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Motion of a SpringIn Exercise 59-64, a 32-pound weight stretched a spring 2 3 foot from its natural length. Use the given information to find a formula for the position of the weight as a function of time. The weight is raised 2 3 foot above equilibrium and released with an initial downward velocity of 1 2 foot per second.

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Multivariable Calculus

11th Edition
Ron Larson + 1 other
Publisher: Cengage Learning
ISBN: 9781337275378
Chapter 16.2, Problem 61E
Textbook Problem
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Motion of a SpringIn Exercise 59-64, a 32-pound weight stretched a spring 2 3 foot from its natural length. Use the given information to find a formula for the position of the weight as a function of time.

The weight is raised 2 3 foot above equilibrium and released with an initial downward velocity of 1 2 foot per second.

To determine

The formula for the position of weight as a function of time when a 32-pounds weight stretchesa spring 23 foot from its natural length. The weight is raised 23 foot above the equilibrium and released with an initial downward velocity of 12 foot per second.

Explanation of Solution

Given information: The weight is pulled 23 foot above the equilibrium and released with an initial velocity of 12 foot per second.

The 32-pounds weight stretches a spring 23 foot from its natural length, so by Hooke’s Law,

32=k(23)32×32=k48=k

As the weight w is given by mg, we have

m=wg=3232=1

So, the resulting differential equation for the undamped motion is

d2ydt2+481y=0d2ydt2+48y=0

The characteristicequation for the differential equation d2ydt2+48y=0 is m2+48=0.

The roots of the characteristic equation m2+48=0 are:

m2+48=0m2=48m=48m=±43i

If m1=α+iβ and m2=αiβ are complex zeros of the characteristics equation, then the general solution is y=C1eαtcosβt+C2eαtsinβt.

Thus, the general solution of the equation d2ydt2+48y=0 is y=C1cos43t+C2sin43t

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Chapter 16 Solutions

Multivariable Calculus
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