Problem 52 - Algebra and Trigonometry

Question

The function \(d\) models the distance (in meters) of the bob of a pendulum of mass \(m\) (in kilograms) from its rest position at time \(t\) (in seconds) is given. The bob is released from the left of its rest position and represents a negative direction. (a) Describe the motion of the object. Be sure to give the mass and damping factor. (b) What is the initial displacement of the bob? That is, what is the displacement at \(t=0 ?\) (c) Graph the motion using a graphing utility. (d) What is the displacement of the bob at the start of the second oscillation? (e) What happens to the displacement of the bob as time increases without bound? $$ d(t)=-10 e^{-0.8 t / 50} \cos \left(\sqrt{\left(\frac{2 \pi}{3}\right)^{2}-\frac{0.64}{2500}} t\right) $$

Step-by-Step Solution

Answer

The mass is inferred from context; the damping factor is -0.016. Initial displacement is -10 meters. Displacement approaches zero as time increases.

Step 1: Analyzing the given function

The function given is $$d(t) = -10 e^{-0.8 t / 50} \cos \left(\sqrt{\left(\frac{2 \pi}{3}\right)^{2}-\frac{0.64}{2500}} t\right).$$ This represents the distance of the bob from its rest position with respect to time.

Step 2: Identifying the mass and damping factor

The mass of the bob can be inferred from additional context, but it is not directly given in the function. The damping factor is represented by the term within the exponential function, $$e^{-0.8 t / 50}$$. Specifically, the damping factor is $$-0.8 / 50 = -0.016.$$

Step 3: Determining the initial displacement

The initial displacement at \(t=0\) is found by evaluating the function at \(t=0\): $$d(0)= -10 e^{-0.8 \, (0) / 50} \cos \left(\sqrt{\left(\frac{2 \pi}{3}\right)^{2}-\frac{0.64}{2500}} \, (0)\right).$$ Simplifying this we get: $$d(0) = -10 e^0 \cos(0) = -10 \cdot 1 \cdot 1 = -10.$$ Thus, the initial displacement is -10 meters.

Step 4: Graphing the motion

Using a graphing utility (such as Desmos or a graphing calculator), plot the function $$d(t) = -10 e^{-0.8 t / 50} \cos \left(\sqrt{\left(\frac{2 \pi}{3}\right)^{2}-\frac{0.64}{2500}} t\right).$$ This will show the motion of the bob over time.

Step 5: Finding the displacement at the start of the second oscillation

The start of the second oscillation happens at the first minimum point after the bob returns to the equilibrium position. Using a graph, it can be visually estimated or calculated precisely by solving $$d'(t) = 0$$ and checking for minima.

Step 6: Describing the behavior as time t increases without bound

As \(t\) increases indefinitely, the term $$e^{-0.8 t / 50}$$ trends towards zero, making $$d(t)$$ tend towards zero. This means the displacement decreases and approaches zero.

Key Concepts

Pendulum Motion

Pendulum motion describes the back-and-forth swinging of an object suspended from a fixed point. This motion is periodic, which means it repeats at regular intervals. However, in real-world scenarios, factors like air resistance and friction cause the motion to gradually slow down—a phenomenon known as damping. In this example, we're studying a pendulum where the bob's motion is influenced by damping forces. This is an example of damped harmonic motion, where the oscillations decrease over time. The behavior can be mathematically described by a function that includes both a trigonometric component for the oscillation and an exponential component for the damping.

Damping Factor

The damping factor represents how quickly the motion of the pendulum slows down. In the given function $$d(t)=-10 e^{-0.8 t / 50} \cos \left(\sqrt{\left(\frac{2 \pi}{3}\right)^{2}-\frac{0.64}{2500}} t\right)$$, the term $$e^{-0.8 t / 50}$$ controls this damping. Specifically, the damping factor is the exponent part, which is $$-0.8 / 50$$ or -0.016. This negative value indicates a decaying nature, meaning the pendulum's motion weakens over time. As $$t$$ increases, the magnitude of $$ e^{-0.8 t / 50} $$ approaches zero, reducing the amplitude of oscillations.

Initial Displacement

Initial displacement refers to the starting position of the pendulum bob at time $$t = 0$$. For our function $$d(t)=-10 e^{-0.8 t / 50} \cos \left(\sqrt{\left(\frac{2 \pi}{3}\right)^{2}-\frac{0.64}{2500}} t\right)$$, we find the initial displacement by evaluating $$d(0)$$. \( t = 0 \), so \( d(0) = -10 \cos (0) \)= -10\left(1\right) \= -10)). Hence, the pendulum starts at a -10 meter position from the equilibrium point, indicating it starts to the left of the rest position.

Graphing Trigonometric Functions

Plotting the given function visually represents the oscillatory motion of the pendulum. The function combines exponential and trigonometric components, allowing us to see the decaying oscillations over time. Use tools like Desmos or a scientific graphing calculator to plot $$d(t)=-10 e^{-0.8 t / 50} \cos \left(\sqrt{\left(\frac{2 \pi}{3}\right)^{2}-\frac{0.64}{2500}} t\right)$$.
- The cosine function shows periodic oscillations.
- The exponential term reduces the overall magnitude over time.
- The graph reveals the gradual decrease in amplitude, illustrating damped harmonic motion.

Asymptotic Behavior

Asymptotic behavior refers to the trend of a function as the variable approaches a specific value—in our case, as \( t \) increases without bound. For the given function, the term $$e^{-0.8 t / 50}$$ governs asymptotic behavior. As time \( t \) goes to infinity, $$ e^{-0.8 t / 50}$$ approaches zero. This means the displacement of the pendulum bob approaches zero as time progresses. In other words, the oscillations diminish, and the bob comes closer to resting position over time. This concept beautifully captures the essence of damping in harmonic motion.