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Chapter 4: Problem 45

A damped oscillator The displacement of a particular object as it bounces vertically up and down on a spring is given by \(y(t)=2.5 e^{-t} \cos 2 t,\) where the initial displacement is \(y(0)=2.5\) and \(y=0\) corresponds to the rest position (see figure). a. Find the time at which the object first passes the rest position, \(y=0\) b. Find the time and the displacement when the object reaches its lowest point. c. Find the time at which the object passes the rest position for the second time. d. Find the time and the displacement when the object reaches its high point for the second time.

Short Answer

Expert verified
Answer: The object first passes the rest position at \(t = \frac{\pi}{4}\).

Step by step solution

01

Set y(t) equal to 0

Find the time at which the object first passes the rest position, which corresponds to \(y(t) = 0\). So, we need to solve for \(t\) in the equation \(2.5 e^{-t} \cos 2t = 0\).
02

Solve the equation

To solve the equation, we note that either \(2.5 e^{-t}\) or \(\cos 2t\) must be equal to zero. Since \(2.5 e^{-t}\) can never be equal to zero, we look at the second part: \(\cos 2t = 0\). This occurs for \(2t = \frac{\pi}{2} + n\pi\), where \(n\) is an integer. For the first time when it passes the rest position, we have \(n=0\): $$2t = \frac{\pi}{2} \Rightarrow t = \frac{\pi}{4}.$$ b. Find the time and the displacement when the object reaches its lowest point
03

Find the critical points

To find the lowest point, we need to find the critical points of the displacement function. First, find the derivative of \(y(t)\): $$y'(t) = \frac{d(2.5e^{-t}\cos 2t)}{dt} = -2.5 e^{-t} \cos 2t + 5 e^{-t} \sin 2t.$$
04

Set the derivative equal to zero

Now, set the derivative \(y'(t)\) equal to zero and solve for \(t\): $$0 = -2.5 e^{-t} \cos 2t + 5 e^{-t} \sin 2t.$$ Since \(e^{-t}\) can never be equal to zero, divide both sides by \(e^{-t}\): $$0 = -2.5 \cos 2t + 5 \sin 2t.$$
05

Solve for t

Solve the equation numerically or graphically to find \(t \approx 0.449\) as the time when the object reaches the lowest point. Now, find the displacement at this time: \(y(0.449) \approx -1.814.\) c. Find the time at which the object passes the rest position for the second time
06

Set y(t) equal to 0

Recall that the object passes the rest position when \(y(t) = 0 \Rightarrow 2.5 e^{-t} \cos 2t = 0\). Since we know \(\cos 2t = 0\) for \(2t = \frac{\pi}{2} + n\pi\), to find the second time it passes the rest position, we use \(n=1\): $$2t = \frac{\pi}{2} + \pi \Rightarrow t = \frac{3\pi}{4}.$$ d. Find the time and the displacement when the object reaches its high point for the second time
07

Find the critical points

The high point for the second time occurs between the second and third time when the object passes the rest position. After solving for t as we did previously, we have \(n=2\), so the third time it passes the rest position will be at: $$2t = \frac{\pi}{2} + 2\pi \Rightarrow t = \frac{5\pi}{4}.$$
08

Solve for t

Now, use the derivative \(y'(t)\) equal to zero and solve for \(t\) within the interval \(\left(\frac{3\pi}{4}, \frac{5\pi}{4}\right)\). The solution is approximately \(t \approx 2.592\). Finally, find the displacement at this time: \(y(2.592) \approx 0.441.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points Calculus
The idea of critical points is an essential concept in calculus, especially when we aim to comprehend the behavior of functions. To identify the critical points of a function, we first need to find its derivative. This is because critical points occur where the derivative either equals zero or is undefined.

For the damped oscillator with displacement given by \(y(t)=2.5 e^{-t} \cos 2t\), the derivative \(y'(t)\) is found and set to zero to identify the function's extrema, which in the physical context, represent the object's highest and lowest points in its motion. By equating \(y'(t)\) to zero, we essentially locate the times where the object changes direction鈥攖he moments it begins moving upward or downward after reaching its peak displacements.
Derivative Applications
Derivatives are not just abstract mathematical concepts; they have real-world applications, one of them being the analysis of motion. When we calculate the derivative of a position function, we obtain a velocity function. In turn, differentiating the velocity function gives us the acceleration function.

In our problem, the function \(y(t)\) represents the displacement of a vertically oscillating object. The first derivative \(y'(t)\) offers valuable insight into the object's velocity at any given time. Setting \(y'(t)\) to zero, which indicates that the object has either reached a high or low point in its motion (i.e., where its velocity is zero), is an application of derivatives that reveals moments of transient rest within continuous motion.
Solving Trigonometric Equations
Trigonometric equations play a key role in understanding oscillatory motion. In the given problem, the part of the displacement function involving \(\cos 2t\) calls for solving a trigonometric equation when setting the displacement to zero. The zeros of \(\cos\) occur at \(\frac{\pi}{2} + n\pi\), with \(n\) being an integer. This defines the points in time where the oscillator passes through the rest position.

It's important to highlight that understanding how to manipulate and solve these equations is crucial, not only in a mathematical sense but also for interpreting physical phenomena. The periodicity of trigonometric functions, as dictated by their fundamental identities, allows us to find solutions over specific intervals, which correspond to the physical timing of the oscillator's motion through its equilibrium point.

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