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Lesson Name: Uniform Circular Motion and Simple Harmonic Motion

Instructional Block

Uniform Circular Motion and Simple Harmonic Motion 


 

LEARNING OBJECTIVES 

 

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

  • Compare simple harmonic motion with uniform circular motion.

 

 

 

Figure 1 The horses on this merry-go-round exhibit uniform circular motion. (credit: Wonderlane, Flickr)

 

There is an easy way to produce simple harmonic motion by using uniform circular motion. Figure 2 shows one way of using this method. A ball is attached to a uniformly rotating vertical turntable, and its shadow is projected on the floor as shown. The shadow undergoes simple harmonic motion. Hooke’s law usually describes uniform circular motions (ω constant) rather than systems that have large visible displacements. So observing the projection of uniform circular motion, as in Figure 2, is often easier than observing a precise large-scale simple harmonic oscillator. If studied in sufficient depth, simple harmonic motion produced in this manner can give considerable insight into many aspects of oscillations and waves and is very useful mathematically. In our brief treatment, we shall indicate some of the major features of this relationship and how they might be useful.

 

 

 Figure 2 The shadow of a ball rotating at constant angular velocity \(\omega\)  on a turntable goes back and forth in precise simple harmonic motion.

 

Figure 3 shows the basic relationship between uniform circular motion and simple harmonic motion. The point P travels around the circle at constant angular velocity \(\omega\). The point P is analogous to an object on the merry-go-round. The projection of the position of P onto a fixed axis undergoes simple harmonic motion and is analogous to the shadow of the object. At the time shown in the figure, the projection has position x and moves to the left with velocity \(v\). The velocity of the point P around the circle equals \(\overline{v}_{\mathrm{max}}\).The projection of \(\overline{v}_{\mathrm{max}}\) on the \(x-\mathrm{axis}\) is the velocity \(v\) of the simple harmonic motion along the \(x-\mathrm{axis}\).

 

 

 

Figure 3 A point P moving on a circular path with a constant angular velocity \(\omega\) is undergoing uniform circular motion. Its projection on the \(x\)-axis undergoes simple harmonic motion. Also shown is the velocity of this point around the circle, \(\overline{v}_{\mathrm{max}}\) , and its projection, which is  v . Note that these velocities form a similar triangle to the displacement triangle.

 

To see that the projection undergoes simple harmonic motion, note that its position \(x\) is given by

 

 

where \(\theta = \omega t, \omega\) is the constant angular velocity, and \(X\) is the radius of the circular path. Thus,

 

 

The angular velocity \(\omega\) is in radians per unit time; in this case \(2 \pi\) radians is the time for one revolution \(T\). That is, \(\omega = 2 \pi /T\). Substituting this expression for \(\omega\), we see that the position \(x\) is given by:

 

 

This expression is the same one we had for the position of a simple harmonic oscillator in Simple Harmonic Motion: A Special Periodic Motion. If we make a graph of position versus time as in Figure 4, we see again the wavelike character (typical of simple harmonic motion) of the projection of uniform circular motion onto the \(x\)-axis.

 

 

 

Figure 4 The position of the projection of uniform circular motion performs simple harmonic motion, as this wavelike graph of \(x\) versus \(t\)  indicates.

 

Now let us use Figure 4 to do some further analysis of uniform circular motion as it relates to simple harmonic motion. The triangle formed by the velocities in the figure and the triangle formed by the displacements \((X, x, \thinspace \mathrm{and} \thinspace \sqrt{X^2-x^2})\) are similar right triangles. Taking ratios of similar sides, we see that

 

\(\frac{v}{v_{\mathrm{max}}} = \frac{ \sqrt{X^2- x^2}}{X} = \sqrt{1-\frac{x^2}{X^2}}.\)

 

We can solve this equation for the speed \(v\) or

 

 

This expression for the speed of a simple harmonic oscillator is exactly the same as the equation obtained from conservation of energy considerations in Energy and the Simple Harmonic Oscillator. You can begin to see that it is possible to get all of the characteristics of simple harmonic motion from an analysis of the projection of uniform circular motion.

Finally, let us consider the period \(T\) of the motion of the projection. This period is the time it takes the point P to complete one revolution. That time is the circumference of the circle \(2 \pi X\) divided by the velocity around the circle, \(v_{\mathrm{max}}\)max. Thus, the period \(T\) is

We know from conservation of energy considerations that

Solving this equation for \(X/v_{\mathrm{max}}\) gives

Substituting this expression into the equation for \(T\) yields

Thus, the period of the motion is the same as for a simple harmonic oscillator. We have determined the period for any simple harmonic oscillator using the relationship between uniform circular motion and simple harmonic motion.

Some modules occasionally refer to the connection between uniform circular motion and simple harmonic motion. Moreover, if you carry your study of physics and its applications to greater depths, you will find this relationship useful. It can, for example, help to analyze how waves add when they are superimposed.

 

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Check All That Apply

Which two of the following quantities are required for the uniform circular motion? Choose 2 correct options.




Answer Rubric % Of Choosen
Torque In-Correct 0 %
Circular path Correct 0 %
Distance In-Correct 0 %
Constant speed Correct 0 %


Multiple Choice

A particle moves in a uniform circular motion. If its period is doubled, what will happen to its angular velocity?

Answer Rubric % Of Choosen
stays the same In-Correct 0 %
will be equal to the square root of the period In-Correct 0 %
reduced to half Correct 0 %
doubled In-Correct 0 %
No hints found for this question

Multiple Choice

Which statement best describes the similarity between uniform circular motion and projectile motion?

Answer Rubric % Of Choosen
In both motions, the acceleration's direction is changing continuously. In-Correct 0 %
In both motions, the direction of the object's acceleration is the same at all times. In-Correct 0 %
The magnitude of the object's acceleration in both kinds of motion is changing at all times. In-Correct 0 %
The magnitude of the object's acceleration in uniform circular motion and projectile motion is equal at all times.  Correct 0 %
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Multiple Choice

If a CD rotates in a clockwise direction within a horizontal player, where does the angular velocity vector point? 

Answer Rubric % Of Choosen
Down Correct 0 %
Tangent to CD In-Correct 0 %
Towards the center of the CD In-Correct 0 %
Upwards In-Correct 0 %

Down


Numeric

Calculate the period in seconds of a moving particle with a constant angular velocity of 39.8 \(rad \over s \).

Is correct? Answer (midpoint) Rounding Margin Answer Range Units Wrong Answer Feedback
Correct 0.165 0.01 [0.155,0.175] -
\(\omega = {2\pi \over T}\)

 \(T = {2\pi \over \omega}= {2\pi \space rad \over 39.8 \space {rad \over s}}\)

Answer: 0.1579


Numeric

What is the maximum velocity, in meters per second, of an 85.0-kg person bouncing on a bathroom scale having a force constant of \(1.50\times10^6\;N/m\), if the amplitude of the bounce is 0.200 cm?

Is correct? Answer (midpoint) Rounding Margin Answer Range Units Wrong Answer Feedback
Correct 0.275 0.01 [0.265,0.285] -

\(v_{max}=\sqrt\frac{k}{m}X\)

\(v_{max}=\sqrt\frac{1.50\times10^6\;N/m}{85.0\;kg}(0.200\times10^{-2}\; m)\)

Answer: 0.266

Numeric + Units

What is the maximum energy stored in the spring?

Is correct? Answer (midpoint) Rounding Margin Answer Range Units Wrong Answer Feedback
Correct 3.05 0.1 [2.95,3.15]

\(J\)

\(PE=\frac{1}{2}kX^2\)

\(PE=\frac{1}{2}(1.50\times10^6\; N/m)(0.200\times10^{-2}\;m)^2\)

Answer: 3 J



Numeric

A novelty clock has a 0.0100-kg mass object bouncing on a spring that has a force constant of 1.25 N/m. What is the maximum velocity, in meters per second, of the object if the object bounces 3.00 cm above and below its equilibrium position?

Is correct? Answer (midpoint) Rounding Margin Answer Range Units Wrong Answer Feedback
Correct 0.345 0.01 [0.335,0.355] -

\(v_{max}=\sqrt{\frac{k}{m}}X\)

\(v_{max}=\sqrt{\frac{1.25\;N/m}{0.0100\;kg}}(0.0300\;m)\)

Answer: 0.3354

Numeric + Units

How many joules of kinetic energy does the object have at its maximum velocity?

Is correct? Answer (midpoint) Rounding Margin Answer Range Units Wrong Answer Feedback
Correct 0.000565 1.0E-5 [0.000555,0.000575]

\(J\)

\(KE=\frac{1}{2}mv^2\)

\(KE=\frac{1}{2}(0.0100\;kg)(0.3354\; m/s)^2\)

Answer: 5.61E-04 J



Multiple Choice

You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system.

Answer Rubric % Of Choosen
You could increase the longitude of the spring in which the object is oscillating. In-Correct 0 %
You could decrease the mass of the object that is oscillating. In-Correct 0 %
You could increase the mass of the object that is oscillating. Correct 0 %
You could decrease the longitude of the spring in which the object is oscillating. In-Correct 0 %
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Numeric

A ladybug sits 12.0 cm from the center of a Beatles music album spinning at 33.33 rpm. What is the maximum velocity of its shadow on the wall behind the turntable, if illuminated parallel to the record by the parallel rays of the setting Sun?

Express your answer in meters per second.

Is correct? Answer (midpoint) Rounding Margin Answer Range Units Wrong Answer Feedback
Correct 0.425 0.01 [0.415,0.435] -

\(\omega = 33.33\;\text{ rev/min }\big(\frac{1\; \text{min}}{60\; s} \big)\big( \frac{2\pi\text{ rad}}{1\text{ rev }}\big)=3.49\; \text{rad/s}\)

\(v_{max}=r\omega\)

\(v_{max}=(0.120\;m)(4.49\text{ rad/s})\)

Answer: 0.419


Numeric

An electrical engine has its maximum performance at 7088 rpm. Compute the period in seconds of the engine at such condition.

Is correct? Answer (midpoint) Rounding Margin Answer Range Units Wrong Answer Feedback
Correct 0.00855 0.0001 [0.00845,0.00865] -

\(T = {2 \pi \over \omega}\)

\(T = {2 \pi \over (7088)({2\pi \over 60})} = {60\;s \over 7088}\)

Answer: 0.0085


Instructional Block

Damped Harmonic Motion 


 

LEARNING OBJECTIVES 

 

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

  • Compare and discuss underdamped and overdamped oscillating systems.
  • Explain critically damped systems.

 

 

 

Figure 1 In order to counteract dampening forces, this mom needs to keep pushing the swing. (credit: Mohd Fazlin Mohd Effendy Ooi, Flickr)

 

A guitar string stops oscillating a few seconds after being plucked. To keep a child happy on a swing, you must keep pushing. Although we can often make friction and other non-conservative forces negligibly small, completely undamped motion is rare. In fact, we may even want to damp oscillations, such as with car shock absorbers.

For a system that has a small amount of damping, the period and frequency are nearly the same as for simple harmonic motion, but the amplitude gradually decreases as shown in Figure 1. This occurs because the non-conservative damping force removes energy from the system, usually in the form of thermal energy. In general, energy removal by non-conservative forces is described as 

 

 

where \(W_{\mathrm{nc}}\)nc is work done by a non-conservative force (here the damping force). For a damped harmonic oscillator, \(W_{\mathrm{nc}}\) is negative because it removes mechanical energy (KE + PE) from the system.

 

 

Figure 2 In this graph of displacement versus time for a harmonic oscillator with a small amount of damping, the amplitude slowly decreases, but the period and frequency are nearly the same as if the system were completely undamped.

 

If you gradually increase the amount of damping in a system, the period and frequency begin to be affected, because damping opposes and hence slows the back and forth motion. (The net force is smaller in both directions.) If there is very large damping, the system does not even oscillate—it slowly moves toward equilibrium. Figure 3 shows the displacement of a harmonic oscillator for different amounts of damping. When we want to damp out oscillations, such as in the suspension of a car, we may want the system to return to equilibrium as quickly as possible Critical damping is defined as the condition in which the damping of an oscillator results in it returning as quickly as possible to its equilibrium position The critically damped system may overshoot the equilibrium position, but if it does, it will do so only once. Critical damping is represented by Curve A in Figure 3. With less-than critical damping, the system will return to equilibrium faster but will overshoot and cross over one or more times. Such a system is underdamped; its displacement is represented by the curve in Figure 2. Curve B in Figure 3 represents an overdamped system. As with critical damping, it too may overshoot the equilibrium position, but will reach equilibrium over a longer period of time. 

 

 

 

Figure 3 Displacement versus time for a critically damped harmonic oscillator (A) and an overdamped harmonic oscillator (B). The critically damped oscillator returns to equilibrium at \(X = 0\) in the smallest time possible without overshooting.

 

Critical damping is often desired, because such a system returns to equilibrium rapidly and remains at equilibrium as well. In addition, a constant force applied to a critically damped system moves the system to a new equilibrium position in the shortest time possible without overshooting or oscillating about the new position. For example, when you stand on bathroom scales that have a needle gauge, the needle moves to its equilibrium position without oscillating. It would be quite inconvenient if the needle oscillated about the new equilibrium position for a long time before settling. Damping forces can vary greatly in character. Friction, for example, is sometimes independent of velocity (as assumed in most places in this text). But many damping forces depend on velocity—sometimes in complex ways, sometimes simply being proportional to velocity.

 

MAKING CONNECTIONS: DAMPED OSCILLATOR 

 

Consider a mass attached to a spring. This system oscillates when in air because air exerts almost no force on the spring. Now put this system in a liquid, say, water. You will see that the system hardly oscillates when in water. When the system is submerged in water an external force acts on the oscillator. This force is exerted by the liquid against the motion of the spring-mass oscillator and is responsible for the inhibition of oscillations. A force that inhibits oscillations is called a “damping force,” and the system that experiences it is called a “damped oscillator." A damped oscillator sees a change in its energy. With time the total energy of the oscillator, which would be its mechanical energy, decreases. Since energy has to be conserved, the energy gets converted into thermal energy, which is stored in the water and the spring.

 

 

EXAMPLE 1 DAMPING AN OSCILLATORY MOTION: FRICTION ON AN OBJECT CONNECTED TO A SPRING

 

Damping oscillatory motion is important in many systems, and the ability to control the damping is even more so. This is generally attained using non-conservative forces such as the friction between surfaces, and viscosity for objects moving through fluids. The following example considers friction. Suppose a 0.200-kg object is connected to a spring as shown in Figure 4, but there is simple friction between the object and the surface, and the coefficient of friction \(\mu _{k}\) is equal to 0.0800. (a) What is the frictional force between the surfaces? (b) What total distance does the object travel if it is released 0.100 m from equilibrium, starting at \(v = 0\)? The force constant of the spring is \(k = 50.0 \thinspace \mathrm{N/m}\).

 

 

Figure 4 The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface.

 

Strategy

This problem requires you to integrate your knowledge of various concepts regarding waves, oscillations, and damping. To solve an integrated concept problem, you must first identify the physical principles involved. Part (a) is about the frictional force. This is a topic involving the application of Newton’s Laws. Part (b) requires an understanding of work and conservation of energy, as well as some understanding of horizontal oscillatory systems.

Now that we have identified the principles we must apply in order to solve the problems, we need to identify the knowns and unknowns for each part of the question, as well as the quantity that is constant in Part (a) and Part (b) of the question. 

Solution a 

  1. Choose the proper equation: Friction is \(f = \mu _{k} mg\).
  2. Identify the known values.
  3. Enter the known values into the equation:
  4. Calculate and convert units: \(f = 0.157 \thinspace \mathrm{N}.\)

Discussion a

The force here is small because the system and the coefficients are small.

Solution b

Identify the known:

  • The system involves elastic potential energy as the spring compresses and expands, friction that is related to the work done, and the kinetic energy as the body speeds up and slows down. 
  • Energy is not conserved as the mass oscillates because friction is a non-conservative force.
  • The motion is horizontal, so gravitational potential energy does not need to be considered. 
  • Because the motion starts from rest, the energy in the system is initially \(\mathrm{PE_{el,i}} = (1/2)kX^2\). This energy is removed by work done by friction \(W_{\mathrm{nc}} = - fd\), where \(d\) is the total distance traveled and \(f = \mu _{k}mg\) is the force of friction. When the system stops moving, the friction force will balance the force exerted by the spring, so  \(\mathrm{PE_{el,f}} = (1/2)kx^2 \) where  \(x\) is the final position and is given by 
    \(kx = \mu _{k}mg.\)
    \(x = \frac{\mu _{k} mg}{k}\)
  1. By equating the work done to the energy removed, solve for the distance \(d\) .
  2. The work done by the non-conservative forces equals the initial, stored elastic potential energy. Identify the correct equation to use:
  3. Recall that \(W_{\mathrm{nc}} = -fd.\).
  4. Enter the friction as \(f = \mu _{k} mg\) into \(W_{\mathrm{nc}} = -fd\), thus
  5. Combine these two equations to find
  6. Solve the equation for \(d\) :
  7. Enter the known values into the resulting equation:
    \(\cdot \left ( (0.100 \thinspace \mathrm{m})^2- \left ( \frac{(0.0800)(0.200 \thinspace \mathrm{kg})(9.80 \thinspace \mathrm{m/s^2})}{50.0 \thinspace \mathrm{N/m}} \right )^2 \right )\)
  8. Calculate \(d\) and convert units:

Discussion b

This is the total distance traveled back and forth across \(x = 0\), which is the undamped equilibrium position. The number of oscillations about the equilibrium position will be more than \(d/X = (1.59 \thinspace \mathrm{m})/ (0.100 \thinspace \mathrm{m}) = 15.9\) because the amplitude of the oscillations is decreasing with time. At the end of the motion, this system will not return to \(x = 0\) for this type of damping force, because static friction will exceed the restoring force. This system is underdamped. In contrast, an overdamped system with a simple constant damping force would not cross the equilibrium position \(x = 0\) a single time. For example, if this system had a damping force 20 times greater, it would only move 0.0484 m toward the equilibrium position from its original 0.100-m position.

This worked example illustrates how to apply problem-solving strategies to situations that integrate the different concepts you have learned. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknowns using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life.

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Free Response

Describe the difference between overdamping, underdamping, and critical damping.

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Free Response

Give an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.)

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Free Response

How would a car bounce after a bump under each of these conditions?

  • overdamping
  • underdamping
  • critical damping
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Free Response

Most harmonic oscillators are damped and, if undriven, eventually come to a stop. How is this observation related to the second law of thermodynamics?

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Numeric

The amplitude of a lightly damped oscillator decreases by 30.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

Is correct? Answer (midpoint) Rounding Margin Answer Range Units Wrong Answer Feedback
Correct 5.95 0.1 [5.85,6.05] -

The energy depends on the square of the amplitude, \(PE=\frac{1}{2}kx^2\).

Therefore, a 3.0% decrease in the amplitude means that x changes to 0.97x, so PE changes to (0.97)2PE, which means 5.9% of the mechanical energy is lost in each cycle.

Answer: 5.9


Multiple Choice

Why are completely undamped harmonic oscillators so rare?

Answer Rubric % Of Choosen
Energy often comes into play whenever an object is moving. The energy causes damping in a harmonic oscillator. In-Correct 0 %
Gravitation force often comes into play whenever an object is moving. Gravitation force causes damping in a harmonic oscillator. In-Correct 0 %
Angular velocity often comes into play whenever an object is moving. The increase in angular velocity causes damping in a harmonic oscillator. In-Correct 0 %
Friction often comes into play whenever an object is moving. Friction causes damping in a harmonic oscillator. Correct 0 %
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