Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Suppose the coupled mass-spring system of Problem (Figure 5.31) is hung vertically from support (with mass $m_{2}$ above $m_{1}$ ), as in Section 4.10, page 226.

(a) Argue that at equilibrium, the lower spring is stretched a distance $l_{1}$ from its natural length $l_{2}$ , given by $l_{1} = m_{1} g / k_{1}$ .
(b) Argue that at equilibrium, the upper spring is stretched a distance $l_{2} = (m_{1} + m_{2}) g / k_{2}$ .
(c) Show that if $x_{1}$ and $x_{2}$ are redefined to be displacements from the equilibrium positions of the masses $m_{1}$ and $m_{2}$ , then the equations of motion are identical with those derived in Problem 1.

The system of differential equations for the displacement x and y is

$m_{1} x'' = k_{1} (y - x) - m_{1} g m_{2} y'' = - k_{1} (y - x) - k_{2} y - m_{2} g$

New equilibrium is $(x_{0}, y_{0}) = (- \frac{m_{1} k_{1} g + m_{2} k_{1} g + m_{1} k_{2} g}{k_{1} k_{2}}, - \frac{(m_{1} + m_{2}) g}{k_{2}})$

(a) $l_{1} = |x_{0} - y_{0}| = m_{1} g / k_{1}$ ,

(b) $l_{2} = |y_{0}| = (m_{1} + m_{2}) g / k_{2}$ ,

(c) Substituting $ξ = x - x_{0}$ and $η = y - y_{0}$ into the system of differential equations for displacement we obtain a system in terms of $ξ$ and $η$ which is identical to the system we derived in Problem 1.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Finding the differential equation

First, one needs to find the system of differential equations for the displacement x and y. Assume that $x, y < 0$ and $x < y$ . In this case, one has that $F_{1} = - m_{1} g, F_{2} = k_{1} (y - x), F_{3} = - k_{2} (y - x), F_{4} = - m_{2} g, F_{5} = - k_{2} y$

and $m_{1} x'' = F_{1} + F_{2}, m_{2} y'' = F_{3} + F_{4} + F_{5}$

But this equation must be true for any x and y, so the system of differential equations for the displacement x and y is:

$\begin{array}{rcl} m_{1} x''' & = & k_{1} (y - x) - m_{1} g \\ m_{2} y'' & = & - k_{1} (y - x) - k_{2} y - m_{2} g \end{array}$

Finding the length of the springs

At the equilibrium $(x, y) = (x_{0}, y_{0})$ , one has that $m_{1} x'' = 0$ and $m_{2} y'' = 0$ . So, to find the change in the length of the springs, one needs to find a new equilibrium first, and to do so one will solve the following system for $x_{0}$ and $y_{0}$ :

$0 = k_{1} (y_{0} - x_{0}) - m_{1} g 0 = - k_{1} (y_{0} - x_{0}) - k_{2} y_{0} - m_{2} g$

Adding those two equations together one will get

$\begin{array}{rcl} 0 & = & - m_{1} g - k_{2} y_{0} - m_{2} g \\ y_{0} & = & - \frac{(m_{1} + m_{2}) g}{k_{2}} \end{array}$

Finding x0

Now we need to find $x_{0}$ . Multiplying the first equation by $x_{0}$ and the second by and then adding them together one will get

$\begin{array}{rcl} 0 & = & m_{2} k_{1} (y_{0} - x_{0}) - m_{1} m_{2} g \partial \\ 0 & = & m_{1} k_{1} (y_{0} - x_{0}) + m_{1} k_{2} y_{0} + m_{1} m_{2} g \\ 0 & = & m_{2} k_{1} (y_{0} - x_{0}) + m_{1} k_{1} (y_{0} - x_{0}) + m_{1} k_{2} y_{0} \\ 0 & = & k_{1} (y_{0} - x_{0}) (m_{1} + m_{2}) + m_{1} k_{2} y_{0} \\ k_{1} x_{0} (m_{1} + m_{2}) & = & k_{1} y_{0} + m_{1} k_{2} y_{0} \end{array}$

Finding a new equilibrium

Substituting the value for $y_{0}$ into the previous equation, one will get

$\begin{array}{rcl} k_{1} x_{0} (m_{1} + m_{2}) & = & - \frac{k_{1}}{k_{2}} {(m_{1} + m_{2})}^{2} g - m_{1} (m_{1} + m_{2}) g \\ k_{1} x_{0} & = & - \frac{m_{1} k_{1} g + m_{2} k_{1} g + m_{1} k_{2} g}{k_{2}} \\ x_{0} & = & - \frac{m_{1} k_{1} g + m_{2} k_{1} g + m_{1} k_{2} g}{k_{1} k_{2}} \end{array}$

The new equilibrium is $(x_{0}, y_{0}) = (- \frac{m_{1} k_{1} g + m_{2} k_{1} g + m_{1} k_{2} g}{k_{1} k_{2}}, - \frac{(m_{1} + m_{2}) g}{k_{2}})$ .

Finding the length I1

one can see from the Figure above that the lower spring is stretched by $l_{1} = |x_{0} - y_{0}|$ which is

$\begin{array}{rcl} l_{1} & = & |x_{0} - y_{0}| \\ = & |- \frac{m_{1} k_{1} g + m_{2} k_{1} g + m_{1} k_{2} g}{k_{1} k_{2}} + \frac{(m_{1} + m_{2}) g}{k_{2}}| \\ = & |- \frac{m_{1} k_{1} g + m_{2} k_{1} g + m_{1} k_{2} g - m_{1} k_{1} g - m_{2} k_{1} g}{k_{1} k_{2}}| \\ = & |- \frac{m_{1} k_{2} g}{k_{1} k_{2}}| \\ = & \frac{m_{1} g}{k_{1}} \end{array}$

Finding the length I2

From the Figure above we can conclude that the upper spring is stretched only by $l_{2} = |y_{0}|$ , so

$\begin{array}{rcl} l_{2} & = & |- \frac{(m_{1} + m_{2}) g}{k_{2}}| \\ = & \frac{(m_{1} + m_{2}) g}{k_{2}} \end{array}$

Introduce new variables

$ξ = x - x_{0}$ and $η = y - y_{0}$

One has that $x = ξ + x_{0}, y = η + y_{0}$ and $x'' = ξ'', y'' = η''$ .

Substituting the values into the differential equation

Substituting this into the system of differential equations one will get

$\begin{array}{rcl} m_{1} ξ'' & = & k_{1} (η + y_{0} - ξ - x_{0}) - m_{1} g \\ m_{2} η'' & = & - k_{1} (η + y_{0} - ξ - x_{0}) - k_{2} (η + y_{0}) - m_{2} g \\ m_{1} ξ'' & = & k_{1} (η - ξ - \frac{(m_{1} + m_{2}) g}{k_{2}} + \frac{(m_{1} + m_{2}) k_{1} g + m_{1} k_{2} g}{k_{1} k_{2}}) - m_{1} g \\ m_{2} η'' & = & - k_{1} (η - ξ - \frac{(m_{1} + m_{2}) g}{k_{2}} + \frac{(m_{1} + m_{2}) k_{1} g + m_{1} k_{2} g}{k_{1} k_{2}}) - k_{2} (η - \frac{(m_{1} + m_{2}) g}{k_{2}}) - m_{2} g \\ m_{1} ξ'' & = & k_{1} (η - ξ + \frac{m_{1} k_{2} g}{k_{1} k_{2}}) - m_{1} g \\ m_{2} η'' & = & - k_{1} (η - ξ + \frac{m_{1} k_{2} g}{k_{1} k_{2}}) - k_{2} η + (m_{1} + m_{2}) g - m_{2} g \\ \Rightarrow & m_{1} ξ'' = k_{1} (η - ξ) + m_{1} g - m_{1} g \\ m_{2} η'' & = & - k_{1} (η - ξ) - m_{1} g - k_{2} η + (m_{1} + m_{2}) g = m_{2} g \\ \Rightarrow & m_{1} ξ'' = k_{1} (η - ξ) \\ m_{2} η'' & = & - k_{1} (η - ξ) - k_{2} η \end{array}$

One sees that the differential equations of motion in terms of $ξ = x - x_{0}$ and $η = y - y_{0}$ are identical to the equations in terms of and derived in Problem 1 when one had masses attached to the horizontal springs.

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Finding the differential equation

Finding the length of the springs

Finding x0

Finding a new equilibrium

Finding the length I1

Finding the length I2

Substituting the values into the differential equation

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Finding the differential equation

Finding the length of the springs

Finding x0

Finding a new equilibrium

Finding the length I1

Finding the length I2

Substituting the values into the differential equation

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