Let $x(t)$ denote the displacement of the mass from equilibrium. Note that for spring-mass systems of this type, it is customary to adopt the convention that down is positive. Thus, a positive displacement indicates the mass is below the equilibrium point, whereas a negative displacement indicates the mass is above equilibrium. Displacement is usually given in feet in the English system or meters in the metric system.

Consider the forces acting on the mass. The force of gravity is given by $mg$. In the English system, mass is in slugs and the acceleration resulting from gravity is in feet per second squared. The acceleration resulting from gravity is constant, so in the English system, $g=32$ ft/sec². Recall that 1 slug-foot/sec² is a pound, so the expression mg can be expressed in pounds. Metric system units are kilograms for mass and m/sec² for gravitational acceleration. In the metric system, we have $g=9.8$ m/sec².

According to Hooke’s law, the restoring force of the spring is proportional to the displacement and acts in the opposite direction from the displacement, so the restoring force is given by $-k(s+x)$. The spring constant is given in pounds per foot in the English system and in newtons per meter in the metric system.

Now, by Newton’s second law, the sum of the forces on the system (gravity plus the restoring force) is equal to mass times acceleration, so we have

$\large{\begin{aligned} mx^{\prime\prime}&=-k(s+x)+mg \\ &=-ks-kx+mg \end{aligned}}$.

However, by the way we have defined our equilibrium position, $mg=ks$, the differential equation becomes

$\large{mx^{\prime\prime}+kx=0}$.

It is convenient to rearrange this equation and introduce a new variable, called the angular frequency, $\omega$. Letting $\omega=\sqrt{k/m}$, we can write the equation as

$\large{x^{\prime\prime}+\omega^2x=0}$.

This differential equation has the general solution

$\large{x(t)=c_1\cos\omega t+c_2\sin\omega t}$,

which gives the position of the mass at any point in time. The motion of the mass is called simple harmonic motion. The period of this motion (the time it takes to complete one oscillation) is $T=\frac{2\pi}{\omega}$ and the frequency is $f=\frac{1}{T}=\frac{\omega}{2\pi}$ (Figure 2).

Writing the general solution in the form $x(t)=c_1\cos(\omega t)+c_2\sin(\omega t)$ has some advantages. It is easy to see the link between the differential equation and the solution, and the period and frequency of motion are evident. This form of the function tells us very little about the amplitude of the motion, however. In some situations, we may prefer to write the solution in the form

$\large{x(t)=A\sin(\omega t+\phi)}$.

Although the link to the differential equation is not as explicit in this case, the period and frequency of motion are still evident. Furthermore, the amplitude of the motion, $A$, is obvious in this form of the function. The constant $\phi$ is called a phase shift and has the effect of shifting the graph of the function to the left or right.

To convert the solution to this form, we want to find the values of $A$ and $\phi$ such that

$\large{c_1\cos(\omega t)+c_2\sin(\omega t)=A\sin(\omega t+\phi)}$.

We first apply the trigonometric identity

$\large{\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta}$

to get

$\begin{aligned} c_1\cos(\omega t)+c_2\sin(\omega t)&=A(\sin(\omega t)\cos\phi+\cos(\omega t)\sin\phi) \\ &=A\sin\phi(\cos(\omega t))+A\cos\phi(\sin(\omega t)) \end{aligned}$.

Thus,

$\large{c_1=A\sin\phi\text{ and }c_2=A\cos\phi}$.

If we square both of these equations and add them together, we get

$\large{\begin{aligned} c_1^2+c_2^2&=A^2\sin^2\phi+A^2\cos^2\phi \\ &=A^2(\sin^2\phi+\cos^2\phi) \\ &=A^2 \end{aligned}}$.

Thus,

$\large{A=\sqrt{c_1^2+c_2^2}}$.

Now, to find $\phi$, go back to the equations for $c_1$ and $c_2$, but this time, divide the first equation by the second equation to get

$\large{\begin{aligned} \frac{c_1}{c_2}&=\frac{A\sin\phi}{A\cos\phi} \\ &=\tan\phi \end{aligned}}$.

Then,

$\large{\tan\phi=\frac{c_1}{c_2}}$.

We summarize this finding in the following theorem.

theorem: solution to the equation for simple harmonic motion

---

The function $x(t)=c_1\cos(\omega t)+c_2\sin(\omega t)$ can be written in the form $x(t)=A\sin(\omega t+\phi)$, where $A=\sqrt{c_1^2+c_2^2}$ and $\tan\phi=\frac{c_1}{c_2}$.

Note that when using the formula $\tan\phi=\frac{c_1}{c_2}$ to find $\phi$, we must take care to ensure $\phi$ is in the right quadrant (Figure 3).

Figure 3. A graph of vertical displacement versus time for simple harmonic motion with a phase change.

 

Example: expressing the solution with a phase shift

Express the following functions in the form $A(\sin\omega t+\phi)$. What is the frequency of motion? The amplitude?

1. $x(t)=2\cos(3t)+\sin(3t)$
2. $x(t)=3\cos(2t)-2\sin(2t)$

Show Solution

1. We have
   

   $A=\sqrt{c_1^2+c_2^2}=\sqrt{2^2+1^2}=\sqrt5$

   and

   $\tan\phi=\frac{c_1}{c_2}=\frac21=2$.

   
   Note that both $c_1$ and $c_2$ are positive, so $\phi$ is in the first quadrant. Thus,
   

   $\phi\approx1.107$ rad,

   
   so we have
   

   $x(t)=2\cos(3t)+\sin(3t)=\sqrt5\sin(3t+1.107)$.

   
   The frequency is $\frac{\omega}{2\pi}=\frac3{2\pi}\approx0.477$. The amplitude is $\sqrt5$.
   We have

   $A=\sqrt{c_1^2+c_2^2}=\sqrt{3^2+2^2}=\sqrt{13}$

   and
   

   $\tan\phi=\frac{c_1}{c_2}=\frac{3}{-2}=-\frac32$.

   
   Note that $c_1$ is positive but $c_2$ is negative, so $\phi$ is in the fourth quadrant. Thus,
   

   $\phi\approx-0.983$ rad,

   
   so we have
   

   $\begin{aligned} x(t)&=3\cos(2t)-2\sin(2t) \\ &=\sqrt{13}\sin(2t-0.983) \end{aligned}$.

   
   The frequency is $\frac{\omega}{2\pi}=\frac2{2\pi}\approx0.318$. The amplitude is $\sqrt{13}$.

try it

Express the function $x(t)=\cos(4t)+4\sin(4t)$ in the form $A\sin(\omega t+\phi)$. What is the frequency of motion? The amplitude?

Show Solution

$x(t)=\sqrt{17}\sin(4t+0.245)$, $\text{frequency}=\frac4{2\pi}\approx0.637, \ A=\sqrt{17}$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 7.16” here (opens in new window).

Damped Vibrations

With the model just described, the motion of the mass continues indefinitely. Clearly, this doesn’t happen in the real world. In the real world, there is almost always some friction in the system, which causes the oscillations to die off slowly—an effect called damping. So now let’s look at how to incorporate that damping force into our differential equation.

Physical spring-mass systems almost always have some damping as a result of friction, air resistance, or a physical damper, called a dashpot (a pneumatic cylinder; see Figure 4).

Figure 4. A dashpot is a pneumatic cylinder that dampens the motion of an oscillating system.

Because damping is primarily a friction force, we assume it is proportional to the velocity of the mass and acts in the opposite direction. So the damping force is given by $-bx\prime$ for some constant $b>0$. Again applying Newton’s second law, the differential equation becomes

$\large{mx^{\prime\prime}+bx^\prime+kx=0}$.

Then the associated characteristic equation is

$\large{m\lambda^2+b\lambda+k=0}$.

Applying the quadratic formula, we have

$\large{\lambda=\frac{-b\pm\sqrt{b^2-4mk}}{2m}}$.

Just as in Second-Order Linear Equations we consider three cases, based on whether the characteristic equation has distinct real roots, a repeated real root, or complex conjugate roots.

Case 1: $b^{2}>4mk$

In this case, we say the system is overdamped. The general solution has the form

$\large{x(t)=c_1e^{\lambda_1t}+c_2e^{\lambda_2t}}$,

where both $\lambda_1$ and $\lambda_2$ are less than zero. Because the exponents are negative, the displacement decays to zero over time, usually quite quickly. Overdamped systems do not oscillate (no more than one change of direction), but simply move back toward the equilibrium position. Figure 5 shows what typical critically damped behavior looks like.

Figure 5. Behavior of an overdamped spring-mass system, with no change in direction (a) and only one change in direction (b).

Example: overdamped spring-mass system

A $16$-lb mass is attached to a $10$-ft spring. When the mass comes to rest in the equilibrium position, the spring measures $15$ ft $4$ in. The system is immersed in a medium that imparts a damping force equal to $\frac{5}{2}$ times the instantaneous velocity of the mass. Find the equation of motion if the mass is pushed upward from the equilibrium position with an initial upward velocity of $5$ ft/sec. What is the position of the mass after $10$ sec? Its velocity?

Show Solution

The mass stretches the spring $5$ ft $4$ in., or $\frac{16}{3}$ ft. Thus, $16=\left(\frac{16}3\right)k$, so $k=3$. We also have $m=\frac{16}{32}=\frac12$, so the differential equation is

$\large{\frac12x^{\prime\prime}+\frac52x^\prime+3x=0}$.

Multiplying through by $2$ gives $x^{\prime\prime}+5x^\prime+6x=0$, which has the general solution

$\large{x(t)=c_1e^{-2t}+c_2e^{-3t}}$.

Applying the initial conditions, $x(0)=0$ and $x^\prime(0)=-5$, we get

$\large{x(t)=-5e^{-2t}+5e^{-3t}}$.

After $10$ sec the mass is at position

$\large{x(10)=-5e^{-20}+5e^{-30}\approx-1.0305\times10^{-8}\approx0}$,

so it is, effectively, at the equilibrium position. We have $x^\prime(t)=10e^{-2t}-15e^{-3t}$, so after $10$ sec the mass is moving at a velocity of

$\large{x^\prime(10)=10e^{-20}-15e^{-30}\approx2.061\times10^{-8}\approx0}$.

After only $10$ sec, the mass is barely moving.

try it

A$2$-kg mass is attached to a spring with spring constant $24$ N/m. The system is then immersed in a medium imparting a damping force equal to $16$ times the instantaneous velocity of the mass. Find the equation of motion if it is released from rest at a point $40$ cm below equilibrium.

Show Solution

$x(t)=0.6e^{-2t}-0.2e^{-6t}$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 7.17” here (opens in new window).

Case 2: $b^{2}=4mk$

In this case, we say the system is critically damped. The general solution has the form

$\large{x(t)=c_1e^{\lambda_1t}+c_2e^{\lambda_2t}}$,

where $\lambda_1$ is less than zero. The motion of a critically damped system is very similar to that of an overdamped system. It does not oscillate. However, with a critically damped system, if the damping is reduced even a little, oscillatory behavior results. From a practical perspective, physical systems are almost always either overdamped or underdamped (case 3, which we consider next). It is impossible to fine-tune the characteristics of a physical system so that $b^{2}$ and $4mk$ are exactly equal. Figure 6 shows what typical critically damped behavior looks like.

Figure 6. Behavior of a critically damped spring-mass system. The system graphed in part (a) has more damping than the system graphed in part (b).

Example: critically damped spring-mass system

A $1$-kg mass stretches a spring $20$ cm. The system is attached to a dashpot that imparts a damping force equal to $14$ times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from equilibrium with an upward velocity of $3$ m/sec.

Show Solution

We have $mg=1(9.8)=0.2k$, so $k=49$. Then, the differential equation is

$\large{x^{\prime\prime}+14x^\prime+49x=0}$,

which has general solution

$\large{x(t)=c_1e^{-7t}+c_2te^{-7t}}$.

Applying the initial conditions $x(0)=0$ and $x^\prime(0)=-3$ gives

$\large{x(t)=-3te^{-7t}}$.

try it

A $1$-lb weight stretches a spring $6$ in., and the system is attached to a dashpot that imparts a damping force equal to half the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point $6$ in. below equilibrium.

Show Solution

$x(t)=\frac12e^{-8t}+4te^{-8t}$

Case 3: $b^{2}<4mk$

In this case, we say the system is underdamped. The general solution has the form

$\large{x(t)=e^{\alpha t}(c_1\cos(\beta t)+\sin(\beta t))}$,

where $\alpha$ is less than zero. Underdamped systems do oscillate because of the sine and cosine terms in the solution. However, the exponential term dominates eventually, so the amplitude of the oscillations decreases over time. Figure 7 shows what typical underdamped behavior looks like.

Figure 7. Behavior of an underdamped spring-mass system.

Note that for all damped systems, $\displaystyle\lim_{t\to\infty}x(t)=0$. The system always approaches the equilibrium position over time.

Example: underdamped spring-mass system

A $16$-lb weight stretches a spring $3.2$ ft. Assume the damping force on the system is equal to the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point $9$ in. below equilibrium.

Show Solution

We have $k=\frac{16}{3.2}=5$ and $m=\frac{16}{32}=\frac12$, so the differential equation is

$\frac12x^{\prime\prime}+x^\prime+5x=0$, or $x^{\prime\prime}+2x^\prime+10x=0$.

This equation has the general solution

$x(t)=e^{-t}(c_1\cos(3t)+c_2\sin(3t))$.

Applying the initial conditions, $x(0)=\frac{3}{4}$ and $x^\prime(0)=0$, we get

$x(t)=e^{-t}\left(\frac34\cos(3t)+\frac14\sin(3t)\right)$.

try it

A $1$-kg mass stretches a spring $49$ cm. The system is immersed in a medium that imparts a damping force equal to four times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point $24$ cm above equilibrium.

Show Solution

$x(t)=-0.24e^{-2t}\cos(4t)-0.12e^{-2t}\sin(4t)$

Example: chapter opener: modeling a motorcycle suspension system

Figure 8. (credit: modification of work by nSeika, Flickr)

For motocross riders, the suspension systems on their motorcycles are very important. The off-road courses on which they ride often include jumps, and losing control of the motorcycle when they land could cost them the race.

This suspension system can be modeled as a damped spring-mass system. We define our frame of reference with respect to the frame of the motorcycle. Assume the end of the shock absorber attached to the motorcycle frame is fixed. Then, the “mass” in our spring-mass system is the motorcycle wheel. We measure the position of the wheel with respect to the motorcycle frame. This may seem counterintuitive, since, in many cases, it is actually the motorcycle frame that moves, but this frame of reference preserves the development of the differential equation that was done earlier. As with earlier development, we define the downward direction to be positive.

When the motorcycle is lifted by its frame, the wheel hangs freely and the spring is uncompressed. This is the spring’s natural position. When the motorcycle is placed on the ground and the rider mounts the motorcycle, the spring compresses and the system is in the equilibrium position (Figure 9).

Figure 9. We can use a spring-mass system to model a motorcycle suspension.

This system can be modeled using the same differential equation we used before:

$mx^{\prime\prime}+bx^\prime+kx=0$.

A motocross motorcycle weighs $204$ lb, and we assume a rider weight of $180$ lb. When the rider mounts the motorcycle, the suspension compresses $4$ in., then comes to rest at equilibrium. The suspension system provides damping equal to $240$ times the instantaneous vertical velocity of the motorcycle (and rider).

1. Set up the differential equation that models the behavior of the motorcycle suspension system.
2. We are interested in what happens when the motorcycle lands after taking a jump. Let time $t=0$ denote the time when the motorcycle first contacts the ground. If the motorcycle hits the ground with a velocity of $10$ ft/sec downward, find the equation of motion of the motorcycle after the jump.
3. Graph the equation of motion over the first second after the motorcycle hits the ground.

Show Solution

 

$\begin{aligned} mg&=ks \\ 384&=k\left(\frac13\right) \\ k&=1,152 \end{aligned}$.

We also have

$\begin{aligned} W&=mg \\ 384&=m(32) \\ m&=12 \end{aligned}$.

Therefore, the differential equation that models the behavior of the motorcycle suspension is

$12x^{\prime\prime}+240x^\prime+1,152x=0$.

Dividing through by$12$, we get

$x^{\prime\prime}+20x^\prime+96x=0$.

1. The differential equation found in part a. has the general solution
   

   $x(t)=c_1e^{-8t}+c_2e^{-12t}$.

   
   Now, to determine our initial conditions, we consider the position and velocity of the motorcycle wheel when the wheel first contacts the ground. Since the motorcycle was in the air prior to contacting the ground, the wheel was hanging freely and the spring was uncompressed. Therefore the wheel is 4 in. ($\frac{1}{3}$ ft) below the equilibrium position (with respect to the motorcycle frame), and we have $x(0)=\frac{1}{3}$. According to the problem statement, the motorcycle has a velocity of 10 ft/sec downward when the motorcycle contacts the ground, so $x^\prime(0)=10$. Applying these initial conditions, we get $c_1=\frac{7}{2}$ and $c_2=-(\frac{19}{6})$, so the equation of motion is
   $x(t)=\frac72e^{-8t}-\frac{19}6e^{-12t}$. 
   

   • The graph is shown in Figure 10.

    

   

   Figure 10. Graph of the equation of motion over a time of one second.

Activity: landing vehicle

NASA is planning a mission to Mars. To save money, engineers have decided to adapt one of the moon landing vehicles for the new mission. However, they are concerned about how the different gravitational forces will affect the suspension system that cushions the craft when it touches down. The acceleration resulting from gravity on the moon is $1.6$ m/sec², whereas on Mars it is $3.7$ m/sec².

The suspension system on the craft can be modeled as a damped spring-mass system. In this case, the spring is below the moon lander, so the spring is slightly compressed at equilibrium, as shown in Figure 11.

Figure 11. The landing craft suspension can be represented as a damped spring-mass system. (credit “lander”: NASA)

We retain the convention that down is positive. Despite the new orientation, an examination of the forces affecting the lander shows that the same differential equation can be used to model the position of the landing craft relative to equilibrium:

$mx^{\prime\prime}+bx^\prime+kx=0$,

where $m$ is the mass of the lander, $b$ is the damping coefficient, and $k$ is the spring constant.

1. The lander has a mass of $15,000$ kg and the spring is $2$ m long when uncompressed. The lander is designed to compress the spring $0.5$ m to reach the equilibrium position under lunar gravity. The dashpot imparts a damping force equal to $48,000$ times the instantaneous velocity of the lander. Set up the differential equation that models the motion of the lander when the craft lands on the moon.
2. Let time $t=0$ denote the instant the lander touches down. The rate of descent of the lander can be controlled by the crew, so that it is descending at a rate of $2$ m/sec when it touches down. Find the equation of motion of the lander on the moon.
3. If the lander is traveling too fast when it touches down, it could fully compress the spring and “bottom out.” Bottoming out could damage the landing craft and must be avoided at all costs. Graph the equation of motion found in part 2. If the spring is $0.5$ m long when fully compressed, will the lander be in danger of bottoming out?
4. Assuming NASA engineers make no adjustments to the spring or the damper, how far does the lander compress the spring to reach the equilibrium position under Martian gravity?
5. If the lander crew uses the same procedures on Mars as on the moon, and keeps the rate of descent to $2$ m/sec, will the lander bottom out when it lands on Mars?
6. What adjustments, if any, should the NASA engineers make to use the lander safely on Mars?

Forced Vibrations

The last case we consider is when an external force acts on the system. In the case of the motorcycle suspension system, for example, the bumps in the road act as an external force acting on the system. Another example is a spring hanging from a support; if the support is set in motion, that motion would be considered an external force on the system. We model these forced systems with the nonhomogeneous differential equation

$mx^{\prime\prime}+bx^\prime+kx=f(t)$,

where the external force is represented by the $f(t)$ term. As we saw in Nonhomogenous Linear Equations, differential equations such as this have solutions of the form

$x(t)=c_1x_1(t)+c_2x_2(t)+x_p(t)$,

where $c_1x_1(t)+c_2x_2(t)$ is the general solution to the complementary equation and $x_p(t)$ is a particular solution to the nonhomogeneous equation. If the system is damped, $\displaystyle\lim_{t\to\infty}c_1x_1(t)+c_2x_2(t)=0$. Since these terms do not affect the long-term behavior of the system, we call this part of the solution the transient solution. The long-term behavior of the system is determined by $x_p(t)$ so we call this part of the solution the steady-state solution.