   Chapter 3.4, Problem 83E

Chapter
Section
Textbook Problem

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 iss(t) = 2e–1.5t sin 2πtwhere 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 ≤ 1 ≤ 2.

To determine

To find: The velocity after t seconds.

Explanation

Given:

The equation of motion of a point on such a spring is s(t)=2e1.5tsin2πt.

Derivative rule:

(1) Constant Multiple Rule: ddx[cf(x)]=cddx[f(x)]

(2) Product Rule: ddx(fg)=fddx(g)+gddx(f)

Recall:

If x(t) is a displacement of a particle and the time t is in seconds, then the velocity of the particle is v(t)=dsdt.

Calculation:

Obtain the velocity at time t.

v(t)=ddx(s(t))=ddx(2e1.5tsin2πt)

Apply the product rule (2) and the constant multiple rule (1),

v(t)=2e1.5tddx(sin2πt)+sin2πtddx(2e1.5t)=2e1.5t(cos2πt)+2sin2πt(1

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