Contents

Hooke’s Law: Stress and Strain Revisited

In this figure a hand holding a ruler tightly at the bottom is shown. The other hand pulls the top of the ruler and then releases it. Then the ruler starts vibrating, and oscillates around the equilibrium position. A vertical line is shown to mark the equilibrium position. A curved double-headed arrow shows the span of the oscillation.

Figure 16.2 When displaced from its vertical equilibrium position, this plastic ruler oscillates back and forth because of the restoring force opposing displacement. When the ruler is on the left, there is a force to the right, and vice versa.

Newton’s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure 16.2. The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until dissipative forces dampen the motion. These forces remove mechanical energy from the system, gradually reducing the motion until the ruler comes to rest.

The simplest oscillations occur when the restoring force is directly proportional to displacement. When stress and strain were covered in Newton’s Third Law of Motion, the name was given to this relationship between force and displacement was Hooke’s law:

F = - kx. F = - kx. size 12{F= -  ital "kx"} {}

16.1

Here, $F F size 12{F} {}$ is the restoring force, $x x size 12{x} {}$ is the displacement from equilibrium or deformation, and $k k size 12{k} {}$ is a constant related to the difficulty in deforming the system. The minus sign indicates the restoring force is in the direction opposite to the displacement.

A series of illustrations of vibrating plastic rulers is shown demonstrating Hooke’s law.

Figure 16.3 (a) The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. (b) The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. (c) The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. (d) Now the ruler has momentum to the left. (e) In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself.

The force constant $k k size 12{k} {}$ is related to the rigidity (or stiffness) of a system—the larger the force constant, the greater the restoring force, and the stiffer the system. The units of $k k size 12{k} {}$ are newtons per meter (N/m). For example, $k k size 12{k} {}$ is directly related to Young’s modulus when we stretch a string. Figure 16.4 shows a graph of the absolute value of the restoring force versus the displacement for a system that can be described by Hooke’s law—a simple spring in this case. The slope of the graph equals the force constant $k k size 12{k} {}$ in newtons per meter. A common physics laboratory exercise is to measure restoring forces created by springs, determine if they follow Hooke’s law, and calculate their force constants if they do.

The given figure a is the graph of restoring force versus displacement. The displacement is given by x in meters along x axis, with scales from zero to point zero five zero, then to point one zero, then forward. The restoring force is given by F in unit newton along y axis, with scales from zero to two point zero to four point zero to forward. The graph line starts from zero and goes to upward to point where x is greater than point one zero and F is greater than four point zero with intersection dots at equal distances on the slope line. The slope is depicted by K which is given by rise along y-axis upon run along x axis . The values of mass in kilogram, weight in newtons, and displacement in meters are given along with the graph in a tabular format. In the figure b a horizontal weight bar is shown with three weight measuring springs tied to its lower part, hanging in the downward vertical direction. The first bar has no mass hanging through it, showing zero displacement, as x is equal to zero. It is the least stretched spring downward. The second spring has mass m one tied to it which exerts a force w one, on the spring, which causes displacement in the spring shown here to be x one. Similarly, the third spring is most stretched downward with a mass m two hanging through it with force w two and displacement x two. The values of mass in kg, weight in newtons and displacement in meters are given with the graph in a tabular format.

Figure 16.4 (a) A graph of absolute value of the restoring force versus displacement is displayed. The fact that the graph is a straight line means that the system obeys Hooke’s law. The slope of the graph is the force constant

k k size 12{k} {}

. (b) The data in the graph were generated by measuring the displacement of a spring from equilibrium while supporting various weights. The restoring force equals the weight supported, if the mass is stationary.

Example 16.1

How Stiff Are Car Springs?

The figure shows the left side of a hatchback car’s back area, showing the font of its rear wheel. There is an arrow on road pointing its head toward this wheel.

Figure 16.5 The mass of a car increases due to the introduction of a passenger. This affects the displacement of the car on its suspension system. (credit: exfordy on Flickr)

What is the force constant for the suspension system of a car that settles 1.20 cm when an 80.0-kg person gets in?

Strategy

Consider the car to be in its equilibrium position $x = 0 x = 0 size 12{x=0} {}$ before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position $x = - 1 . 20 \times 10^{- 2} m x = - 1 . 20 \times 10^{- 2} size 12{x= - 1 "." "20" times "10" rSup { size 8{ - 2} } m} {}$ . At that point, the springs supply a restoring force $F F size 12{F} {}$ equal to the person’s weight $w = mg = (80.0) (9.80) = 784 w = mg = (80.0) (9.80) = 784 size 12{w= ital "mg"= left ("80" "." 0`"kg" right ) left (9 "." "80"`"m/s" rSup { size 8{2} } right )="784"`N} {}$ . We take this force to be $F F size 12{F} {}$ in Hooke’s law. Knowing $F F size 12{F} {}$ and $x x size 12{x} {}$ , we can then solve the force constant $k k size 12{k} {}$ .

Solution

Solve Hooke’s law, $F = - kx F = - kx size 12{F= - ital "kx"} {}$ , for $k k size 12{k} {}$ :
$k = - \frac{F}{x} . k = - \frac{F}{x} . size 12{k= - { {F} over {x} } } {}$
16.2

Substitute known values and solve $k k size 12{k} {}$ :

$\begin{array}{l} k & = & - \frac{784 N}{- 1 . 20 \times 10^{- 2} m} \\ = & 6 . 53 \times 10^{4} N/m . \end{array} \begin{array}{l} k & = & - \frac{784}{- 1 . 20 \times 10^{- 2}} \\ = & 6 . 53 \times 10^{4} \end{array} alignl { stack {
size 12{k= - { {"784"" N"} over { - 1 "." "20" times "10" rSup { size 8{ - 2} } " m"} } } {} #
```=6 "." "53" times "10" rSup { size 8{4} } " N/m" {}
} } {}$
16.3

Discussion

Note that $F F size 12{F} {}$ and $x x size 12{x} {}$ have opposite signs because they are in opposite directions—the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in if it were not for damping (due to frictional forces) provided by shock absorbers. Bouncing cars are a sure sign of bad shock absorbers.

Energy in Hooke’s Law of Deformation

In order to produce a deformation, work must be done. That is, a force must be exerted through a distance, whether you pluck a guitar string or compress a car spring. If the only result is deformation, and no work goes into thermal, sound, or kinetic energy, then all the work is initially stored in the deformed object as some form of potential energy. The potential energy stored in a spring is ${PE}_{el} = \frac{1}{2} {kx}^{2} {PE}_{el} = \frac{1}{2} {kx}^{2} size 12{"PE" rSub { size 8{"el"} } = { {1} over {2} } ital "kx" rSup { size 8{2} } } {}$ . Here, we generalize the idea to elastic potential energy for a deformation of any system that can be described by Hooke’s law. Hence,

{PE}_{el} = \frac{1}{2} {kx}^{2}, {PE}_{el} = \frac{1}{2} {kx}^{2}, size 12{"PE" size 8{"el"}= {  {1}  over  {2} }  ital "kx" rSup { size 8{2} } } {}

16.4

where ${PE}_{el} {PE}_{el} size 12{"PE" rSub { size 8{"el"} } } {}$ is the elastic potential energy stored in any deformed system that obeys Hooke’s law and has a displacement $x x size 12{x} {}$ from equilibrium and a force constant $k k size 12{k} {}$ .

It is possible to find the work done in deforming a system in order to find the energy stored. This work is performed by an applied force $F_{app} F_{app} size 12{F rSub { size 8{"app"} } } {}$ . The applied force is exactly opposite to the restoring force (action-reaction), and so $F_{app} = kx F_{app} = kx size 12{F rSub { size 8{ ital "app"} } = ital "kx"} {}$ . Figure 16.6 shows a graph of the applied force versus deformation $x x size 12{x} {}$ for a system that can be described by Hooke’s law. Work done on the system is force multiplied by distance, which equals the area under the curve or $(1 / 2) {kx}^{2} (1 / 2) {kx}^{2} size 12{ $ 1/2 $ ital "kx" rSup { size 8{2} } } {}$ (Method A in the figure). Another way to determine the work is to note that the force increases linearly from 0 to $kx kx size 12{ ital "kx"} {}$ , so that the average force is $(1 / 2) kx (1 / 2) kx size 12{ $ 1/2 $ ital "kx"} {}$ , the distance moved is $x x size 12{x} {}$ , and thus $W = F_{app} d = [(1 / 2) kx] (x) = (1 / 2) {kx}^{2} W = F_{app} d = [(1 / 2) kx] (x) = (1 / 2) {kx}^{2} size 12{W=F rSub { size 8{ ital "app"} } "." d= \[ $ 1/2 $ ital "kx" \] $ x $ = $ 1/2 $ ital "kx" rSup { size 8{2} } } {}$ (Method B in the figure).

The graph here represents applied force, given along y-axis, versus deformation or displacement, given along x axis. The slope is linear slanting and the slope area is covered between x axis and the slope, given by F is equal to k multiplied by x, where k is constant and x is displacement. The force applied along y-axis is given by half of k multiplied by x. Along with the graph, two methods are provided to calculate weight, W. The first method gives the solution by multiplying half of b multiplied by h, whereas in the second we can get the solution by multiplying f with x.

Figure 16.6 A graph of applied force versus distance for the deformation of a system that can be described by Hooke’s law is displayed. The work done on the system equals the area under the graph or the area of the triangle, which is half its base multiplied by its height, or

W = (1 / 2) {kx}^{2} W = (1 / 2) {kx}^{2} size 12{W= \( 1/2 \)  ital "kx" rSup { size 8{2} } } {}

Example 16.2

Calculating Stored Energy: A Tranquilizer Gun Spring

We can use a toy gun’s spring mechanism to ask and answer two simple questions: (a) How much energy is stored in the spring of a tranquilizer gun that has a force constant of 50.0 N/m and is compressed 0.150 m? (b) If you neglect friction and the mass of the spring, at what speed will a 2.00-g projectile be ejected from the gun?

The figure a shows an artistic impression of a tranquilizer gun, which shows the inside of it revealing the gun spring and a panel just below it, in the outside area, attached to the spring. This stage shows the gun before it is cocked, and the spring is uncompressed covering the entire inside area. The figure b shows the gun with the spring in the compressed mode. The spring has been compressed to a distance x, where x distance shows the vacant area inside the gun through which the spring has been compressed. The panel is also moving along the spring. And a bullet of mass m is shown at the front of the compressed spring. The spring here has elastic potential energy, represented by P E sub e l. The figure c is the third stage of the above two stages of the gun. The spring here is released from the compressed stage releasing the bullet in the outer forward direction with velocity V and the spring’s potential energy is converted into kinetic energy, represented here by K E.

Figure 16.7 (a) In this image of the gun, the spring is uncompressed before being cocked. (b) The spring has been compressed a distance

x x size 12{x} {}

, and the projectile is in place. (c) When released, the spring converts elastic potential energy

{PE}_{el} {PE}_{el} size 12{"PE" rSub { size 8{"el"} } } {}

into kinetic energy.

Strategy for a

(a): The energy stored in the spring can be found directly from elastic potential energy equation, because $k k size 12{k} {}$ and $x x size 12{x} {}$ are given.

Solution for a

Entering the given values for $k k size 12{k} {}$ and $x x size 12{x} {}$ yields

\begin{array}{l} {PE}_{el} & = & \frac{1}{2} {kx}^{2} = \frac{1}{2} (50.0 N/m) {(0.150 m)}^{2} = 0 . 563 N \cdot m \\ = & 0.563 J \end{array} \begin{array}{l} {PE}_{el} & = & \frac{1}{2} {kx}^{2} = \frac{1}{2} (50.0) {(0.150)}^{2} = 0 . 563 \\ = & 0.563 \end{array} alignl { stack {
 size 12{"PE" size 8{"el"}= {  {1}  over  {2} }  ital "kx" rSup { size 8{2} } = {  {1}  over  {2} }  left ("50" "." 0" N/m" right ) left (0 "." "150"" m" right ) rSup { size 8{2} } =0 "." "563"N cdot M}  {} #
=0 "." "563"J {}
} } {}

16.5

Strategy for b

Because there is no friction, the potential energy is converted entirely into kinetic energy. The expression for kinetic energy can be solved for the projectile’s speed.

Solution for b

Identify known quantities:
${KE}_{f} = {PE}_{el} or 1 / 2 {mv}^{2} = (1 / 2) {kx}^{2} = {PE}_{el} = 0 . 563 J {KE}_{f} = {PE}_{el} KE_f = PE_el 1 / 2 {mv}^{2} = (1 / 2) {kx}^{2} = {PE}_{el} = 0 . 563 size 12{1/2 ital "mv" rSup { size 8{2} } = $ 1/2 $ ital "kx" rSup { size 8{2} } = ital "PE" rSub { size 8{e1} } =0 "." "563"J} {}$
16.6
Solve for $v v size 12{v} {}$ :
$v = {[\frac{2 {PE}_{el}}{m}]}^{1 / 2} = {[\frac{2 (0.563 J)}{0.002 kg}]}^{1 / 2} = 23 . 7 {(J/kg)}^{1 / 2} v = {[\frac{2 {PE}_{el}}{m}]}^{1 / 2} = {[\frac{2 (0.563)}{0.002}]}^{1 / 2} = 23 . 7 {(J/kg)}^{1 / 2} size 12{v= left [ { {2"PE" size 8{"el"}} over {m} } right ] rSup { size 8{1/2} } = left [ { {2 left (0 "." "563"" J" right )} over {0 "." "002"" kg"} } right ] rSup { size 8{1/2} } ="23" "." 7`` left ("J/kg" right ) rSup { size 8{ { {1} over {2} } } } } {}$
16.7
Convert units: $23.7 m / s 23.7 23.7 m/s$

Discussion

(a) and (b): This projectile speed is impressive for a tranquilizer gun (more than 80 km/h). The numbers in this problem seem reasonable. The force needed to compress the spring is small enough for an adult to manage, and the energy imparted to the dart is small enough to limit the damage it might do. Yet, the speed of the dart is great enough for it to travel an acceptable distance.

Check Your Understanding

Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations of the ruler. In what way could you modify this simple experiment to increase the rigidity of the system?

Show solution

Solution

You could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.

Check Your Understanding

If you apply a deforming force on an object and let it come to equilibrium, what happened to the work you did on the system?

Show solution

Solution

It was stored in the object as potential energy.