We begin this first section by examining the motion of a mass m attached to a spring. The word spring in this discussion will refer to a flexible coil, or helical, spring. See the Remarks at the end of this section.

Hooke’s Law

Suppose a spring/mass system consists of a flexible spring suspended vertically from a rigid support with a mass m attached to its free end. The amount of stretch s, or elongation, of the spring will of course depend on the mass; masses with different weights stretch the spring by differing amounts. Hooke’s law, devised by the English natural philosopher and polymath Robert Hooke (1635–1703) around 1660, states that the spring itself exerts a restoring force F opposite to the direction of elongation and proportional to the amount of elongation s. Simply stated, F = −ks, where k > 0 is a constant of proportionality called the spring constant or stiffness of the spring. Thus a spring is essentially characterized by the number k. Using |F| = k|s|, we see that if a mass weighing 10 lb stretches a spring ft, then 10 = k() implies k = 20 lb/ft. Necessarily then, a mass weighing, say, 16 lb stretches the same spring ft.

Newton’s Second Law

After a mass m is attached to a spring, it stretches the spring by an amount s and attains a position of equilibrium at which its weight W is balanced by the restoring force ks. Recall that weight is defined by W = mg, where mass is measured in slugs, kilograms, or grams and g = 32 ft/s², 9.8 m/s², or 980 cm/s², respectively. As indicated in FIGURE 3.8.1(b), the condition of equilibrium is mg = ks or mg − ks = 0. If the mass is displaced by an amount x from its equilibrium position, the restoring force of the spring is then k(x + s). Assuming that there are no retarding forces acting on the system and assuming that the mass vibrates free of other external forces—free motion—we can equate Newton’s second law with the net, or resultant, force of the restoring force and the weight:

(1)

The negative sign in (1) indicates that the restoring force of the spring acts opposite to the direction of motion. Furthermore, we can adopt the convention that displacements measured below the equilibrium position are positive. See FIGURE 3.8.2.

DE of Free Undamped Motion

By dividing (1) by the mass m we obtain the second-order differential equation d²x/dt² + (k/m)x = 0 or

, (2)

where ω² = k/m. Equation (2) is said to describe simple harmonic motion or free undamped motion. Two obvious initial conditions associated with (2) are x(0) = x₀, the amount of initial displacement, and x′(0) = x₁, the initial velocity of the mass. For example, if x₀ > 0, x₁ < 0, the mass starts from a point below the equilibrium position with an imparted upward velocity. When x₁ = 0 the mass is said to be released from rest. For example, if x₀ < 0, x₁ = 0, the mass is released from rest from a point units above the equilibrium position.

Solution and Equation of Motion

To solve equation (2) we note that the solutions of the auxiliary equation m² + ω² = 0 are the complex numbers m₁ = ωi, m₂ = −ωi. Thus from (8) of Section 3.3 we find the general solution of (2) to be

x(t) = c₁ cos ωt + c₂ sin ωt. (3)

The period of free vibrations described by (3) is T = 2π/ω, and the frequency is f = 1/T = ω/2π. For example, for x(t) = 2 cos 3t − 4 sin 3t the period is 2π/3 and the frequency is 3/2π. The former number means that the graph of x(t) repeats every 2π/3 units; the latter number means that there are three cycles of the graph every 2π units or, equivalently, that the mass undergoes 3/2π complete vibrations per unit time. In addition, it can be shown that the period 2π/ω is the time interval between two successive maxima of x(t). Keep in mind that a maximum of x(t) is a positive displacement corresponding to the mass’s attaining a maximum distance below the equilibrium position, whereas a minimum of x(t) is a negative displacement corresponding to the mass’s attaining a maximum height above the equilibrium position. We refer to either case as an extreme displacement of the mass. Finally, when the initial conditions are used to determine the constants c₁ and c₂ in (3), we say that the resulting particular solution or response is the equation of motion.

EXAMPLE 1 Free Undamped Motion

Alternative Form of x(t)

When c₁ ≠ 0 and c₂ ≠ 0, the actual amplitude A of free vibrations is not obvious from inspection of equation (3). For example, although the mass in Example 1 is initially displaced foot beyond the equilibrium position, the amplitude of vibrations is a number larger than . Hence it is often convenient to convert an equation of simple harmonic motion of the form given in (3) into the form of a shifted sine function (6) or a shifted cosine function (6′). In both (6) and (6′) the number is the amplitude of free vibrations, and is a phase angle. Note carefully that is defined in a slightly different manner in (7) and (7′).

To verify (6), we expand by the addition formula for the sine function:

(8)

It follows from FIGURE 3.8.3 that if is defined by

then (8) becomes

EXAMPLE 2 Alternative Form of Solution (5)

Graphical Interpretation

FIGURE 3.8.4(a) illustrates the mass in Example 2 going through approximately two complete cycles of motion. Reading left to right, the first five positions marked with black dots in the figure correspond, respectively,

• to the initial position of the mass below the equilibrium position,
• to the mass passing through the equilibrium position for the first time heading upward,
• to the mass at its extreme displacement above the equilibrium position,
• to the mass passing through the equilibrium position for the second time heading downward, and
• to the mass at its extreme displacement below the equilibrium position.

The dots on the graph of (9) given in Figure 3.8.4(b) also agree with the five positions just given. Note, however, that in Figure 3.8.4(b) the positive direction in the tx-plane is the usual upward direction and so is opposite to the positive direction indicated in Figure 3.8.4(a). Hence the blue graph representing the motion of the mass in Figure 3.8.4(b) is the mirror image through the t-axis of the red dashed curve in Figure 3.8.4(a).

Form (6) is very useful since it is easy to find values of time for which the graph of x(t) crosses the positive t-axis (the line x = 0). We observe that sin(ωt + ϕ) = 0 when ωt + ϕ = nπ, where n is a nonnegative integer.

Double Spring Systems

Suppose two parallel springs, with constants k₁ and k₂, are attached to a common rigid support and then to a metal plate of negligible mass. A single mass m is attached to the center of the plate in the double-spring arrangement as shown in FIGURE 3.8.5. If the mass is displaced from its equilibrium position, the displacement x is the same for both springs and so the net restoring force of the spring in (1) is simply −k₁x − k₂x = −(k₁ + k₂)x. We say that

is the effective spring constant of the system.

On the other hand, suppose two springs supporting a single mass m are in series, that is, the springs are attached end-to-end as shown in FIGURE 3.8.6. In this case, a displacement x of the mass from its equilibrium consists of the sum x = x₁ + x₂, where x₁ and x₂ are the displacements of the respective springs. But the restoring force is the same for both springs so if k_eff is the effective spring constant of the system we have

From k₁x₁ = k₂x₂ we see x₁ = (k₂/k₁)x₂ and so − k_eff (x₁ + x₂) = − k₂x₂ is the same as

Solving the last equation for k_eff yields

So in either of the above cases, the differential equation of motion is (1) with k replaced by k_eff. See Problems 13–18 in Exercises 3.8.

Systems with Variable Spring Constants

In the model discussed above, we assumed an ideal world, a world in which the physical characteristics of the spring do not change over time. In the nonideal world, however, it seems reasonable to expect that when a spring/mass system is in motion for a long period the spring would weaken; in other words, the “spring constant” would vary, or, more specifically, decay with time. In one model for an aging spring, the spring constant k in (1) is replaced by the decreasing function K(t) = ke^−αt, k > 0, α > 0. The linear differential equation mx″ + ke^−αtx = 0 cannot be solved by the methods considered in this chapter. Nevertheless, we can obtain two linearly independent solutions using the methods in Chapter 5. See Problem 19 in Exercises 3.8.

When a spring/mass system is subjected to an environment in which the temperature is rapidly decreasing, it might make sense to replace the constant k with K(t) = kt,k > 0, a function that increases with time. The resulting model, mx″ + ktx = 0, is a form of Airy’s differential equation. Like the equation for an aging spring, Airy’s equation can be solved by the methods of Chapter 5. See Problem 20 in Exercises 3.8.

The concept of free harmonic motion is somewhat unrealistic since the motion described by equation (1) assumes that there are no retarding forces acting on the moving mass. Unless the mass is suspended in a perfect vacuum, there will be at least a resisting force due to the surrounding medium. As FIGURE 3.8.7 shows, the mass could be suspended in a viscous medium or connected to a dashpot damping device.

DE of Free Damped Motion

In the study of mechanics, damping forces acting on a body are considered to be proportional to a power of the instantaneous velocity. In particular, we shall assume throughout the subsequent discussion that this force is given by a constant multiple of dx/dt. When no other external forces are impressed on the system, it follows from Newton’s second law that

(10)

where β is a positive damping constant and the negative sign is a consequence of the fact that the damping force acts in a direction opposite to the motion.

Dividing (10) by the mass m, we find the differential equation of free damped motion is d²x/dt² + (β/m) dx/dt + (k/m)x = 0 or

(11)

where

(12)

The symbol 2λ is used only for algebraic convenience, since the auxiliary equation is m² + 2λm + ω² = 0 and the corresponding roots are then

We can now distinguish three possible cases depending on the algebraic sign of λ² − ω². Since each solution contains the damping factor e^−λt, λ > 0, the displacements of the mass become negligible over a long period of time.

Case I: λ² − ω² > 0 In this situation the system is said to be overdamped because the damping coefficient β is large when compared to the spring constant k. The corresponding solution of (11) is x(t) = c₁ + c₂ or

(13)

Equation 13 represents a smooth and nonoscillatory motion. FIGURE 3.8.8 shows two possible graphs of x(t).

Case II: λ² − ω² = 0 The system is said to be critically damped because any slight decrease in the damping force would result in oscillatory motion. The general solution of (11) is x(t) = c₁ + c₂t or

x(t) = e^−λt(c₁ + c₂t). (14)

Some graphs of typical motion are given in FIGURE 3.8.9. Notice that the motion is quite similar to that of an overdamped system. It is also apparent from (14) that the mass can pass through the equilibrium position at most one time.

Case III: λ² − ω² < 0 In this case the system is said to be underdamped because the damping coefficient is small compared to the spring constant. The roots m₁ and m₂ are now complex:

Thus the general solution of equation (11) is

(15)

As indicated in FIGURE 3.8.10, the motion described by (15) is oscillatory, but because of the coefficient e^−λt, the amplitudes of vibration → 0 as t → ∞.

EXAMPLE 3 Overdamped Motion

It is readily verified that the solution of the initial-value problem

is

(16)

The problem can be interpreted as representing the overdamped motion of a mass on a spring. The mass starts from a position 1 unit below the equilibrium position with a downward velocity of 1 ft/s.

To graph x(t) we find the value of t for which the function has an extremum—that is, the value of time for which the first derivative (velocity) is zero. Differentiating (16) gives x′(t) = −e^−t + e^−4t so that x′(t) = 0 implies e^3t = or t = ln = 0.157. It follows from the first derivative test, as well as our intuition, that x(0.157) = 1.069 ft is actually a maximum. In other words, the mass attains an extreme displacement of 1.069 feet below the equilibrium position.

We should also check to see whether the graph crosses the t-axis, that is, whether the mass passes through the equilibrium position. This cannot happen in this instance since the equation x(t) = 0, or e^3t = , has the physically irrelevant solution t = ln = −0.305.

The graph of x(t), along with some other pertinent data, is given in FIGURE 3.8.11. ≡

EXAMPLE 4 Critically Damped Motion

An 8-pound weight stretches a spring 2 feet. Assuming that a damping force numerically equal to two times the instantaneous velocity acts on the system, determine the equation of motion if the weight is released from the equilibrium position with an upward velocity of 3 ft/s.

SOLUTION

From Hooke’s law we see that 8 = k(2) gives k = 4 lb/ft and that W = mg gives m = = slug. The differential equation of motion is then

(17)

The auxiliary equation for (17) is m² + 8m + 16 = (m + 4)² = 0 so that m₁ = m₂ = −4. Hence the system is critically damped and

x(t) = c₁e^−4t + c₂te^−4t. (18)

Applying the initial conditions x(0) = 0 and x′(0) = −3, we find, in turn, that c₁ = 0 and c₂ = −3. Thus the equation of motion is

x(t) = −3te^−4t. (19)

To graph x(t) we proceed as in Example 3. From x′(t) = −3e^−4t(1 − 4t) we see that x′(t) = 0 when t = . The corresponding extreme displacement is x() = −3()e⁻¹ = −0.276 ft. As shown in FIGURE 3.8.12, we interpret this value to mean that the weight reaches a maximum height of 0.276 foot above the equilibrium position. ≡

EXAMPLE 5 Underdamped Motion

Alternative Form of x(t)

In a manner identical to the procedure used on page 158, we can write any solution

in the alternative form

, (23)

where A = and the phase angle ϕ is determined from the equations

The coefficient Ae^−λt is sometimes called the damped amplitude of vibrations. Because (23) is not a periodic function, the number 2π is called the quasi period and 2π is the quasi frequency. The quasi period is the time interval between two successive maxima of x(t). You should verify, for the equation of motion in Example 5, that A = 23 and ϕ = 4.391. Therefore an equivalent form of (22) is

DE of Driven Motion with Damping

Suppose we now take into consideration an external force f(t) acting on a vibrating mass on a spring. For example, f(t) could represent a driving force causing an oscillatory vertical motion of the support of the spring. See FIGURE 3.8.13. The inclusion of f(t) in the formulation of Newton’s second law gives the differential equation of driven or forced motion:

(24)

Dividing (24) by m gives

(25)

where F(t) = f(t)/m and, as in the preceding section, 2λ = β/m, ω² = k/m. To solve the latter nonhomogeneous equation we can use either the method of undetermined coefficients or variation of parameters.

EXAMPLE 6 Interpretation of an Initial-Value Problem

Transient and Steady-State Terms

When F is a periodic function, such as F(t) = F₀ sin γt or F(t) = F₀ cos γt, the general solution of (25) for λ > 0 is the sum of a nonperiodic function x_c(t) and a periodic function x_p(t). Moreover, x_c(t) dies off as time increases; that is, lim_t→∞ x_c(t) = 0. Thus for a long period of time, the displacements of the mass are closely approximated by the particular solution x_p(t). The complementary function x_c(t) is said to be a transient term or transient solution, and the function x_p(t), the part of the solution that remains after an interval of time, is called a steady-state term or steady-state solution. Note therefore that the effect of the initial conditions on a spring/mass system driven by F is transient. In the particular solution (28), e^−3t( cos t − sin t) is a transient term and x_p(t) = − cos 4t + sin 4t is a steady-state term. The graphs of these two terms and the solution (28) are given in FIGURES 3.8.14(a) and 3.8.14(b), respectively.

EXAMPLE 7 Transient/Steady-State Solutions

DE of Driven Motion Without Damping

With a periodic impressed force and no damping force, there is no transient term in the solution of a problem. Also, we shall see that a periodic impressed force with a frequency near or the same as the frequency of free undamped vibrations can cause a severe problem in any oscillatory mechanical system.

EXAMPLE 8 Undamped Forced Motion

Pure Resonance

Although equation (30) is not defined for γ = ω, it is interesting to observe that its limiting value as γ → ω can be obtained by applying L’Hôpital’s rule. This limiting process is analogous to “tuning in” the frequency of the driving force (γ/2π) to the frequency of free vibrations (ω/2π). Intuitively we expect that over a length of time we should be able to increase the amplitudes of vibration substantially. For γ = ω we define the solution to be

(31)

As suspected, when t → ∞ the displacements become large; in fact, | x(t_n) | → ∞ when t_n = nπ/ω, n = 1, 2, …. The phenomenon we have just described is known as pure resonance. The graph given in FIGURE 3.8.16 shows typical motion in this case.

In conclusion, it should be noted that there is no actual need to use a limiting process on (30) to obtain the solution for γ = ω. Alternatively, equation (31) follows by solving the initial-value problem

+ ω²x = F₀ sin ωt, x(0) = 0, x′(0) = 0

directly by the methods of undetermined coefficients or variation of parameters.

If the displacements of a spring/mass system were actually described by a function such as (31), the system would necessarily fail. Large oscillations of the mass would eventually force the spring beyond its elastic limit. One might argue too that the resonating model presented in Figure 3.8.16 is completely unrealistic because it ignores the retarding effects of ever-present damping forces. Although it is true that pure resonance cannot occur when the smallest amount of damping is taken into consideration, large and equally destructive amplitudes of vibration (although bounded as t → ∞) can occur. See Problem 49 in Exercises 3.8.

LRC-Series Circuits

As mentioned in the introduction to this chapter, many different physical systems can be described by a linear second-order differential equation similar to the differential equation of forced motion with damping:

(32)

If i(t) denotes current in the LRC-series electrical circuit shown in FIGURE 3.8.17, then the voltage drops across the inductor, resistor, and capacitor are as shown in Figure 1.3.5. By Kirchhoff’s second law, the sum of these voltages equals the voltage E(t) impressed on the circuit; that is,

(33)

But the charge q(t) on the capacitor is related to the current i(t) by i = dq/dt, and so (33) becomes the linear second-order differential equation

(34)

The nomenclature used in the analysis of circuits is similar to that used to describe spring/mass systems.

If E(t) = 0, the electrical vibrations of the circuit are said to be free. Since the auxiliary equation for (34) is Lm² + Rm + 1/C = 0, there will be three forms of the solution with R ≠ 0, depending on the value of the discriminant R² − 4L/C. We say that the circuit is

In each of these three cases the general solution of (34) contains the factor e^−Rt/2L, and so q(t) → 0 as t → ∞. In the underdamped case when q(0) = q₀, the charge on the capacitor oscillates as it decays; in other words, the capacitor is charging and discharging as t → ∞. When E(t) = 0 and R = 0, the circuit is said to be undamped and the electrical vibrations do not approach zero as t increases without bound; the response of the circuit is simple harmonic.

EXAMPLE 9 Underdamped Series Circuit

When there is an impressed voltage E(t) on the circuit, the electrical vibrations are said to be forced. In the case when R ≠ 0, the complementary function q_c(t) of (34) is called a transient solution. If E(t) is periodic or a constant, then the particular solution q_p(t) of (34) is a steady-state solution.

EXAMPLE 10 Steady-State Current

The quantities X = Lγ − 1/(Cγ) and Z = defined in Example 10 are called, respectively, the reactance and impedance of the circuit. Both the reactance and the impedance are measured in ohms.

3.8 Exercises Answers to selected odd-numbered problems begin on page ANS-6.

3.8.1 Spring/Mass Systems: Free Undamped Motion

1. A mass weighing 4 pounds is attached to a spring whose spring constant is 16 lb/ft. What is the period of simple harmonic motion?
2. A 20-kilogram mass is attached to a spring. If the frequency of simple harmonic motion is 2/π cycles/s, what is the spring constant k? What is the frequency of simple harmonic motion if the original mass is replaced with an 80-kilogram mass?
3. A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 3 inches above the equilibrium position. Find the equation of motion.
4. Determine the equation of motion if the mass in Problem 3 is initially released from the equilibrium position with an initial downward velocity of 2 ft/s.
5. A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the equilibrium position.
   
   1. Find the position of the mass at the times t = π/12, π/8, π/6, π/4, and 9π/32s.
   2. What is the velocity of the mass when t = 3π/16s? In which direction is the mass heading at this instant?
   3. At what times does the mass pass through the equilibrium position?
6. A force of 400 newtons stretches a spring 2 meters. A mass of 50 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10 m/s. Find the equation of motion.
7. Another spring whose constant is 20 N/m is suspended from the same rigid support but parallel to the spring/mass system in Problem 6. A mass of 20 kilograms is attached to the second spring, and both masses are initially released from the equilibrium position with an upward velocity of 10 m/s.
   
   1. Which mass exhibits the greater amplitude of motion?
   2. Which mass is moving faster at t = π/4s? At π/2s?
   3. At what times are the two masses in the same position? Where are the masses at these times? In which directions are they moving?
8. A mass weighing 32 pounds stretches a spring 2 feet.
   
   1. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 2 ft/s.
   2. How many complete cycles will the mass have made at the end of 4π seconds?
9. A mass weighing 8 pounds is attached to a spring. When set in motion, the spring/mass system exhibits simple harmonic motion.
   
   1. Determine the equation of motion if the spring constant is 1 lb/ft and the mass is initially released from a point 6 inches below the equilibrium position with a downward velocity of
   2. Express the equation of motion in the form of a shifted sine function given in (6).
   3. Express the equation of motion in the form of a shifted cosine function given in
10. A mass weighing 10 pounds stretches a spring foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point foot above the equilibrium position with a downward velocity of
    
    1. Express the equation of motion in the form of a shifted sine function given in (6).
    2. Express the equation of motion in the form of a shifted cosine function given in
    3. Use one of the solutions obtained in parts (a) and (b) to determine the times the mass attains a displacement below the equilibrium position numerically equal to the amplitude of motion.
11. A mass weighing 64 pounds stretches a spring 0.32 foot. The mass is initially released from a point 8 inches above the equilibrium position with a downward velocity of 5 ft/s.
    
    1. Find the equation of motion.
    2. What are the amplitude and period of motion?
    3. How many complete cycles will the mass have completed at the end of 3π seconds?
    4. At what time does the mass pass through the equilibrium position heading downward for the second time?
    5. At what time does the mass attain its extreme displacement on either side of the equilibrium position?
    6. What is the position of the mass at t = 3s?
    7. What is the instantaneous velocity at t = 3s?
    8. What is the acceleration at t = 3s?
    9. What is the instantaneous velocity at the times when the mass passes through the equilibrium position?
    10. At what times is the mass 5 inches below the equilibrium position?
    11. At what times is the mass 5 inches below the equilibrium position heading in the upward direction?
12. A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of ft/s. Find the times for which the mass is heading downward at a velocity of 3 ft/s.
13. A mass weighing 20 pounds stretches one spring 6 inches and another spring 2 inches. The springs are attached to a common rigid support and then to a metal plate of negligible mass as shown in Figure 3.8.5; the mass is attached to the center of the plate in the double-spring arrangement. Determine the effective spring constant of this system. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s.
14. A certain weight stretches one spring foot and another spring foot. The two springs are attached in parallel to a common rigid support in a manner indicated in Problem 13 and Figure 3.8.5. The first weight is set aside, and a mass weighing 8 pounds is attached to the double-spring arrangement and the system is set in motion. If the period of motion is second, determine how much the first mass weighs.
15. Solve Problem 13 again, but this time assume that the springs are in series as shown in Figure 3.8.6.
16. Solve Problem 14 again, but this time assume that the springs are in series as shown in Figure 3.8.6.
17. Find the effective spring constant of the parallel-spring system shown in Figure 3.8.5 when both springs have the spring constant k. Give a physical interpretation of this result.
18. Find the effective spring constant of the series-spring system shown in Figure 3.8.6 when both springs have the spring constant k. Give a physical interpretation of this result.
19. A model of a spring/mass system is 4x″ + e^−0.1tx = 0. By inspection of the differential equation only, discuss the behavior of the system over a long period of time.
20. A model of a spring/mass system is 4x″ + tx = 0. By inspection of the differential equation only, discuss the behavior of the system over a long period of time.

3.8.2 Spring/Mass Systems: Free Damped Motion

In Problems 21–24, the given figure represents the graph of an equation of motion for a damped spring/mass system. Use the graph to determine

1. whether the initial displacement is above or below the equilibrium position, and
2. whether the mass is initially released from rest, heading downward, or heading upward.

21. 

22. 

23. 

24. 

25. A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 8 ft/s. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant?
26. A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers damping force numerically equal to times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 5 ft/s. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant?
27. A 1-kilogram mass is attached to a spring whose constant is 16 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 10 times the instantaneous velocity. Determine the equations of motion if
    
    1. the mass is initially released from rest from a point 1 meter below the equilibrium position, and then
    2. the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of 12 m/s.
28. In parts (a) and (b) of Problem 27, determine whether the mass passes through the equilibrium position. In each case find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant?
29. A force of 2 pounds stretches a spring 1 foot. A mass weighing 3.2 pounds is attached to the spring, and the system is then immersed in a medium that offers a damping force numerically equal to 0.4 times the instantaneous velocity.
    
    1. Find the equation of motion if the mass is initially released from rest from a point 1 foot above the equilibrium position.
    2. Express the equation of motion in the form given in (23).
    3. Find the first time at which the mass passes through the equilibrium position heading upward.
30. After a mass weighing 10 pounds is attached to a 5-foot spring, the spring measures 7 feet. This mass is removed and replaced with another mass that weighs 8 pounds. The entire system is placed in a medium that offers a damping force numerically equal to the instantaneous velocity.
    
    1. Find the equation of motion if the mass is initially released from a point foot below the equilibrium position with a downward velocity of 1 ft/s.
    2. Express the equation of motion in the form given in (23).
    3. Find the times at which the mass passes through the equilibrium position heading downward.
    4. Graph the equation of motion.
31. A mass weighing 10 pounds stretches a spring 2 feet. The mass is attached to a dashpot damping device that offers a damping force numerically equal to β (β > 0) times the instantaneous velocity. Determine the values of the damping constant β so that the subsequent motion is (a) overdamped, (b) critically damped, and (c) underdamped.
32. A mass weighing 24 pounds stretches a spring 4 feet. The subsequent motion takes place in a medium that offers a damping force numerically equal to β (β > 0) times the instantaneous velocity. If the mass is initially released from the equilibrium position with an upward velocity of 2 ft/s, show that when β > 3 the equation of motion is
    

    

3.8.3 Spring/Mass Systems: Driven Motion

33. A mass weighing 16 pounds stretches a spring feet. The mass is initially released from rest from a point 2 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force numerically equal to one-half the instantaneous velocity. Find the equation of motion if the mass is driven by an external force equal to f(t) = 10 cos 3t.
34. A mass of 1 slug is attached to a spring whose constant is 5 lb/ft. Initially the mass is released 1 foot below the equilibrium position with a downward velocity of 5 ft/s, and the subsequent motion takes place in a medium that offers a damping force numerically equal to two times the instantaneous velocity.
    
    1. Find the equation of motion if the mass is driven by an external force equal to f(t) = 12 cos 2t + 3 sin 2t.
    2. Graph the transient and steady-state solutions on the same coordinate axes.
    3. Graph the equation of motion.
35. A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at t = 0, an external force equal to f(t) = 8 sin 4t is applied to the system. Find the equation of motion if the surrounding medium offers a damping force numerically equal to eight times the instantaneous velocity.
36. In Problem 35 determine the equation of motion if the external force is f(t) = e^−t sin 4t. Analyze the displacements for t → ∞.
37. When a mass of 2 kilograms is attached to a spring whose constant is 32 N/m, it comes to rest in the equilibrium position. Starting at t = 0, a force equal to f(t) = 68e^−2t cos 4t is applied to the system. Find the equation of motion in the absence of damping.
38. In Problem 37 write the equation of motion in the form x(t) = A sin(ωt + ϕ) + Be^−2t sin(4t + θ). What is the amplitude of vibrations after a very long time?
39. A mass m is attached to the end of a spring whose constant is k. After the mass reaches equilibrium, its support begins to oscillate vertically about a horizontal line L according to a formula h(t). The value of h represents the distance in feet measured from L. See FIGURE 3.8.23.
    
    1. Determine the differential equation of motion if the entire system moves through a medium offering a damping force numerically equal to β(dx/dt).
    2. Solve the differential equation in part (a) if the spring is stretched 4 feet by a weight of 16 pounds and β = 2, h(t) = 5 cos t,x(0) = x′(0) = 0.
       

       

40. A mass of 100 grams is attached to a spring whose constant is 1600 dynes/cm. After the mass reaches equilibrium, its support oscillates according to the formula h(t) = sin 8t, where h represents displacement from its original position. See Problem 39 and Figure 3.8.23.
    
    1. In the absence of damping, determine the equation of motion if the mass starts from rest from the equilibrium position.
    2. At what times does the mass pass through the equilibrium position?
    3. At what times does the mass attain its extreme displacements?
    4. What are the maximum and minimum displacements?
    5. Graph the equation of motion.

In Problems 41 and 42, describe in words a spring/mass system whose model is the given differential equation. Assume that all constants are positive.

In Problems 43 and 44, solve the given initial-value problem.

43. + 4x = −5 sin 2t + 3 cos 2t, x(0) = −1, x′(0) = 1
44. + 9x = 5 sin 3t, x(0) = 2, x′(0) = 0
45. 1. Show that the solution of the initial-value problem
       

       + ω²x = F₀ cos γt, x(0) = 0, x′(0) = 0

       is

    2. Evaluate (cos γt − cos ωt).
46. Compare the result obtained in part (b) of Problem 45 with the solution obtained using variation of parameters when the external force is F₀ cos ωt.
47. 1. Show that x(t) given in part (a) of Problem 45 can be written in the form
       

       

    2. If we define , show that when ∈ is small, an approximate solution is
       

       

       When ∈ is small the frequency γ/2π of the impressed force is close to the frequency ω/2π of free vibrations. When this occurs, the motion is as indicated in FIGURE 3.8.24. Oscillations of this kind are called beats and are due to the fact that the frequency of sin ∈t is quite small in comparison to the frequency of sin γt. The red dashed curves, or envelope of the graph of x(t), are obtained from the graphs of ±(F₀/2∈γ) sin ∈t. Use a graphing utility with various values of F₀, ∈, and γ to verify the graph in Figure 3.8.24.

       

Computer Lab Assignments

48. Can there be beats when a damping force is added to the model in part (a) of Problem 45? Defend your position with graphs obtained either from the explicit solution of the problem
    

    + ω²x = F₀ cos γt, x(0) = 0, x′(0) = 0

    or from solution curves obtained using a numerical solver.

49. 
    
    1. Show that the general solution of
       

       

       is

       

       

       where A = and the phase angles ϕ and θ are, respectively, defined by sin ϕ = c₁/A, cos ϕ = c₂/A and

       

       

    2. The solution in part (a) has the form x(t) = x_c(t) + x_p(t). Inspection shows that x_c(t) is transient, and hence for large values of time, the solution is approximated by x_p(t) = g(γ) sin(γt + θ), where
       

       

       Although the amplitude g(γ) of x_p(t) is bounded as t → ∞, show that the maximum oscillations will occur at the value γ₁ = What is the maximum value of g? The number is said to be the resonance frequency of the system.

    3. When F₀ = 2, m = 1, and k = 4, g becomes
       

       

       Construct a table of the values of γ₁ and g(γ₁) corresponding to the damping coefficients β = 2, β = 1, β = , β = , and β = . Use a graphing utility to obtain the graphs of g corresponding to these damping coefficients. Use the same coordinate axes. This family of graphs is called the resonance curve or frequency response curve of the system. What is γ₁ approaching as β → 0? What is happening to the resonance curve as β → 0?

50. Consider a driven undamped spring/mass system described by the initial-value problem
    

    + ω²x = F₀ sinⁿ γt, x(0) = 0, x′(0) = 0.

    
    
    1. For n = 2, discuss why there is a single frequency γ₁/2π at which the system is in pure resonance.
    2. For n = 3, discuss why there are two frequencies γ₁/2π and γ₂/2π at which the system is in pure resonance.
    3. Suppose ω = 1 and F₀ = 1. Use a numerical solver to obtain the graph of the solution of the initial-value problem for n = 2 and γ = γ₁ in part (a). Obtain the graph of the solution of the initial-value problem for n = 3 corresponding, in turn, to γ = γ₁ and γ = γ₂ in part (b).

3.8.4 Series Circuit Analogue

51. Find the charge on the capacitor in an LRC-series circuit at t = 0.01s when L = 0.05h, R = 2Ω, C = 0.01F, E(t) = 0V, q(0) = 5C, and i(0) = 0A. Determine the first time at which the charge on the capacitor is equal to zero.
52. Find the charge on the capacitor in an LRC-series circuit when L = h, R = 20Ω, C = F, E(t) = 0V, q(0) = 4C, and i(0) = 0A. Is the charge on the capacitor ever equal to zero?

In Problems 53 and 54, find the charge on the capacitor and the current in the given LRC-series circuit. Find the maximum charge on the capacitor.

53. L = h, R = 10Ω, C = F, E(t) = 300V, q(0) = 0C, i(0) = 0A
54. L = 1h, R = 100Ω, C = 0.0004F, E(t) = 30V, q(0) = 0C, i(0) = 2A
55. Find the steady-state charge and the steady-state current in an LRC-series circuit when L = 1h, R = 2Ω, C = 0.25F, and E(t) = 50 cos t V.
56. Show that the amplitude of the steady-state current in the LRC-series circuit in Example 10 is given by E₀/Z, where Z is the impedance of the circuit.
57. Use Problem 56 to show that the steady-state current in an LRC-series circuit when L = h, R = 20Ω, C = 0.001F, and E(t) = 100 sin 60t V is given by i_p(t) = 4.160 sin(60t − 0.588).
58. Find the steady-state current in an LRC-series circuit when L = h, R = 20Ω, C = 0.001F, and E(t) = 100 sin 60t + 200 cos 40t V.
59. Find the charge on the capacitor in an LRC-series circuit when L = h, R = 10Ω, C = 0.01F, E(t) = 150V, q(0) = 1C, and i(0) = 0A. What is the charge on the capacitor after a long time?
60. Show that if L, R, C, and E₀ are constant, then the amplitude of the steady-state current in Example 10 is a maximum when γ = 1/. What is the maximum amplitude?
61. Show that if L, R, E₀, and γ are constant, then the amplitude of the steady-state current in Example 10 is a maximum when the capacitance is C = 1/Lγ².
62. Find the charge on the capacitor and the current in an LC-circuit when L = 0.1h, C = 0.1F, E(t) = 100 sin γt V, q(0) = 0C, and i(0) = 0A.
63. Find the charge on the capacitor and the current in an LC-circuit when E(t) = E₀ cos γt V, q(0) = q₀ C, and i(0) = i₀ A.
64. In Problem 63, find the current when the circuit is in resonance.