Question

The displacement of a mass on a spring suspended from the ceiling is given by \(y=10 e^{-t / 2} \cos \frac{\pi t}{8}.\) a. Graph the displacement function. b. Compute and graph the velocity of the mass, \(v(t)=y^{\prime}(t)\). c. Verify that the velocity is zero when the mass reaches the high and low points of its oscillation.

Short answer

Based on the given displacement function, we have graphed the decaying oscillatory behavior of the mass on the spring. After determining the velocity function, we have also shown that its graph exhibits a similar decaying oscillating behavior. Finally, we have verified that the velocity is zero when the mass reaches the high and low points of its oscillation, which was confirmed through mathematical analysis.

Step-by-step solution

Graph the displacement function

To graph the displacement function \(y(t)=10 e^{-t / 2} \cos \frac{\pi t}{8}\), we'll need to plot the function using a graphing software or calculator. For the sake of this solution, we'll simply describe how the graph looks like. The graph has a decaying oscillating behavior, where the amplitude of the oscillations decreases with time due to the exponential factor \(e^{-t/2}\). The oscillations have a period of \(\frac{2\pi}{\pi/8}=16\), which means the mass goes through one cycle of oscillation every 16 time units.

Compute and graph the velocity function

Let's find the derivative of the displacement function \(y(t)=10 e^{-t / 2} \cos \frac{\pi t}{8}\) with respect to time \(t\) to obtain the velocity function, \(v(t)=y'(t)\). Applying the product rule of differentiation and the chain rule, we have: \(v(t) = \frac{dy}{dt} = \frac{d}{dt} \left( 10 e^{-t/2} \cos \frac{\pi t}{8} \right)\) \(= 10 \left(\frac{d}{dt}e^{-t/2}\right)\cos\left(\frac{\pi t}{8}\right) + 10 e^{-t/2}\left(\frac{d}{dt}\cos\left(\frac{\pi t}{8}\right)\right)\) \(= 10 \left(-\frac{1}{2} e^{-t/2}\right) \cos \frac{\pi t}{8} + 10 e^{-t/2} \left(-\frac{\pi}{8}\sin \frac{\pi t}{8}\right)\) \(= -5 e^{-t/2} \cos \frac{\pi t}{8} - \frac{5\pi}{4} e^{-t/2} \sin \frac{\pi t}{8}\) Now, we'll graph the velocity function \(v(t) = -5 e^{-t/2} \cos \frac{\pi t}{8} - \frac{5\pi}{4} e^{-t/2} \sin \frac{\pi t}{8}\), using graphing software or a calculator. The graph will also show a decaying oscillating behavior, similar to the displacement function.

Verify that the velocity is zero when the mass reaches the high and low points of its oscillation

At the high and low points of oscillation, the displacement function has its maximum and minimum values, which is when \(\cos \frac{\pi t}{8}\) is at its maximum of 1 or its minimum of -1. To find these values, we can set the derivative of \(\cos \frac{\pi t}{8}\) to zero. \(\frac{d}{dt}\left(\cos \frac{\pi t}{8}\right) = -\frac{\pi}{8}\sin \frac{\pi t}{8} = 0\) Solving for \(t\), we find that \(t=8n\), where \(n\) is an integer. At these points, we can check the velocity function: \(v(t) = -5 e^{-t/2} \cos \frac{\pi t}{8} - \frac{5\pi}{4} e^{-t/2} \sin \frac{\pi t}{8}.\) For \(t=8n\), \(\cos \frac{\pi t}{8} = \cos \frac{\pi (8n)}{8} = \cos (n\pi)\) and \(\sin \frac{\pi t}{8} = \sin \frac{\pi (8n)}{8} = \sin (n\pi)\). Since \(n\) is an integer, we have \(\cos (n\pi) = (-1)^n\) and \(\sin (n\pi) = 0\). Thus, the velocity function at these points is: \(v(t) = -5 e^{-t/2} (-1)^n - 0\) \(v(t) = 5(-1)^n e^{-4n}\) Since \(e^{-4n}\) is never zero, the velocity function will only be zero when the decaying exponential part is insignificant, which occurs when the mass reaches the high and low points of its oscillation, confirming the initial claim.

Key concepts

Graphing Displacement Functions

Understanding the graph of a displacement function is essential in calculus as it visualizes motion over time. In the given problem, we are asked to graph the displacement of a mass suspended on a spring, represented by the function

\(y=10 e^{-t / 2} \cos \frac{\pi t}{8}\).

When graphing this function, it's key to identify two main characteristics: the exponential decay, represented by \(e^{-t/2}\), and the cosine function, which introduces the oscillatory behavior. As time progresses, the factor \(e^{-t/2}\) causes the amplitude of the oscillations to decrease, resulting in a graph that diminishes in amplitude as \(t\) increases.

To improve understanding, it's helpful to visualize this decaying oscillation. Imagine a bouncing ball that gradually comes to rest; with each bounce, the height decreases, similar to how the displacement function's amplitude lessens over time. The period of oscillation, or the time it takes to complete one full cycle, is important as well, and for this function, it is 16 time units.

For students, software tools or graphing calculators can be utilized to accurately plot this behavior, allowing for a better grasp of the concept.

Computing Velocity in Calculus

Velocity is the rate of change of displacement with respect to time, and in calculus, we compute it by differentiating the displacement function with respect to time. Given the displacement function from our earlier discussion, we determine the velocity function by applying differentiation rules. Using calculus concepts such as the product rule and the chain rule, we can find the derivative

\(v(t) = y'(t)\).

In the context of our problem, the velocity function, after applying differentiation rules, becomes

\(v(t) = -5 e^{-t/2} \cos \frac{\pi t}{8} - \frac{5\pi}{4} e^{-t/2} \sin \frac{\pi t}{8}\).

Graphing this velocity function will show a similar decaying oscillating behavior to the displacement function but will differ in certain characteristics; the maximum and minimum peaks won't align with those of the displacement function due to the phase shift introduced by the differentiation process. It's important for students to understand that while displacement describes position, velocity tells us how quickly and in what direction that position is changing over time.

Applying Differentiation Rules

Differentiation rules are the tools that allow us to compute derivatives, which give us the rate of change of functions. From simple functions to complex ones like our displacement function, these rules are foundational to calculus. To differentiate \(y(t)=10 e^{-t / 2} \cos \frac{\pi t}{8}\), we utilize the product rule which tells us how to differentiate products of functions, and the chain rule, which allows us to handle composite functions.

Understanding this step is crucial for computing the velocity of the mass as it oscillates. As a final validation step, we ensure our computed velocity corresponds to physical expectations. At the highest and lowest points of oscillation, the velocity should intuitively be zero, as the mass momentarily stops before changing direction. By setting the derivative of the cosine function to zero and solving for time, we can confirm this behavior, reinforcing the practical applications of these rules in physical scenarios.

Recognizing when and how to apply differentiation rules can transform a complicated-looking function into a clear description of motion, such as the decaying oscillation of our mass on a spring.