Particle Motion: Amplitude, Velocity, Period & Frequency

Hey guys! Let's dive into the exciting world of particle motion and break down how to analyze it. We're going to tackle a problem where the position of a particle is described by a mathematical expression. This might sound intimidating, but trust me, it's super manageable once we understand the key concepts. We'll focus on finding the amplitude, angular velocity, period, and frequency of the motion. So, buckle up and let's get started!

Understanding the Basics of Particle Motion

Before we jump into the specific problem, let's quickly recap some fundamental concepts. When we talk about particle motion, we're essentially describing how a tiny object moves over time. This movement can be simple, like a straight line, or more complex, like oscillations or vibrations. In our case, we're dealing with simple harmonic motion (SHM), which is a specific type of oscillatory motion. Think of a pendulum swinging back and forth or a mass bouncing on a spring – these are classic examples of SHM. The key characteristic of SHM is that the restoring force is directly proportional to the displacement from equilibrium. This leads to a sinusoidal pattern in the particle's position over time, which is exactly what we see in our problem.

To truly grasp the concepts, let's define some crucial terms related to particle motion and, specifically, Simple Harmonic Motion (SHM). Understanding these definitions is essential for tackling problems like the one we have. Amplitude, in simple terms, is the maximum displacement of the particle from its equilibrium position. Imagine a swing; the amplitude is how far it goes from the center point. Angular velocity, denoted by ω (omega), tells us how fast the particle is oscillating in radians per second. It's like measuring the speed of rotation. Period, represented by T, is the time it takes for one complete oscillation or cycle. Think of it as the time it takes for the swing to go back and forth once. Frequency, denoted by f, is the number of oscillations or cycles per unit of time, usually measured in Hertz (Hz), which is cycles per second. It's how often the swing completes a full cycle. Finally, the expression of motion is the mathematical equation that describes the particle's position as a function of time.

Knowing these definitions inside and out is the first step in mastering particle motion problems. With these concepts in mind, we're well-prepared to tackle the problem at hand and dissect the given expression to extract meaningful information about the particle's motion.

Problem Statement: Decoding the Particle's Dance

Alright, let's get down to business! We're given the expression for the position of a particle as: x = 4 sin(3πt + π). Remember, x is in meters, t is in seconds, and the angles are in radians. Our mission, should we choose to accept it (and we do!), is to determine the following:

a) The amplitude of the motion. b) The angular velocity. c) The period and frequency of the motion. d) The expression of (we'll need to clarify what this refers to, but most likely it's asking for the velocity and/or acceleration equations).

This expression, x = 4 sin(3πt + π), might look a bit intimidating at first glance, but don't worry! It's actually a standard form for describing simple harmonic motion. The key is to recognize the different parts and what they represent. The sine function itself tells us that the motion is oscillatory. The numbers in front of and inside the sine function hold the secrets to amplitude, angular velocity, and more. So, let's start cracking the code!

To make our lives easier, let's compare this given equation to the general equation for SHM: x = A sin(ωt + φ). Here, A represents the amplitude, ω represents the angular velocity, and φ (phi) represents the phase constant. By comparing our specific equation to this general form, we can directly identify the values we need to answer the first few parts of the problem. This is a powerful technique that simplifies the process of analyzing SHM. Now, let's dive into solving each part step-by-step!

Solving for Amplitude (a)

First up, let's find the amplitude of the particle's motion. This is arguably the easiest part, thanks to our general SHM equation. Remember, the general form is x = A sin(ωt + φ), where A represents the amplitude. Now, let's compare this to our given equation: x = 4 sin(3πt + π). Can you spot the amplitude? It's the number right in front of the sine function!

In our case, the number in front of the sine function is 4. Therefore, the amplitude (A) is 4 meters. That's it! Amplitude represents the maximum displacement of the particle from its equilibrium position. So, in this scenario, the particle moves a maximum of 4 meters away from its resting point in either direction. This is a direct and straightforward application of comparing the given equation with the general form of SHM. You see, these problems aren't so scary once you break them down!

Finding the amplitude was a breeze, right? It's often the most straightforward part. The key is to remember what each component of the general SHM equation represents. Now that we've conquered the amplitude, let's move on to the next challenge: finding the angular velocity. We'll use a similar approach, comparing our given equation to the general form, but this time, we'll focus on a different part of the equation. So, let's keep the momentum going and tackle angular velocity!

Determining Angular Velocity (b)

Alright, time to find the angular velocity! We're still using the same given equation, x = 4 sin(3πt + π), and our trusty general SHM equation, x = A sin(ωt + φ). Remember, ω (omega) represents the angular velocity in the general equation. So, where do we look in our given equation to find ω? Think about what's inside the sine function.

In our given equation, we have 3πt inside the sine function. Comparing this to the ωt term in the general equation, we can clearly see that 3π corresponds to ω. Therefore, the angular velocity (ω) is 3π radians per second. Awesome! Angular velocity tells us how quickly the particle is oscillating. A higher angular velocity means the particle is moving back and forth more rapidly. In this case, the particle is oscillating at a rate of 3π radians per second, which is a pretty zippy pace!

Notice how we're building on our understanding of the general SHM equation. By recognizing the correspondence between the terms in the general equation and the terms in our specific equation, we can easily extract the values we need. This is a powerful strategy for solving these types of problems. We've found the amplitude and the angular velocity – we're on a roll! Now, let's tackle the period and frequency, which are closely related to the angular velocity we just calculated.

Calculating Period and Frequency (c)

Now, let's calculate the period and frequency of the motion. These two are closely related, and once we find one, the other is a piece of cake! Remember, the period (T) is the time it takes for one complete oscillation, and the frequency (f) is the number of oscillations per second. We already found the angular velocity (ω) to be 3π radians per second. The key relationship that connects these quantities is: ω = 2πf = 2π/T. This is a crucial formula to remember for SHM problems!

Let's start by finding the period (T). We can rearrange the formula ω = 2π/T to solve for T: T = 2π/ω. Plugging in our value for ω (3π radians per second), we get: T = 2π / (3π) = 2/3 seconds. Excellent! So, the period of the motion is 2/3 seconds. This means it takes 2/3 of a second for the particle to complete one full oscillation.

Now, let's find the frequency (f). We can use the relationship f = 1/T. Since we know T = 2/3 seconds, we have: f = 1 / (2/3) = 3/2 Hz. Fantastic! The frequency is 3/2 Hz, or 1.5 Hz. This means the particle completes 1.5 oscillations every second. Notice the inverse relationship between period and frequency – a shorter period means a higher frequency, and vice versa.

We've successfully calculated both the period and the frequency using our knowledge of angular velocity and the fundamental relationships between these quantities. This demonstrates how interconnected these concepts are in SHM. We're making great progress! We've determined the amplitude, angular velocity, period, and frequency. Now, let's move on to the final part of the problem: finding the expression of, which likely refers to the velocity and acceleration equations.

Deriving the Velocity and Acceleration Expressions (d)

Finally, let's derive the velocity and acceleration expressions for the particle's motion. We have the position expression: x = 4 sin(3πt + π). Remember from calculus that velocity is the time derivative of position, and acceleration is the time derivative of velocity (or the second time derivative of position). So, we'll use our calculus skills to find these expressions.

First, let's find the velocity (v). We need to differentiate x(t) with respect to time (t): v(t) = dx/dt = d[4 sin(3πt + π)]/dt. Using the chain rule, we get: v(t) = 4 * cos(3πt + π) * (3π) = 12π cos(3πt + π). There's our velocity expression! Notice how the derivative of the sine function is a cosine function. The 12π represents the maximum velocity of the particle.

Now, let's find the acceleration (a). We need to differentiate v(t) with respect to time (t): a(t) = dv/dt = d[12π cos(3πt + π)]/dt. Again, using the chain rule, we get: a(t) = 12π * [-sin(3πt + π)] * (3π) = -36π² sin(3πt + π). And there's our acceleration expression! Notice how the derivative of the cosine function is a negative sine function. The -36π² represents the maximum acceleration of the particle, and the negative sign indicates that the acceleration is in the opposite direction to the displacement, which is a characteristic of SHM.

We've successfully derived the expressions for both velocity and acceleration. These expressions tell us how the particle's velocity and acceleration change over time. They're crucial for a complete understanding of the particle's motion. By taking the derivatives of the position function, we've unlocked even more information about the particle's dance!

Conclusion: Mastering Particle Motion Analysis

Wow, we've covered a lot! We started with the position expression x = 4 sin(3πt + π) and successfully determined the amplitude, angular velocity, period, frequency, and the expressions for velocity and acceleration. We used the general form of the SHM equation, our knowledge of calculus, and some key relationships between these quantities. This problem demonstrates the power of breaking down complex problems into smaller, manageable steps.

The key takeaways from this exercise are:

• Understanding the general form of the SHM equation (x = A sin(ωt + φ)) is crucial for identifying amplitude, angular velocity, and phase constant.
• The relationship ω = 2πf = 2π/T connects angular velocity, frequency, and period.
• Calculus (differentiation) allows us to derive velocity and acceleration expressions from the position expression.

By mastering these concepts and techniques, you'll be well-equipped to tackle a wide range of particle motion problems. Remember to practice, practice, practice! The more you work with these equations and concepts, the more comfortable and confident you'll become. So, keep up the great work, and happy problem-solving!

I hope this breakdown was helpful, guys! If you have any questions or want to explore more particle motion problems, feel free to ask. Let's keep learning and growing together!