# Spring Problems

A simple spring system consists of a mass $$${m}$$$ attached to the lower end of a spring that is itself suspended vertically from a mounting. The system is in its equilibrium position when it is at rest.

The mass is set in motion by one or more of the following means: displacing the mass from its equilibrium position, providing it with an initial velocity, or subjecting it to an external force $$${F}{\left({t}\right)}$$$.

**Hooke's Law**: The restoring force $$${F}$$$ of a spring is equal and opposite to the forces applied to the spring and is proportional to the extension (contraction) $$${l}$$$ of the spring as a result of the applied force; that is, $$${F}=-{k}{l}$$$, where $$${k}$$$ denotes the constant of proportionality, generally called the spring constant.

**Example 1.** A steel ball weighing 128 lb. is suspended from a spring, whereupon the spring is stretched 2 ft. from its natural length. The applied force responsible for the 2-ft. displacement is the weight of the ball, 128 lb. Thus, $$${F}=-{128}\text{lb}$$$. Hooke's law then gives $$$-{128}=-{k}{\left({2}\right)}$$$, or $$${k}={64}\frac{{{l}{b}}}{{{f{{t}}}}}$$$.

For convenience, we choose the downward direction as the positive direction and take the origin to be the center of gravity of the mass in the equilibrium position. We assume that the mass of the spring is negligible and that air resistance, when present, is proportional to the velocity of the mass. Thus, at any time $$${t}$$$, there are three forces acting upon the system: $$${F}{\left({t}\right)}$$$, measured in the positive direction; a restoring force given by Hooke's law as $$${F}_{{s}}=-{k}{x}$$$, $$${k}>{0}$$$; and a force due to air resistance given by $$${F}_{{a}}=-{a}\frac{{{d}{x}}}{{{d}{t}}}'$$$, $$${a}>{0}$$$, where $$${a}$$$ is the constant of proportionality. Note that the restoring force $$${F}_{{s}}$$$ always acts in the direction that will tend to return the system to the equilibrium position: if the mass is below the equilibrium position, $$${x}$$$ is positive, and $$$-{k}{x}$$$ is negative; whereas if the mass is above the equilibrium position, $$${x}$$$ is negative, and $$$-{k}{x}$$$ is positive. Also, note that because $$${a}>{0}$$$, the force $$${F}_{{a}}$$$ due to air resistance acts in the opposite direction of the velocity and thus tends to retard, or damp, the motion of the mass.

It now follows from Newton's second law that $$${m}\frac{{{{d}}^{{2}}{x}}}{{{d}{{t}}^{{2}}}}=-{k}{x}-{a}\frac{{{d}{x}}}{{{d}{t}}}+{F}{\left({t}\right)}$$$, or $$$\frac{{{{d}}^{{2}}{x}}}{{{d}{{t}}^{{2}}}}+\frac{{a}}{{m}}\frac{{{d}{x}}}{{{d}{t}}}+\frac{{k}}{{m}}{x}=\frac{{{F}{\left({t}\right)}}}{{m}}$$$.

If the system starts at $$${t}={0}$$$ with an initial velocity $$${v}_{{0}}$$$ and from the initial position $$${x}_{{0}}$$$, we also have the initial conditions $$${x}{\left({0}\right)}={x}_{{0}},\ {x}'{\left({0}\right)}={v}_{{0}}$$$.

The force of gravity does not explicitly appear in the differential equation, but it is present nonetheless. We automatically compensated for this force by measuring the distance from the equilibrium position of the spring. If one wishes to exhibit gravity explicitly, then distance has to be measured from the bottom end of the natural length of the spring. That is, the motion of a vibrating spring can be given by $$$\frac{{{{d}}^{{2}}{x}}}{{{d}{{t}}^{{2}}}}+\frac{{a}}{{m}}\frac{{{d}{x}}}{{{d}{t}}}+\frac{{k}}{{m}}{x}={g{+}}\frac{{{F}{\left({t}\right)}}}{{m}}$$$ if the origin, $$${x}={0}$$$, is the terminal point of the unstretched spring before the mass $$${m}$$$ is attached.

**Example 2.** A mass of 2 kg is suspended from a spring with a known spring constant of 10 N/m and is allowed to come to rest. It is then set in motion by giving it an initial velocity of 150 cm/sec. Find an expression for the motion of the mass, assuming no air resistance.

Here, the externally applied force on the mass $$${F}{\left({t}\right)}={0}$$$, and there is no resistance from the surrounding medium, $$${a}={0}$$$. The mass and the spring constants are given as m = 2 kg and k = 10 N/m respectively; so, the differential equation becomes $$$\frac{{{{d}}^{{2}}{x}}}{{{d}{{t}}^{{2}}}}+{5}{x}={0}$$$. The roots of its characteristic equation $$${{r}}^{{2}}+{5}={0}$$$ are purely imaginary; hence, its solution is

$$${x}{\left({t}\right)}={c}_{{1}}{\cos{{\left(\sqrt{{{5}}}{t}\right)}}}+{c}_{{2}}{\sin{{\left(\sqrt{{{5}}}{t}\right)}}}$$$.

At $$${t}={0}$$$, the position of the ball is at the equilibrium position $$${x}_{{0}}={0}$$$ m. Applying this initial condition, we find that $$${x}{\left({0}\right)}={0}={c}_{{1}}{\cos{{\left(\sqrt{{{5}}}\cdot{0}\right)}}}+{c}_{{2}}{\sin{{\left(\sqrt{{{5}}}\cdot{0}\right)}}}$$$, or $$${c}_{{1}}={0}$$$.

So, $$${x}{\left({t}\right)}={c}_{{2}}{\sin{{\left(\sqrt{{{5}}}{t}\right)}}}$$$. Thus, $$${x}'=\sqrt{{{5}}}{c}_{{2}}{\cos{{\left(\sqrt{{{5}}}{t}\right)}}}$$$.

Applying the second initial condition $$${x}'{\left({0}\right)}={150}{c}\frac{{m}}{{s}}={1.5}\frac{{m}}{{s}}$$$ gives that $$${x}'{\left({0}\right)}={1.5}=\sqrt{{{5}}}{c}_{{2}}{\cos{{\left(\sqrt{{{5}}}\cdot{0}\right)}}}$$$, or $$${c}_{{2}}=\frac{{1.5}}{\sqrt{{{5}}}}$$$.

Finally, $$${x}{\left({t}\right)}=\frac{{1.5}}{\sqrt{{{5}}}}{\sin{{\left(\sqrt{{{5}}}{t}\right)}}}$$$.

Let's take a look at one more example.

**Example 3.** A 128-lb. weight is attached to a spring whereupon the spring is stretched 2 ft. and allowed to come to rest. The weight is set in motion from rest by displacing the spring 6 in. (which is equal to 0.5 ft.) above its equilibrium position and also by applying an external force $$${F}{\left({t}\right)}={8}{\sin{{\left({4}{t}\right)}}}$$$. Find the subsequent motion of the weight if the surrounding medium offers a negligible resistance.

Here, $$${W}={128}$$$. We need to find $$${m}$$$ using $$${W}$$$. Since $$${W}={m}{g{}}$$$ and $$${g{=}}{32}\frac{{f{{t}}}}{{{s}}^{{2}}}$$$, we have that $$${m}=\frac{{W}}{{{g}}}=\frac{{128}}{{32}}={4}$$$. From Hooke's law, $$${W}=-{k}{l}$$$, where $$${l}={2}$$$.

So, $$${k}=-\frac{{W}}{{{l}}}=-\frac{{128}}{{2}}={64}\frac{{{l}{b}}}{{{f{{t}}}}}$$$.

There is no air resistance; so, $$${a}={0}$$$.

Thus, the given differential eqaution is $$$\frac{{{{d}}^{{2}}{x}}}{{{d}{{t}}^{{2}}}}+{16}{x}={2}{\sin{{\left({4}{t}\right)}}}$$$.

The solution of the corresponding homogeneous equation is $$${x}_{{h}}={c}_{{1}}{\cos{{\left({4}{t}\right)}}}+{c}_{{2}}{\sin{{\left({4}{t}\right)}}}$$$.

To find the particular solution, use the method of undetermined coefficients. We can't assume that $$${x}_{{p}}={A}{\cos{{\left({4}{t}\right)}}}+{B}{\sin{{\left({4}{t}\right)}}}$$$ because these terms are already in the homogeneous solution; so, assume that $$${x}_{{p}}={A}{t}{\cos{{\left({4}{t}\right)}}}+{B}{t}{\sin{{\left({4}{t}\right)}}}$$$.

Then, $$${x}_{{p}}'={A}{\cos{{\left({4}{t}\right)}}}-{4}{A}{t}{\sin{{\left({4}{t}\right)}}}+{B}{\sin{{\left({4}{t}\right)}}}+{4}{B}{t}{\cos{{\left({4}{t}\right)}}}$$$, and $$${x}_{{p}}''=-{4}{A}{\sin{{\left({4}{t}\right)}}}-{4}{A}{\sin{{\left({4}{t}\right)}}}-{16}{A}{t}{\cos{{\left({4}{t}\right)}}}+{4}{B}{\cos{{\left({4}{t}\right)}}}+{4}{B}{\cos{{\left({4}{t}\right)}}}-{16}{B}{t}{\sin{{\left({4}{t}\right)}}}$$$

Plugging these values into the equation gives:

$$$-{4}{A}{\sin{{\left({4}{t}\right)}}}-{4}{A}{\sin{{\left({4}{t}\right)}}}-{16}{A}{t}{\cos{{\left({4}{t}\right)}}}+{4}{B}{\cos{{\left({4}{t}\right)}}}+{4}{B}{\cos{{\left({4}{t}\right)}}}-{16}{B}{t}{\sin{{\left({4}{t}\right)}}}+{16}{\left({A}{t}{\cos{{\left({4}{t}\right)}}}+{B}{t}{\sin{{\left({4}{t}\right)}}}\right)}=-{8}{A}{\sin{{\left({4}{t}\right)}}}+{8}{B}{\cos{{\left({4}{t}\right)}}}$$$

Equating the like terms with $$${8}{\sin{{\left({4}{t}\right)}}}$$$ gives

$$${\left\{\begin{array}{c}-{8}{A}={2}\\{8}{B}={0}\\ \end{array}\right.}$$$,

Which has the solution $$${A}=-\frac{{1}}{{4}},\ {B}={0}$$$. So, $$${x}_{{p}}=-\frac{{1}}{{4}}{t}{\cos{{\left({4}{t}\right)}}}$$$.

Therefore, $$${x}={x}_{{h}}+{x}_{{p}}={c}_{{1}}{\cos{{\left({4}{t}\right)}}}+{c}_{{2}}{\sin{{\left({4}{t}\right)}}}-\frac{{1}}{{4}}{t}{\cos{{\left({4}{t}\right)}}}$$$.

Now, find $$${x}'{\left({t}\right)}$$$: $$${x}'{\left({t}\right)}=-{4}{c}_{{1}}{\sin{{\left({4}{t}\right)}}}+{4}{c}_{{2}}{\cos{{\left({4}{t}\right)}}}-\frac{{1}}{{4}}{\cos{{\left({4}{t}\right)}}}+{t}{\sin{{\left({4}{t}\right)}}}$$$.

Applying the initial conditions $$${x}{\left({0}\right)}=-{0.5}$$$, $$${x}'{\left({0}\right)}={0}$$$ gives

$$${\left\{\begin{array}{c}{x}{\left({0}\right)}=-{0.5}={c}_{{1}}{\cos{{\left({4}\cdot{0}\right)}}}+{c}_{{2}}{\sin{{\left({4}\cdot{0}\right)}}}-\frac{{1}}{{4}}\cdot{0}\cdot{\cos{{\left({4}\cdot{0}\right)}}}\\{x}'{\left({0}\right)}={0}=-{4}{c}_{{1}}{\sin{{\left({4}\cdot{0}\right)}}}+{4}{c}_{{2}}{\cos{{\left({4}\cdot{0}\right)}}}-\frac{{1}}{{4}}{\cos{{\left({4}\cdot{0}\right)}}}+{0}\cdot{\sin{{\left({4}\cdot{0}\right)}}}\\ \end{array}\right.}$$$,

Or

$$${\left\{\begin{array}{c}{c}_{{1}}=-{0.5}\\{4}{c}_{{2}}-\frac{{1}}{{4}}={0}\\ \end{array}\right.}$$$,

Which has the solution $$${c}_{{1}}=-\frac{{1}}{{2}},\ {c}_{{2}}=\frac{{1}}{{16}}$$$.

Finally, $$${x}{\left({t}\right)}=-\frac{{1}}{{2}}{\cos{{\left({4}{t}\right)}}}+\frac{{1}}{{16}}{\sin{{\left({4}{t}\right)}}}-\frac{{1}}{{4}}{t}{\cos{{\left({4}{t}\right)}}}$$$.

Note that $$${\left|{x}\right|}\ \to\infty$$$ as $$${t}\to\infty$$$. This phenomenon is called pure resonance. It is due to the forcing function $$${F}{\left({t}\right)}$$$ having the same circular frequency as that of the associated free undamped system.