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Chapter 3: Problem 83

The motion of a spring that is subject to a frictional force or a damping force (such as a shock absorber in a car) is often modeled by the product of an exponential function and a sine or cosine function. Suppose the equation of motion of a point on such a spring is $$s ( t ) = 2 e ^ { - 1.5 t } \sin 2 \pi t$$ where \(s\) is measured in centimeters and \(t\) in seconds. Find the velocity after \(t\) seconds and graph both the position and velocity functions for \(0 \leq t \leq 2\) .

Short Answer

Expert verified
The velocity function is \( v(t) = -3e^{-1.5t}\sin(2\pi t) + 4\pi e^{-1.5t}\cos(2\pi t) \).

Step by step solution

01

Differentiate the Position Function

The position function given is \( s(t) = 2 e^{-1.5t} \sin(2\pi t) \). To find the velocity function \( v(t) \), we need to differentiate \( s(t) \) with respect to \( t \). Use the product rule, which states that \( (uv)' = u'v + uv' \), where \( u(t) = 2e^{-1.5t} \) and \( v(t) = \sin(2\pi t) \).
02

Apply the Product Rule

Differentiate \( u(t) = 2e^{-1.5t} \):\( u'(t) = -3e^{-1.5t} \).Differentiate \( v(t) = \sin(2\pi t) \):\( v'(t) = 2\pi \cos(2\pi t) \).Apply the product rule:\[ v(t) = u'(t)v + uv'(t) = (-3e^{-1.5t})\sin(2\pi t) + (2e^{-1.5t})(2\pi \cos(2\pi t)) \].
03

Simplify the Velocity Expression

Combine the terms to express the velocity function:\[ v(t) = -3e^{-1.5t}\sin(2\pi t) + 4\pi e^{-1.5t}\cos(2\pi t) \].This is the expression for the velocity function in terms of \( t \).
04

Determine the Velocity at Specific Times

To find the velocity after some specific time, simply substitute the desired \( t \) value into the velocity function. For example, to find the velocity at \( t = 1 \), substitute \( t = 1 \) into the equation for \( v(t) \).
05

Graph Position and Velocity Functions

Using a graphing tool (like Desmos, Geogebra, or a graphing calculator), plot the functions \( s(t) = 2e^{-1.5t}\sin(2\pi t) \) and \( v(t) = -3e^{-1.5t}\sin(2\pi t) + 4\pi e^{-1.5t}\cos(2\pi t) \) for \( 0 \leq t \leq 2 \). Observe the damped oscillatory motion in \( s(t) \) and velocity behavior in \( v(t) \).

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Key Concepts

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

Product Rule
The product rule is an essential tool in differential calculus, particularly when dealing with functions involving the product of two functions. In our context, we have a position function which is the product of an exponential function and a trigonometric function. The product rule formula is:
  • If you have a function that is the product of two functions, \( u(t) \) and \( v(t) \), then its derivative is given by \( (uv)' = u'v + uv' \).
In the exercise, \( u(t) = 2e^{-1.5t} \) and \( v(t) = \sin(2\pi t) \). By differentiating these individually:
  • \( u'(t) = -3e^{-1.5t} \) because the derivative of an exponential function involves the chain rule.
  • \( v'(t) = 2\pi \cos(2\pi t) \) because the derivative of \( \sin( \text{something}) \) is \( \cos( \text{something}) \) times the derivative of the argument, which is \( 2\pi \).
Finally, combining these using the product rule gives the velocity function, a crucial step for understanding motion represented by the original function.
Damped Oscillations
Damped oscillations describe the behavior of a system in which the amplitude of the oscillation decreases over time, often due to a damping force like friction. In our example, the function \( s(t) = 2e^{-1.5t} \sin(2\pi t) \) models such behavior. The damping is represented by:
  • The exponential part, \( e^{-1.5t} \), which decreases over time. This represents the "damping" effect, leading to a gradual reduction in amplitude.
  • The trigonometric part, \( \sin(2\pi t) \), represents the oscillatory motion of the system.
Together, these components create a graph that shows oscillations with a decreasing amplitude as time progresses, a key aspect of systems experiencing damping forces like shock absorbers in vehicles.
Exponential Functions
Exponential functions are a fundamental concept in mathematics, characterized by their constant rate of growth or decay. They appear in many scientific models, particularly those involving growth and decay processes. In the position function given:
  • \( e^{-1.5t} \) represents an exponential decay. The negative exponent indicates that the function decreases over time, reflecting a decrease in the amplitude of the oscillation as seen in damped motion.
  • This exponential function affects the entire product, reducing the overall value of the position function \( s(t) \) as time increases.
Exponential functions are powerful in modeling scenarios where quantities decrease multiplicatively over time, such as radioactive decay, and are crucial in understanding the behavior of the damped oscillations in this exercise.
Velocity Function
The velocity function is derived from the position function by differentiation and represents the rate of change of position with respect to time. In this model, the velocity function is obtained by applying the product rule:
  • The original position function is \( s(t) = 2e^{-1.5t} \sin(2\pi t) \).
  • The differentiated form gives \( v(t) = -3e^{-1.5t} \sin(2\pi t) + 4\pi e^{-1.5t} \cos(2\pi t) \).
This function captures how fast the position changes and combines both terms:
  • The \( -3e^{-1.5t} \sin(2\pi t) \) component reflects the damping and original oscillation impact on velocity.
  • \( 4\pi e^{-1.5t} \cos(2\pi t) \) represents the velocity's oscillatory nature and how quickly it adjusts.
Understanding this function is fundamental to analyzing the dynamics of the system, including predicting future movement or analyzing past behavior of oscillating systems under damping influence.

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