Mass-Spring System: Finding Displacement And Frequency

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mass-spring systems in physics and mathematics. If you've ever wondered how things oscillate, how far they stretch, or how fast they go back and forth, you've come to the right place. We're going to tackle a specific problem that will unravel these concepts for you, focusing on finding the displacement and frequency of oscillation. So, grab your calculators and let's get nerdy!

Understanding the Basics: The Mass-Spring System

Alright, let's set the scene. Imagine you have an object, say a weight, attached to a spring. When this object is just hanging there, not moving, it's in a state called equilibrium. This is our baseline, our zero point. Now, if you pull or push this object, the spring will resist, trying to pull it back to that equilibrium position. This resistance is what causes the object to oscillate, or move back and forth. The amount the spring stretches or compresses from its equilibrium position is called displacement, and how quickly it completes a full cycle of motion is its frequency. These are the core concepts we'll be exploring today.

Part A: Calculating Displacement Over Time

So, the problem states we have an object that stretches a spring by a certain amount when it's at equilibrium. This initial stretch is important because it tells us about the forces at play – the weight of the object is balanced by the spring's restoring force. However, for our calculations of motion after equilibrium, we often set this equilibrium position as our reference point, y=0. We're given that at time t=0 (the moment we start observing), the object is initially displaced 5 inches above equilibrium. This means our starting displacement, y(0), is +5 inches. Since the problem mentions units of feet per second later, it's crucial to be consistent. Let's convert everything to feet. So, 5 inches is 5/12 feet.

Furthermore, we're told it's given an upward velocity of 6 ft/s at this initial moment. In our coordinate system where positive y is typically 'up', this means our initial velocity, y'(0), is +6 ft/s. The motion of a mass-spring system, when there's no damping (like friction) and no external driving force, is described by a second-order linear homogeneous differential equation. For a vertical spring-mass system, this equation looks like $my'' + ky = 0$, where 'm' is the mass and 'k' is the spring constant. This equation leads to oscillatory solutions of the form $y(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t)$, where $\omega = \sqrt{k/m}$ is the angular frequency. However, the problem implicitly gives us enough information through the equilibrium stretch and initial conditions to bypass needing to explicitly find 'm' and 'k' if we're clever. The equilibrium stretch itself relates 'm' and 'k'. Let the equilibrium stretch be $L$. Then $mg = kL$. So $m/k = L/g$. Our angular frequency squared, $\omega^2 = k/m = g/L$. We need 'L' in feet. The problem states 'an object stretches a spring inches in equilibrium'. This is a bit ambiguous – does it mean 1 inch, or some unspecified number of inches? Assuming the problem implies a specific, but unstated, equilibrium stretch. Let's re-read carefully: 'An object stretches a spring inches in equilibrium.' This is the most crucial part of the problem statement that seems to be missing a numerical value. It's likely intended to provide the value of $L$ in inches. Let's assume, for the sake of demonstrating the method, that the equilibrium stretch is $L$ inches. Then $\omega = \sqrt{g/L_{ft}}$, where $L_{ft} = L_{in}/12$. If the problem meant '1 inch', then $L=1$ inch, and $L_{ft} = 1/12$ ft. In this case, $\omega = \sqrt{32.2 / (1/12)} = \sqrt{32.2 \times 12} = \sqrt{386.4} \approx 19.66$ rad/s.

Let's proceed assuming the equilibrium stretch is 1 inch ($L=1$ inch), so $\omega \approx 19.66$ rad/s. Our displacement equation is $y(t) = C_1 \cos(19.66t) + C_2 \sin(19.66t)$. We use the initial conditions to find $C_1$ and $C_2$.

At t=0, $y(0) = 5/12$ ft. So, $(5/12) = C_1 \cos(0) + C_2 \sin(0) = C_1$. Thus, $C_1 = 5/12$.

Now we need the velocity, $y'(t)$. Differentiating $y(t)$: $y'(t) = -19.66 C_1 \sin(19.66t) + 19.66 C_2 \cos(19.66t)$.

At t=0, $y'(0) = 6$ ft/s. So, $6 = -19.66 C_1 \sin(0) + 19.66 C_2 \cos(0) = 19.66 C_2$.

Solving for $C_2$: $C_2 = 6 / 19.66 \approx 0.305$ ft.

Therefore, the displacement equation for $t > 0$ is approximately $y(t) = (5/12) \cos(19.66t) + 0.305 \sin(19.66t)$ feet.

Important Note: If the equilibrium stretch was meant to be a different value than 1 inch, you would simply recalculate $\omega$ using that value for $L$ in feet, and the constants $C_1$ and $C_2$ would remain the same as they depend only on the initial displacement and velocity. The problem statement is indeed missing this crucial numerical value for the equilibrium stretch. We've proceeded assuming it's 1 inch for demonstration purposes. If you have the exact value, please plug it in!

Part B: Determining the Frequency of Oscillation

Now, let's talk about the frequency of oscillation. This tells us how many full cycles the object completes in one second. In our general solution $y(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t)$, the Greek letter omega, $\omega$, represents the angular frequency in radians per second. The frequency, usually denoted by 'f', is related to the angular frequency by the simple formula: $f = \omega / (2\pi)$.

Using our assumed value of $\omega \approx 19.66$ rad/s (derived from an assumed 1-inch equilibrium stretch), we can now calculate the frequency:

$f = 19.66 / (2\pi) \approx 19.66 / (2 \times 3.14159) \approx 19.66 / 6.28318 \approx 3.13$ Hz (Hertz).

So, the frequency of oscillation is approximately 3.13 Hz. This means that for every second that passes, the object completes about 3.13 full back-and-forth motions around its equilibrium position. It's pretty neat to see how these initial conditions and the physical properties of the system (represented by the spring constant and mass, which are encapsulated in $\omega$) dictate the entire motion!

Wrapping It Up

And there you have it, folks! We've successfully calculated the displacement of the object over time and determined its frequency of oscillation. Remember, the key takeaway here is understanding how initial conditions (displacement and velocity) and the system's inherent properties (like the spring's stiffness and the object's mass, which together determine $\omega$) govern the oscillatory behavior. It's a fundamental concept in physics that pops up in everything from musical instruments to earthquake analysis. Keep practicing these problems, and don't be afraid to revisit the basic differential equations and their solutions. We'll catch you in the next article with more awesome math and physics insights! Stay curious!

Keywords: mass-spring system, equilibrium, displacement, oscillation, frequency, angular frequency, differential equation, initial conditions, physics, mathematics, harmonic motion