuits to modeling population growth. So, keep exploring, keep questioning, and keep applying your newfound knowledge to the world around you! You've got this!