Problem 51 - 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? $$ d(t)=-15 e^{-0.9 t / 30} \cos \left(\sqrt{\left(\frac{\pi}{3}\right)^{2}-\frac{0.81}{900}} t\right) $$

Step-by-Step Solution

Answer

The bob oscillates with decreasing amplitude due to the damping factor. The initial displacement is -15 meters. The displacement approaches zero as time increases.

Step 1: Describe the motion of the object

The bob’s motion is described by the function \( d(t) = -15 e^{-0.9 t / 30} \cos \left(\sqrt{\left( \frac{\pi}{3} \right)^2 - \frac{0.81}{900}} t \right) \). The negative sign indicates that the initial motion is in the negative direction. The mass \( m \) and the damping factor are required. From the problem, the damping factor is \(-0.9 \) and the mass is given but not needed for the description of motion.

Step 2: Initial displacement at \( t = 0 \)

To find the initial displacement, substitute \( t = 0 \) into the given function: \( d(0) = -15 e^{-0.9 \cdot 0 / 30} \cos \left( \sqrt{\left( \frac{\pi}{3} \right)^2 - \frac{0.81}{900}} \cdot 0 \right) \). This simplifies to \( d(0) = -15 \cdot 1 \cdot \cos(0) \), resulting in \( d(0) = -15 \) meters.

Step 3: Graph the motion using a graphing utility

Use a graphing calculator or software to plot the function \( d(t) = -15 e^{-0.9 t / 30} \cos \left( \sqrt{\left( \frac{\pi}{3} \right)^2 - \frac{0.81}{900}} t \right) \). This graph will show the damped oscillation of the bob over time.

Step 4: Displacement at the start of the second oscillation

Find the time it takes for the first oscillation. The angular frequency is given by \( \omega = \sqrt{\left( \frac{\pi}{3} \right)^2 - \frac{0.81}{900}} \). The period \( T \) is given by \( T = 2 \pi / \omega \). Calculate \( \omega \) and then \( T \), and plug \( T \) back into the displacement equation to get \( d(T) \).

Step 5: Long-term behavior of the bob's displacement

As time \( t \) increases without bound, the exponential term \( e^{-0.9 t / 30} \) approaches zero. Thus, the displacement \( d(t) \) approaches zero because it is multiplied by this dampening factor.

Key Concepts

damped oscillation

Damped oscillation describes the motion of a pendulum when its amplitude gradually decreases over time due to frictional or resistive forces. In the given function, this is modeled by the term involving the exponential decay, \(-15 e^{-0.9 t / 30}\). The negative exponent indicates a damping effect on the pendulum's motion. As time passes, the amplitude of oscillation reduces until the pendulum eventually comes to a rest. Without the damping force, the pendulum would continue to swing back and forth indefinitely, but with damping, each oscillation is smaller than the previous one.

The role of damping in oscillations is crucial in many real-world applications, such as in car suspensions, where damping prevents excessive oscillation and ensures a smoother ride.

initial displacement

Initial displacement is the starting position of the pendulum bob relative to its resting position at time \(\t = 0\). For this problem, we substitute \(\t = 0\) into the given function. The equation simplifies to:
\(d(0) = -15 e^{-0.9 \cdot 0 / 30} \cos \(\sqrt{\left( \frac{\pi}{3} \right)^2 - \frac{0.81}{900}} \cdot 0 \)\).

The exponential and cosine terms both evaluate to 1 when \(\t = 0\), leading us to: \(d(0) = -15\), meaning the pendulum was initially displaced by -15 meters. This value is crucial for analyzing how far the pendulum will swing in subsequent cycles and for understanding the initial energy in the system.

angular frequency

Angular frequency, denoted by \(\omega\), represents the rate at which the pendulum oscillates. It is a key component in analyzing oscillatory systems. The given exercise provides the formula for \(\omega\) as: \[\omega = \sqrt{\left( \frac{\pi}{3} \right)^2 - \frac{0.81}{900}}\].
Angular frequency is connected to the period \(\( T \)\) of the pendulum's motion through the relationship \[ T = \frac{2\pi}{\omega} \]. Angular frequency helps determine how quickly the pendulum completes one full oscillation and its dynamical behavior.

High angular frequencies imply faster oscillations, while lower angular frequencies indicate slower oscillations. Knowing \(\t\omega \) allows us to predict the time periods and understand the energy propagation within the system.

exponential decay

Exponential decay describes the reduction in the amplitude of the pendulum’s oscillation over time, as modeled by the term \(-15 e^{-0.9 t / 30}\). This mathematical expression shows that the amplitude of the pendulum’s displacement reduces at an exponential rate due to the damping factor of \(-0.9\).

Exponential decay can be generally expressed as \(A(t) = A_0 e^{-kt}\), where \(A_0\) is the initial amplitude, and \(k\) is the decay constant determining how quickly the amplitude decreases. In practical situations, this principle can explain the gradual loss of motion in various systems, such as vibrations in mechanical structures or the cooling of hot objects over time.

Understanding exponential decay is essential for predicting the long-term behavior of oscillating systems and for designing mechanisms that use or counteract this effect.