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# The motion of a spring that is subject to a fictional 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$s(t) = 2e^{-1.5t} \sin 2 \pi t$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 $0 \le t \le 2.$

## 2$e^{-1.5 t}[2 \pi \cos (2 \pi t)-1.5 \sin (2 \pi t)]$

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

all right. So here we have the equation of motion for the point on a spring, and what we want to do is find the velocity equation, and the velocity is the derivative of position. And so what we're going to do is use the product rule. So we have the first and the second. And so for the product rule, we have the first times the derivative of the second. And when we take the derivative of the second, we're using the chain rule. So the derivative of sign is co sign. So if coastline of two pi t and then the derivative of two pi t the inside is two pi, then we have plus the second times the derivative of the first. And when we take the derivative of the first, we use the product, the chain rule as well. We leave the to the constant, and then we have e to the negative 1.5 t. That's the derivative of the outside function times negative 1.5. That's the derivative of the inside function. And then we can simplify this. So we noticed in our unseen cliff I'd form that we have a two and a two pi so we can multiply those and that gives us four pi. And then we noticed in the second term we have a negative 1.5 times a two and we multiply those, and that gives us minus three. So what we want to do now is take both the position equation and the velocity equation and graft them on a calculator. So I grabbed a calculator. We go to y equals. We type in the position function that we were given, and we type in the velocity function as well. And we want to look at them four times between zero and two so we can go to window and we can change our X Men 20 and R X max to to. And then we can fiddle around with different Y values until we find what we like. And I decided to go from negative 7 to 6 on my Y axis. So here we have our graphs. The blue one is the original function. That's the position or equation of motion. And then the red one is the velocity, and we can see that when the position equation is increasing the velocity is positive, and when the position equation is decreasing, the velocity is negative and we see that happening at different points. Along the way we see that when the position equation is more steep, the velocity is farther from zero and when it's more shallow velocities closer to zero.

Oregon State University

#### Topics

Derivatives

Differentiation

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp