What 2 forces are necessary in system to make the claim that elastic potential energy exists?

Learning Objectives

Past the end of this section, you will be able to:

• Depict the energy conservation of the system of a mass and a spring
• Explain the concepts of stable and unstable equilibrium points

To produce a deformation in an object, we must do work. That is, whether you pluck a guitar string or shrink a car's daze cushion, a forcefulness must be exerted through a distance. If the only result is deformation, and no work goes into thermal, sound, or kinetic free energy, then all the work is initially stored in the deformed object as some grade of potential energy.

Consider the case of a cake attached to a spring on a frictionless table, oscillating in SHM. The force of the spring is a conservative strength (which you studied in the chapter on potential energy and conservation of energy), and nosotros tin define a potential energy for it. This potential energy is the energy stored in the spring when the jump is extended or compressed. In this case, the block oscillates in 1 dimension with the force of the leap interim parallel to the motion:

[latex] Due west=\underset{{10}_{i}}{\overset{{10}_{f}}{\int }}{F}_{x}dx=\underset{{x}_{i}}{\overset{{10}_{f}}{\int }}\text{−}kxdx={[-\frac{1}{2}k{x}^{two}]}_{{10}_{i}}^{{10}_{f}}=\text{−}[\frac{i}{2}chiliad{x}_{f}^{two}-\frac{1}{2}k{x}_{i}^{2}]=\text{−}[{U}_{f}-{U}_{i}]=\text{−}\text{Δ}U. [/latex]

When considering the energy stored in a spring, the equilibrium position, marked as [latex] {10}_{i}=0.00\,\text{m,} [/latex] is the position at which the energy stored in the leap is equal to zero. When the leap is stretched or compressed a altitude x, the potential energy stored in the leap is

[latex] U=\frac{ane}{ii}k{x}^{2}. [/latex]

Energy and the Simple Harmonic Oscillator

To study the energy of a simple harmonic oscillator, we need to consider all the forms of energy. Consider the example of a block attached to a leap, placed on a frictionless surface, aquiver in SHM. The potential free energy stored in the deformation of the spring is

[latex] U=\frac{1}{two}k{ten}^{2}. [/latex]

In a simple harmonic oscillator , the energy oscillates between kinetic energy of the mass [latex] K=\frac{one}{ii}thou{v}^{2} [/latex] and potential free energy [latex] U=\frac{1}{ii}k{10}^{2} [/latex] stored in the spring. In the SHM of the mass and spring system, there are no dissipative forces, so the full energy is the sum of the potential free energy and kinetic energy. In this section, we consider the conservation of energy of the system. The concepts examined are valid for all uncomplicated harmonic oscillators, including those where the gravitational strength plays a function.

Consider (Effigy), which shows an oscillating cake attached to a spring. In the example of undamped SHM, the free energy oscillates back and forth between kinetic and potential, going completely from i form of energy to the other as the system oscillates. And then for the unproblematic instance of an object on a frictionless surface attached to a leap, the motion starts with all of the energy stored in the bound as rubberband potential energy. As the object starts to move, the elastic potential free energy is converted into kinetic energy, becoming entirely kinetic energy at the equilibrium position. The energy is then converted dorsum into elastic potential energy past the spring as it is stretched or compressed. The velocity becomes zilch when the kinetic energy is completely converted, and this cycle and so repeats. Understanding the conservation of free energy in these cycles will provide actress insight here and in later applications of SHM, such as alternating circuits.

Effigy 15.10 The transformation of energy in SHM for an object attached to a spring on a frictionless surface. (a) When the mass is at the position [latex] ten=+A [/latex], all the energy is stored as potential energy in the spring [latex] U=\frac{1}{two}g{A}^{2} [/latex]. The kinetic energy is equal to zilch considering the velocity of the mass is zero. (b) As the mass moves toward [latex] ten=\text{−}A [/latex], the mass crosses the position [latex] x=0 [/latex]. At this point, the spring is neither extended nor compressed, and so the potential free energy stored in the leap is nada. At [latex] ten=0 [/latex], the total free energy is all kinetic free energy where [latex] K=\frac{one}{2}g{(\text{−}{v}_{\text{max}})}^{2} [/latex]. (c) The mass continues to move until it reaches [latex] x=\text{−}A [/latex] where the mass stops and starts moving toward [latex] ten=+A [/latex]. At the position [latex] x=\text{−}A [/latex], the total energy is stored as potential energy in the compressed [latex] U=\frac{one}{two}m{(\text{−}A)}^{2} [/latex] and the kinetic energy is zero. (d) As the mass passes through the position [latex] x=0 [/latex], the kinetic energy is [latex] K=\frac{one}{2}m{5}_{\text{max}}^{2} [/latex] and the potential energy stored in the jump is zero. (due east) The mass returns to the position [latex] x=+A [/latex], where [latex] K=0 [/latex] and [latex] U=\frac{1}{2}k{A}^{ii} [/latex].

Consider (Figure), which shows the energy at specific points on the periodic motion. While staying constant, the energy oscillates between the kinetic energy of the block and the potential energy stored in the spring:

[latex] {Due east}_{\text{Total}}=U+M=\frac{ane}{2}one thousand{10}^{ii}+\frac{i}{ii}k{v}^{2}. [/latex]

The motility of the block on a spring in SHM is divers by the position [latex] x(t)=A\text{cos}(\omega t+\varphi ) [/latex] with a velocity of [latex] v(t)=\text{−}A\omega \text{sin}(\omega t+\varphi ) [/latex]. Using these equations, the trigonometric identity [latex] {\text{cos}}^{2}\theta +{\text{sin}}^{ii}\theta =1 [/latex] and [latex] \omega =\sqrt{\frac{k}{g}} [/latex], nosotros can discover the total energy of the system:

[latex] \begin{array}{cc}\hfill {Eastward}_{\text{Total}}& =\frac{ane}{ii}thou{A}^{2}{\text{cos}}^{2}(\omega t+\varphi )+\frac{1}{2}m{A}^{2}{\omega }^{2}{\text{sin}}^{2}(\omega t+\varphi )\hfill \\ & =\frac{i}{2}k{A}^{2}{\text{cos}}^{2}(\omega t+\varphi )+\frac{1}{2}yard{A}^{2}(\frac{chiliad}{chiliad}){\text{sin}}^{2}(\omega t+\varphi )\hfill \\ & =\frac{1}{ii}grand{A}^{2}{\text{cos}}^{2}(\omega t+\varphi )+\frac{1}{2}one thousand{A}^{2}{\text{sin}}^{two}(\omega t+\varphi )\hfill \\ & =\frac{ane}{2}g{A}^{two}({\text{cos}}^{2}(\omega t+\varphi )+{\text{sin}}^{2}(\omega t+\varphi ))\hfill \\ & =\frac{ane}{2}thousand{A}^{2}.\hfill \finish{array} [/latex]

The full energy of the organisation of a block and a bound is equal to the sum of the potential energy stored in the spring plus the kinetic energy of the block and is proportional to the square of the amplitude [latex] {E}_{\text{Full}}=(1\text{/}2)grand{A}^{2}. [/latex] The full free energy of the organisation is constant.

A closer look at the energy of the organization shows that the kinetic energy oscillates like a sine-squared office, while the potential free energy oscillates like a cosine-squared function. All the same, the total energy for the system is constant and is proportional to the aamplitude squared. (Figure) shows a plot of the potential, kinetic, and total energies of the block and spring organization as a role of time. Also plotted are the position and velocity as a function of time. Before time [latex] t=0.0\,\text{s,} [/latex] the block is attached to the spring and placed at the equilibrium position. Piece of work is done on the block by applying an external forcefulness, pulling it out to a position of [latex] x=+A [/latex]. The system at present has potential energy stored in the bound. At time [latex] t=0.00\,\text{s,} [/latex] the position of the block is equal to the amplitude, the potential energy stored in the spring is equal to [latex] U=\frac{1}{2}k{A}^{2} [/latex], and the force on the block is maximum and points in the negative x-direction [latex] ({F}_{S}=\text{−}kA) [/latex]. The velocity and kinetic energy of the block are nada at time [latex] t=0.00\,\text{southward}\text{.} [/latex] At time [latex] t=0.00\,\text{south,} [/latex] the block is released from rest.

Effigy 15.12 Graph of the kinetic energy, potential energy, and total energy of a block aquiver on a spring in SHM. Also shown are the graphs of position versus time and velocity versus time. The full energy remains abiding, simply the free energy oscillates between kinetic energy and potential energy. When the kinetic energy is maximum, the potential energy is zero. This occurs when the velocity is maximum and the mass is at the equilibrium position. The potential free energy is maximum when the speed is aught. The total energy is the sum of the kinetic energy plus the potential energy and it is constant.

Oscillations About an Equilibrium Position

We have just considered the energy of SHM as a function of time. Another interesting view of the simple harmonic oscillator is to consider the free energy as a function of position. (Figure) shows a graph of the energy versus position of a organization undergoing SHM.

Figure 15.xiii A graph of the kinetic energy (red), potential energy (blueish), and total energy (green) of a simple harmonic oscillator. The force is equal to [latex] F=-\frac{dU}{dx} [/latex]. The equilibrium position is shown every bit a black dot and is the betoken where the force is equal to zero. The force is positive when [latex] x<0 [/latex], negative when [latex] ten>0 [/latex], and equal to goose egg when [latex] x=0 [/latex].

The potential energy curve in (Figure) resembles a bowl. When a marble is placed in a bowl, information technology settles to the equilibrium position at the lowest point of the bowl [latex] (ten=0) [/latex]. This happens because a restoring force points toward the equilibrium point. This equilibrium point is sometimes referred to as a fixed point. When the marble is disturbed to a different position [latex] (10=+A) [/latex], the marble oscillates around the equilibrium position. Looking back at the graph of potential energy, the force can be found past looking at the slope of the potential free energy graph [latex] (F=-\frac{dU}{dx}) [/latex]. Since the force on either side of the fixed point points back toward the equilibrium bespeak, the equilibrium signal is called a stable equilibrium point. The points [latex] 10=A [/latex] and [latex] ten=\text{−}A [/latex] are chosen the turning points. (See Potential Energy and Conservation of Energy.)

Stability is an important concept. If an equilibrium betoken is stable, a slight disturbance of an object that is initially at the stable equilibrium point will crusade the object to oscillate effectually that point. The stable equilibrium point occurs because the force on either side is directed toward it. For an unstable equilibrium point, if the object is disturbed slightly, it does not render to the equilibrium point.

Consider the marble in the bowl example. If the bowl is correct-side up, the marble, if disturbed slightly, will oscillate around the stable equilibrium point. If the bowl is turned upside downwards, the marble can exist balanced on the peak, at the equilibrium point where the cyberspace force is zero. Nonetheless, if the marble is disturbed slightly, it volition not return to the equilibrium betoken, merely will instead roll off the bowl. The reason is that the forcefulness on either side of the equilibrium betoken is directed away from that indicate. This bespeak is an unstable equilibrium bespeak.

(Figure) shows three conditions. The offset is a stable equilibrium point (a), the second is an unstable equilibrium betoken (b), and the last is likewise an unstable equilibrium point (c), considering the force on merely one side points toward the equilibrium signal.

Effigy fifteen.fourteen Examples of equilibrium points. (a) Stable equilibrium point; (b) unstable equilibrium point; (c) unstable equilibrium point (sometimes referred to as a half-stable equilibrium betoken).

The process of determining whether an equilibrium bespeak is stable or unstable can exist formalized. Consider the potential energy curves shown in (Figure). The force tin be found by analyzing the gradient of the graph. The force is [latex] F=-\frac{dU}{dx}. [/latex] In (a), the stock-still betoken is at [latex] x=0.00\,\text{m}\text{.} [/latex] When [latex] x<0.00\,\text{m,} [/latex] the force is positive. When [latex] x>0.00\,\text{m,} [/latex] the forcefulness is negative. This is a stable point. In (b), the fixed point is at [latex] x=0.00\,\text{m}\text{.} [/latex] When [latex] 10<0.00\,\text{chiliad,} [/latex] the force is negative. When [latex] x>0.00\,\text{chiliad,} [/latex] the force is also negative. This is an unstable point.

Figure 15.15 Two examples of a potential free energy office. The force at a position is equal to the negative of the slope of the graph at that position. (a) A potential energy part with a stable equilibrium bespeak. (b) A potential energy role with an unstable equilibrium point. This point is sometimes chosen half-stable because the force on one side points toward the stock-still bespeak.

A applied awarding of the concept of stable equilibrium points is the force between 2 neutral atoms in a molecule. If two molecules are in close proximity, separated by a few atomic diameters, they can experience an attractive force. If the molecules motion close enough and then that the electron shells of the other electrons overlap, the force between the molecules becomes repulsive. The attractive force between the two atoms may cause the atoms to form a molecule. The force between the two molecules is not a linear force and cannot be modeled merely every bit ii masses separated past a jump, but the atoms of the molecule can oscillate around an equilibrium betoken when displaced a modest amount from the equilibrium position. The atoms oscillate due the attractive force and repulsive forcefulness between the two atoms.

Consider ane example of the interaction between two atoms known as the van Der Waals interaction. It is across the telescopic of this affiliate to discuss in depth the interactions of the two atoms, but the oscillations of the atoms can be examined by because one case of a model of the potential free energy of the system. One suggestion to model the potential energy of this molecule is with the Lennard-Jones vi-12 potential:

[latex] U(10)=4\epsilon [{(\frac{\sigma }{x})}^{12}-{(\frac{\sigma }{x})}^{half-dozen}]. [/latex]

A graph of this function is shown in (Effigy). The two parameters [latex] \epsilon [/latex] and [latex] \sigma [/latex] are found experimentally.

Figure 15.16 The Lennard-Jones potential energy function for a system of two neutral atoms. If the free energy is beneath some maximum energy, the organisation oscillates near the equilibrium position between the ii turning points.

From the graph, you can run into that there is a potential energy well, which has some similarities to the potential energy well of the potential energy function of the simple harmonic oscillator discussed in (Figure). The Lennard-Jones potential has a stable equilibrium point where the potential energy is minimum and the strength on either side of the equilibrium point points toward equilibrium signal. Note that dissimilar the simple harmonic oscillator, the potential well of the Lennard-Jones potential is not symmetric. This is due to the fact that the force between the atoms is not a Hooke'south constabulary force and is not linear. The atoms can still oscillate around the equilibrium position [latex] {x}_{\text{min}} [/latex] because when [latex] 10<{x}_{\text{min}} [/latex], the force is positive; when [latex] 10>{x}_{\text{min}} [/latex], the force is negative. Find that every bit x approaches cypher, the slope is quite steep and negative, which means that the strength is large and positive. This suggests that it takes a large force to try to push button the atoms close together. As x becomes increasingly large, the slope becomes less steep and the strength is smaller and negative. This suggests that if given a big enough free energy, the atoms can exist separated.

If you are interested in this interaction, observe the strength between the molecules past taking the derivative of the potential energy role. You will see immediately that the force does not resemble a Hooke'south law strength [latex] (F=\text{−}kx) [/latex], but if you are familiar with the binomial theorem:

[latex] {(1+x)}^{n}=i+nx+\frac{n(n-1)}{2!}{x}^{ii}+\frac{n(n-1)(n-two)}{3!}{x}^{iii}+\cdots , [/latex]

the force tin be approximated by a Hooke'due south law strength.

Velocity and Energy Conservation

Getting dorsum to the organisation of a block and a spring in (Figure), one time the cake is released from rest, it begins to move in the negative direction toward the equilibrium position. The potential free energy decreases and the magnitude of the velocity and the kinetic free energy increase. At fourth dimension [latex] t=T\text{/}4 [/latex], the block reaches the equilibrium position [latex] ten=0.00\,\text{thousand,} [/latex] where the force on the block and the potential free energy are zero. At the equilibrium position, the cake reaches a negative velocity with a magnitude equal to the maximum velocity [latex] v=\text{−}A\omega [/latex]. The kinetic free energy is maximum and equal to [latex] K=\frac{1}{2}thou{v}^{2}=\frac{i}{ii}m{A}^{two}{\omega }^{two}=\frac{i}{2}k{A}^{2}. [/latex] At this point, the force on the cake is zero, merely momentum carries the cake, and information technology continues in the negative direction toward [latex] x=\text{−}A [/latex]. As the block continues to move, the force on information technology acts in the positive direction and the magnitude of the velocity and kinetic energy subtract. The potential energy increases equally the spring compresses. At time [latex] t=T\text{/}2 [/latex], the block reaches [latex] x=\text{−}A [/latex]. Here the velocity and kinetic energy are equal to nothing. The strength on the block is [latex] F=+kA [/latex] and the potential free energy stored in the spring is [latex] U=\frac{1}{ii}k{A}^{2} [/latex]. During the oscillations, the total energy is constant and equal to the sum of the potential energy and the kinetic energy of the system,

[latex] {E}_{\text{Full}}=\frac{1}{two}k{ten}^{2}+\frac{1}{2}m{v}^{two}=\frac{1}{2}k{A}^{2}. [/latex]

The equation for the energy associated with SHM tin can exist solved to find the magnitude of the velocity at whatsoever position:

[latex] |v|=\sqrt{\frac{k}{one thousand}({A}^{ii}-{10}^{2})}. [/latex]

The energy in a simple harmonic oscillator is proportional to the square of the aamplitude. When considering many forms of oscillations, you will detect the energy proportional to the amplitude squared.

Check Your Understanding

Why would it hurt more if you snapped your hand with a ruler than with a loose spring, even if the deportation of each organisation is equal?

Show Solution

The ruler is a stiffer system, which carries greater force for the same amount of displacement. The ruler snaps your hand with greater strength, which hurts more.

Check Your Understanding

Identify i mode yous could decrease the maximum velocity of a elementary harmonic oscillator.

Show Solution

You could increment the mass of the object that is oscillating. Other options would exist to reduce the amplitude, or utilise a less potent spring.

Summary

• The simplest blazon of oscillations are related to systems that can exist described past Hooke'due south law, F = −kx, where F is the restoring force, x is the displacement from equilibrium or deformation, and k is the force constant of the arrangement.
• Elastic potential free energy U stored in the deformation of a organisation that can be described past Hooke'due south law is given by[latex] U=\frac{ane}{ii}k{x}^{2}. [/latex]
• Energy in the unproblematic harmonic oscillator is shared between elastic potential energy and kinetic free energy, with the full being abiding:

  [latex] {East}_{\text{Full}}=\frac{1}{2}k{v}^{2}+\frac{1}{two}k{10}^{2}=\frac{i}{2}1000{A}^{2}=\text{constant.} [/latex]

• The magnitude of the velocity as a function of position for the simple harmonic oscillator can be found by using

  [latex] |v|=\sqrt{\frac{m}{k}({A}^{2}-{ten}^{2})}. [/latex]

Conceptual Questions

Describe a system in which rubberband potential energy is stored.

Show Solution

In a car, rubberband potential energy is stored when the shock is extended or compressed. In some running shoes rubberband potential energy is stored in the compression of the material of the soles of the running shoes. In pole vaulting, elastic potential energy is stored in the bending of the pole.

Explain in terms of energy how dissipative forces such as friction reduce the amplitude of a harmonic oscillator. Too explain how a driving mechanism can compensate. (A pendulum clock is such a system.)

The temperature of the atmosphere oscillates from a maximum near noontime and a minimum near sunrise. Would you consider the atmosphere to be in stable or unstable equilibrium?

Show Solution

The overall organisation is stable. In that location may be times when the stability is interrupted by a storm, merely the driving force provided by the lord's day bring the atmosphere dorsum into a stable pattern.

Problems

Fish are hung on a bound scale to determine their mass. (a) What is the force constant of the jump in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load? (b) What is the mass of a fish that stretches the jump 5.50 cm? (c) How far apart are the half-kilogram marks on the calibration?

It is counterbalance-in fourth dimension for the local under-85-kg rugby squad. The bath scale used to appraise eligibility can exist described past Hooke'south law and is depressed 0.75 cm past its maximum load of 120 kg. (a) What is the spring's effective force constant? (b) A histrion stands on the scales and depresses information technology by 0.48 cm. Is he eligible to play on this under-85-kg team?

Show Solution

a. [latex] 1.57\,×\,{ten}^{5}\,\text{N/yard} [/latex]; b. 77 kg, yeah, he is eligible to play

One type of BB gun uses a spring-driven plunger to blow the BB from its butt. (a) Calculate the force constant of its plunger's spring if you lot must compress it 0.150 m to bulldoze the 0.0500-kg plunger to a meridian speed of xx.0 m/s. (b) What forcefulness must exist exerted to compress the spring?

When an 80.0-kg man stands on a pogo stick, the leap is compressed 0.120 g. (a) What is the force constant of the spring? (b) Will the jump exist compressed more than when he hops downward the road?

Show Solution

a. [latex] 6.53\,×\,{10}^{3}\,\text{Due north/thousand} [/latex]; b. yes, when the human is at his everyman point in his hopping the bound volition exist compressed the most

A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a one.95-kg mass hangs from it. (a) What is the force constant of the leap? (b) What is the unloaded length of the bound?

The length of nylon rope from which a mountain climber is suspended has an effective force constant of [latex] 1.40\,×\,{10}^{4}\,\text{N/m} [/latex]. (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? (b) How much would this rope stretch to break the climber's fall if he costless-falls 2.00 m before the rope runs out of slack? (Hint: Use conservation of free energy.) (c) Echo both parts of this trouble in the situation where twice this length of nylon rope is used.

Show Solution

a. i.99 Hz; b. 50.2 cm; c. 0.710 1000

Glossary

elastic potential energy

potential energy stored as a event of deformation of an rubberband object, such as the stretching of a leap

restoring forcefulness

force acting in opposition to the force caused by a deformation

stable equilibrium point

signal where the net strength on a arrangement is zero, but a small-scale displacement of the mass volition cause a restoring forcefulness that points toward the equilibrium bespeak

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