Particle Oscillation: Solving Motion With Acceleration A=k(100-x)

Hey guys! Let's dive into an interesting physics problem involving a particle oscillating between two points. This is the kind of stuff that really makes you think about how things move and interact. We're going to break down the problem step-by-step, making sure we understand each part so we can arrive at the solution with confidence. Our main focus? It’s all about understanding oscillations and how acceleration plays a role in this specific scenario.

Understanding the Problem: Oscillations and Acceleration

So, we've got this particle, right? And it's bouncing back and forth between two points, x=40 mm and x=160 mm. What makes this interesting is that its acceleration isn't constant; it changes depending on where the particle is. We're given the formula a=k(100 - x), where a is the acceleration, x is the position, and k is a constant we need to figure out. The units are important here: acceleration is in mm/s² and position is in mm. We also have some extra info: the particle's speed is 18 mm/s when it's at x=100 mm, and it stops momentarily (zero speed) when it reaches x=40 mm. Think of it like a tiny spring system, where the force (and thus acceleration) changes as the spring stretches or compresses. This relationship between acceleration and position is crucial for us to understand the particle's motion.

The first thing we need to do is wrap our heads around what this equation, a=k(100 - x), is telling us. It's not just a random jumble of letters and numbers. It's a description of how the particle's acceleration changes as it moves along the x-axis. Notice that when x is less than 100 mm, the acceleration is positive, meaning the particle is speeding up in the positive x-direction. When x is greater than 100 mm, the acceleration is negative, meaning the particle is slowing down or speeding up in the negative x-direction. At x = 100 mm, the acceleration is zero, which is a critical point in the particle's motion. This is likely the point of equilibrium, where the particle momentarily experiences no net force. Understanding this relationship is key to unraveling the whole problem. It's not just about plugging numbers into a formula; it's about grasping the physics of the situation. We're talking about the fundamental way the particle responds to its position, and that's where the real insight comes from.

Moreover, consider the significance of the given velocities at specific positions. When the particle is at x = 100 mm, it has a velocity of 18 mm/s. This provides us with a snapshot of the particle's kinetic energy at that particular point in its trajectory. It's a crucial piece of information because it connects the particle's speed to its position, which is exactly what we need to solve for the unknown constant k. The fact that the velocity is zero at x = 40 mm is equally important. This tells us that this point is an extreme of the particle's motion – a turning point where it momentarily stops before changing direction. This zero-velocity condition is a boundary condition that will help us nail down the limits of integration when we start doing some calculus. So, each piece of information we're given is like a puzzle piece, and we need to fit them together carefully to reveal the whole picture of the particle's oscillation. It’s all connected, guys, and that's the beauty of physics!

Solving for the Constant k: A Step-by-Step Approach

Alright, let's get our hands dirty and solve for the constant k. This is where the math comes in, but don't worry, we'll take it slow. We know that acceleration is the derivative of velocity with respect to time (a = dv/dt). But we also know that velocity is the derivative of position with respect to time (v = dx/dt). We need to somehow relate acceleration to position directly, because that's what our equation a=k(100 - x) does. The trick here is to use the chain rule. We can write a = dv/dt as a = (dv/dx) * (dx/dt), which simplifies to a = v * (dv/dx). See what we did there? We've now got acceleration expressed in terms of velocity and the derivative of velocity with respect to position. This is exactly what we need to link our given acceleration equation to the velocity information.

Now we can substitute our acceleration equation into this new form: v * (dv/dx) = k(100 - x). This is a differential equation, but it's a separable one, which means we can rearrange it to get all the v's on one side and all the x's on the other. Multiply both sides by dx and we get v dv = k(100 - x) dx. Now we can integrate both sides. The left side integrates to (1/2)v², and the right side integrates to k(100x - (1/2)x²). Don't forget to add a constant of integration, let's call it C. So, we now have (1/2)v² = k(100x - (1/2)x²) + C. This equation relates the particle's velocity to its position, which is exactly what we need to find k.

To solve for k, we'll use the information we have about the particle's velocity at specific positions. We know that when x = 40 mm, v = 0 mm/s. Plug these values into our equation: (1/2)(0)² = k(100(40) - (1/2)(40)²) + C. This simplifies to 0 = k(4000 - 800) + C, or 0 = 3200k + C. This gives us a relationship between k and C: C = -3200k. Now we use the other piece of information: when x = 100 mm, v = 18 mm/s. Plug these values and our expression for C into the equation: (1/2)(18)² = k(100(100) - (1/2)(100)²) - 3200k. This simplifies to 162 = k(10000 - 5000) - 3200k, which further simplifies to 162 = 5000k - 3200k, or 162 = 1800k. Finally, we can solve for k: k = 162 / 1800 = 0.09. So, we've found the value of the constant k! It's 0.09 mm/s² per mm.

Determining the Particle's Velocity at x=115 mm

Now that we've got k, we can figure out the particle's velocity at any position between 40 mm and 160 mm. The question asks us to find the velocity when x=115 mm. Remember our equation relating velocity and position: (1/2)v² = k(100x - (1/2)x²) + C. We know k is 0.09, and we know C is -3200k, which is -3200 * 0.09 = -288. So, our equation becomes (1/2)v² = 0.09(100x - (1/2)x²) - 288.

Now, plug in x=115 mm: (1/2)v² = 0.09(100(115) - (1/2)(115)²) - 288. Let's simplify this step by step. First, calculate the terms inside the parentheses: 100(115) = 11500 and (1/2)(115)² = (1/2)(13225) = 6612.5. So, we have (1/2)v² = 0.09(11500 - 6612.5) - 288. Next, subtract inside the parentheses: 11500 - 6612.5 = 4887.5. Now we have (1/2)v² = 0.09(4887.5) - 288. Multiply 0.09 by 4887.5: 0. 09 * 4887.5 = 439.875. So, our equation is now (1/2)v² = 439.875 - 288. Subtract 288 from 439.875: 439.875 - 288 = 151.875. This gives us (1/2)v² = 151.875.

Multiply both sides by 2 to get rid of the (1/2): v² = 2 * 151.875 = 303.75. Finally, take the square root of both sides to solve for v: v = ±√303.75. This gives us v ≈ ±17.43 mm/s. The ± sign indicates that the particle could be moving in either direction at x=115 mm. Without more information about the particle's direction of motion at that specific instant, we can't determine the sign. However, the magnitude of the velocity is approximately 17.43 mm/s.

Calculating the Range of Positions: Where Does the Particle Move?

Okay, so we know the particle oscillates between x=40 mm and x=160 mm. But let's confirm that by finding the extreme positions mathematically. Remember, at the extreme positions, the particle's velocity is zero. We'll use our trusty equation (1/2)v² = k(100x - (1/2)x²) + C, with k=0.09 and C=-288. Since v=0 at the extremes, we have 0 = 0.09(100x - (1/2)x²) - 288.

To make things easier, let's get rid of the 0.09 by dividing both sides by it: 0 = 100x - (1/2)x² - 288/0.09. Calculate 288/0.09: 288 / 0.09 = 3200. So, our equation becomes 0 = 100x - (1/2)x² - 3200. Now, let's multiply the entire equation by 2 to get rid of the fraction: 0 = 200x - x² - 6400. Rearrange the terms to get a standard quadratic equation: x² - 200x + 6400 = 0.

Now we can use the quadratic formula to solve for x: x = [-b ± √(b² - 4ac)] / 2a, where a=1, b=-200, and c=6400. Plug in the values: x = [200 ± √((-200)² - 4(1)(6400))] / 2(1). Simplify inside the square root: (-200)² = 40000 and 4(1)(6400) = 25600. So, we have x = [200 ± √(40000 - 25600)] / 2. Subtract inside the square root: 40000 - 25600 = 14400. Now we have x = [200 ± √14400] / 2. The square root of 14400 is 120, so x = [200 ± 120] / 2.

Now we have two possible solutions for x: x₁ = (200 + 120) / 2 = 320 / 2 = 160 mm, and x₂ = (200 - 120) / 2 = 80 / 2 = 40 mm. These are the extreme positions of the particle's motion, exactly as stated in the problem! So, our calculations confirm that the particle oscillates between x=40 mm and x=160 mm. We've used the physics and the math to show it. How cool is that?

Conclusion: Mastering Oscillation Problems

Alright guys, we've tackled a pretty challenging problem involving oscillations and acceleration. We started by understanding the relationship between acceleration and position, then we used calculus to find the constant k and the particle's velocity at a specific point. We even verified the range of the particle's motion. The key takeaways here are the importance of understanding the physics behind the equations, using the chain rule to relate acceleration to velocity and position, and applying boundary conditions to solve for unknowns. These are skills that will serve you well in any physics problem, especially those involving oscillations and harmonic motion. Keep practicing, keep thinking, and you'll be a physics pro in no time!