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In contrast to springs in parallel, if two springs connected in series have very dissimilar stiffnesses, the equivalent stiffness will be dominated by (yet even lower than) the spring with the lower stiffness The same rules for series and parallel connections hold for torsional springs 1144 Equivalent Springs and Mechanical Advantage A spring combined with some mechanical advantage has an equivalent spring constant that is directly analogous to the equivalent mass or inertia discussed previously Instead of equating the kinetic energy to determine the equivalent mass, however, the potential energies of the two systems are equated to give an equivalent stiffness or spring rate Figure 1116a shows a lever arm with a spring attached To get the equivalent system of Fig 1116b, which is referred to the displacement xf, it is necessary to equate the potential energies of the two systems Assuming small displacements, the potential energy of the original system is

This number minus the 130 already there, or 2725-130 = 2595, deer need to be introduced to achieve the 10,000 goal in the prescribed time

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Since for small angles, the relationship between x and xf is x xf = l1 l2 Using this expression for x above gives l Keq = K 1 l2 When l1 > l2, the spring appears stiffer than its physical stiffness, while for l1 < l2, the equivalent stiffness is less than the physical stiffness of the spring because of the mechanical advantage 1145 Massive Springs Up to this point, the springs considered have been ideal springs and have had no mass associated with them In practice, it is often acceptable to model real springs as ideal springs if the mass of other elements in the system dominates the mass of the springs This is not always the case, however, and real springs may need to be modeled as a combination of an ideal spring and a lumped mass The equivalent mass of a massive spring (Fig 1117) can be derived by applying the concepts used in the previous section Since the spring is a continuum, the kinetic energy is given by KE = 1 v 2 ( x )dm 2 spring (1155)

This integral is easily evaluated and we see that it has value [2 ( 97) 3/2 128]/243

If the mass is evenly distributed along the length of the spring,

Notice that the last example would have been considerably more difficult (the integral would have been harder to evaluate) had we expressed the curve in the form y = f (x)

(1156)

Assuming that the spring compresses uniformly, the velocity pro le along the spring can be interpolated between the end points At x = 0, the spring is xed, so the velocity is zero; at x = L, de ne the velocity as vL The velocity at any point x between 0 and L is therefore v= x vL L (1157)

Let f (x) be a non-negative function on the interval [a, b] Imagine rotating the graph of f about the x axis This procedure will generate a surface of revolution, as shown in Figure 831 We will develop a procedure for determining the area of such a surface We partition the interval [a, b] : a = x0 x1 x2 xk 1 xk = b Corresponding to each pair of elements x j 1 , x j in the partition is a portion of curve, as shown in Figure 832 When that piece of curve is rotated about the x axis, we obtain a cylindrical surface Now the area of a true right circular cylinder is 2 r h We do not have a true cylinder, so we proceed as follows We may approximate the radius by f (x j ) And the height of the cylinder can be

Another and more formal definition of the natural logarithm relates the lnx to portions of the area under the curve y = l/t The natural logarithm of any number x is deflnedby l n x =

Putting these together, the expression for kinetic energy gives KE = 1 x 2 2 ms 1 ms v 2 1 ms v 2 L3 1 m L2 vL L dx = 2 L3 L x 2 dx = 2 L3 L 3 = 2 3s vL2 20 0

y = f (x)

(1158)

Equating this expression with the kinetic energy of a single equivalent mass attached at the end of the spring (x = L), the equivalent mass is simply meq = ms 3 (1159)

y = f (x)

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