Chapter 2 Motion Along a Straight Line Position, Displacement, Average Speed

Essential idea: A study of oscillations underpins many areas of physics with simple harmonic motion (SHM) a fundamental oscillation that appears in various natural phenomena.Topic 4: Waves4.1 Oscillations1 2006 By Timothy K. LundNature of science: Models: Oscillations play a great part in our lives, from the tides to the motion of the swinging pendulum that once governed our perception of time. General principles govern this area of physics, from water waves in the deep ocean or the oscillations of a car suspension system. This introduction to the topic reminds us that not all oscillations are isochronous. However, the simple harmonic oscillator is of great importance to physicists because all periodic oscillations can be described through the mathematics of simple harmonic motion.Topic 4: Waves4.1 Oscillations2 2006 By Timothy K. LundUnderstandings: Simple harmonic oscillations Time period, frequency, amplitude, displacement and phase difference Conditions for simple harmonic motion Applications and skills: Qualitatively describing the energy changes taking place during one cycle of an oscillation Sketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 Oscillations3 2006 By Timothy K. LundGuidance: Graphs describing simple harmonic motion should include displacementtime, velocitytime, accelerationtime and accelerationdisplacement Students are expected to understand the significance of the negative sign in the relationship: a = -x.Data booklet reference: T = 1 / f Topic 4: Waves4.1 Oscillations4 2006 By Timothy K. LundInternational-mindedness: Oscillations are used to define the time systems on which nations agree so that the world can be kept in synchronization. This impacts most areas of our lives including the provision of electricity, travel and location-determining devices and all microelectronics. Theory of knowledge: The harmonic oscillator is a paradigm for modelling where a simple equation is used to describe a complex phenomenon. How do scientists know when a simple model is not detailed enough for their requirements?Topic 4: Waves4.1 Oscillations5 2006 By Timothy K. LundUtilization: Isochronous oscillations can be used to measure time Many systems can approximate simple harmonic motion: mass on a spring, fluid in U-tube, models of icebergs oscillating vertically in the ocean, and motion of a sphere rolling in a concave mirror Simple harmonic motion is frequently found in the context of mechanics (see Physics topic 2) Topic 4: Waves4.1 Oscillations6 2006 By Timothy K. LundAims: Aim 6: experiments could include (but are not limited to): mass on a spring; simple pendulum; motion on a curved air track Aim 7: IT skills can be used to model the simple harmonic motion defining equation; this gives valuable insight into the meaning of the equation itself Topic 4: Waves4.1 Oscillations7 2006 By Timothy K. LundEXAMPLE: Oscillations can be driven externally, like a pendulum in a gravitational field.EXAMPLE: Oscillations can be driven internally, like a mass on a spring.OscillationsOscillations are vibrations which repeat themselves.Topic 4: Waves4.1 Oscillationsxv = 0v = vmaxv = 0v = 0v = vmaxv = 0FYI In all oscillations, v = 0 at the extremesand v = vmax in the middle of the motion. 2006 By Timothy K. LundTopic 4: Waves4.1 OscillationsEXAMPLE: Oscillations can be very rapid vibrations such as in a plucked guitar string or a tuning fork.

OscillationsOscillations are vibrations which repeat themselves. 2006 By Timothy K. LundTime period, amplitude and displacementConsider a mass on a spring that is displaced 4 meters to the right and then released.We call the maximum displacement x0 the amplitude. In this example x0 = 4 m.We call the point of zero displacement the equilibrium position. Displacement x is measured from equilibrium.The period T (measured in s) is the time it takes for the mass to make one full oscillation or cycle.For this particular oscillation, the period T is about 24 seconds (per cycle).Topic 4: Waves4.1 Oscillationsx0x equilibrium 2006 By Timothy K. LundEXAMPLE: The cycle of the previous example repeated each 24 s. What are the period and the frequency of the oscillation?SOLUTION:The period is T = 24 s.The frequency is f = 1 / T = 1 / 24 = 0.042 HzTime period and frequencyThe frequency f (measured in Hz or cycles / s) is defined as how many cycles (oscillations, repetitions) occur each second.Since period T is seconds per cycle, frequency must be 1 / T.Topic 4: Waves4.1 Oscillationsf = 1 / Trelation between T and fT = 1 / f 2006 By Timothy K. LundPhase differenceWe can pull the mass to the right and then release it to begin its motion:

Or we could push it to the left and release it:Both motions would have the same values for T and f.However, the resulting motion will have a phase difference of half a cycle.Topic 4: Waves4.1 OscillationsStart stretchedStart compressedThe two motions are half a cycle out of phase.xx 2006 By Timothy K. LundPRACTICE: Two identical mass-spring systems are started in two different ways. What is their phase difference?

SOLUTION:The phase difference is one-quarter of a cycle.Phase differenceTopic 4: Waves4.1 OscillationsStart stretched and then releaseStart unstretched with a push leftxx 2006 By Timothy K. LundPRACTICE: Two identical mass-spring systems are started in two different ways. What is their phase difference?

SOLUTION:The phase difference is three-quarters of a cycle.Topic 4: Waves4.1 OscillationsStart stretched and then releaseStart unstretched with a push rightxxPhase difference 2006 By Timothy K. LundEXAMPLE: A spring having a spring constant of 125 N m-1 is attached to a 5.0-kg mass, stretched +4.0 m as shown, and then released from rest.(a) Using Hookes law, show that the acceleration a of a mass-spring system is related to the springs displacement x by the proportion a -x. SOLUTION:Recall Hookes law: F = -kx (see Topic 2-3).From Newtons second law F = ma we then havema = -kx or a = -(k / m) x.Thus, a -x. Conditions for simple harmonic motion Topic 4: Waves4.1 Oscillationsx 2006 By Timothy K. LundConditions for simple harmonic motion Topic 4: Waves4.1 OscillationsEXAMPLE: A spring having a spring constant of 125 N m-1 is attached to a 5.0-kg mass, stretched +4.0 m as shown, and then released from rest.(b) Tailor your equation to this example, and find the acceleration of the mass when x = -2.0 m.(c) What is the displacement of the mass when the acceleration is -42 ms-2?SOLUTION:(b) a = -(k / m) x = -(125 / 5) x = -25x. Thus a = -25x so a = -25(-2.0) = +50. ms-2.(c) a = -25x -42 = - 25 x x = -42 / -25 = +1.7 m.x 2006 By Timothy K. LundPRACTICE: Show that a mass oscillating on a spring executes simple harmonic motion.

SOLUTION:We already did when we showed that a = -(k / m)x, since this means that a -x.Conditions for simple harmonic motion A very special kind of oscillation that shows up often in the physical world is called simple harmonic motion.In simple harmonic motion (SHM), a and x are related in a very precise way: Namely, a -x.Topic 4: Waves4.1 Oscillationsa -xdefinition of SHMx 2006 By Timothy K. LundConditions for simple harmonic motion

The minus sign in Hookes law, F = -kx, tells us that if the displacement x is positive (right), the spring force F is negative (left). It also tells us that if the displacement x is negative (left), the spring force F is positive (right).Any force that is proportional to the opposite of a displacement is called a restoring force.For any restoring force F -x. Since F = ma we see that ma -x, or a -x.All restoring forces can drive simple harmonic motion (SHM).Topic 4: Waves4.1 Oscillationsa -xdefinition of SHM

F

x0FxxF and x oppose each other. 2006 By Timothy K. LundConditions for simple harmonic motion If we place a pen on the oscillating mass, and pull a piece of paper at a constant speed past the pen, we trace out the displacement vs. time graph of SHM.SHM traces out perfect sinusoidal waveforms.

Note that the period can be found from the graph: Just look for repeating cycles.

Topic 4: Waves4.1 Oscillationsxt 2006 By Timothy K. LundQualitatively describing the energy changes taking place during one cycle of an oscillation Consider the pendulum to the right which is placed in position and held there.Let the green rectangle represent the potential energy of the system.Let the red rectangle represent the kinetic energy of the system.Because there is no motion yet, there is no kinetic energy. But if we release it, the kinetic energy will grow as the potential energy diminishes.A continuous exchange between EK and EP occurs.Topic 4: Waves4.1 Oscillations

2006 By Timothy K. LundQualitatively describing the energy changes taking place during one cycle of an oscillation Consider the mass-spring system shown here. The mass is pulled to the right and held in place.Let the green rectangle represent the potential energy of the system.Let the red rectangle represent the kinetic energy of the system.A continuous exchange between EK and EP occurs.Note that the sum of EK and EP is constant.Topic 4: Waves4.1 Oscillationsx

EK + EP = ET = CONSTrelation between EK and EPFYI If friction and drag are both zero ET = CONST. 2006 By Timothy K. LundQualitatively describing the energy changes taking place during one cycle of an oscillation

If we plot both kinetic energy and potential energy vs. time for either system we would get the following graph:Topic 4: Waves4.1 OscillationsEnergytimex

EK + EP = ET = CONSTrelation between EK and EP 2006 By Timothy K. Lund

EK + EP = ET = CONSTrelation between EK and EP 2006 By Timothy K. LundEXAMPLE: The dis- placement x vs. time t for a 2.5-kg mass on a spring having spring constant k = 4.0 Nm-1 is shown in the sinusoidal graph. (a) Find the period and frequency of the motion.SOLUTION: The period is the time for a complete cycle.From the graph it is T = 6.0 ms = 6.010-3 s.Then f = 1 / T = 1 / 0.006 = 170 Hz.Sketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 Oscillations

2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: The dis- placement x vs. time t for a 2.5-kg mass on a spring having spring constant k = 4.0 Nm-1 is shown in the sinusoidal graph. (b) Find the amplitude of the motion.SOLUTION: The amplitude is the maximum displacement.From the graph it is xMAX = 2.0 mm = 2.010-3 m.

2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: The dis- placement x vs. time t for a 2.5-kg mass on a spring having spring constant k = 4.0 Nm-1 is shown in the sinusoidal graph. (c) Sketch the graph of x vs. t for the situation where the amplitude is cut in half.SOLUTION: For SHM, the period is independent of the amplitude.

2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: The dis- placement x vs. time t for a 2.5-kg mass on a spring having spring constant k = 4.0 Nm-1 is shown in the sinusoidal graph. (c) The blue graph shows an equivalent system in SHM. What is the phase difference between the red and blue?SOLUTION: We see that it is T / 6 (= 360/ 6 = 60 = 2/ 6 rad).

2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: The displacement x vs. time t for a system undergoing SHM is shown here.

Sketch in red the velocity vs. time graph.SOLUTION: At the extremes, v = 0. At x = 0, v = vMAX. The slope determines sign of vMAX.

x-blackv-red (different scale)t(+)( -)(+)( -)(+)xv = 0 v = 0v = vMAX -2.00.02.0 2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: The displacement x vs. time t for a system undergoing SHM is shown here.

Sketch in blue the acceleration vs. time graph.SOLUTION: Since a -x, a is just a reflection of x.Note: x is a sine, v is a cosine, and a is a sine wave.

x-blackv-red (different scale)txv = 0 v = 0v = vMAX -2.00.02.0a-blue (different scale) 2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesEXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring.(a) Determine the maximum velocity of the mass.SOLUTION: When the kinetic energy is maximum, the velocity is also maximum. Thus 4.0 = (1/ 2)mvMAX2 so that4.0 = (1/ 2)(.125)vMAX2 vMAX = 8.0 ms-1.

Topic 4: Waves4.1 Oscillationsx-2.00.02.0 2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesEXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring.(b) Sketch EP and determine the total energy of the system.SOLUTION: Since EK + EP = ET = CONST, and since EP = 0 when EK = EK,MAX, it must be that ET = EK,MAX = 4.0 J.Thus the EP graph will be the inverted EK graph.

Topic 4: Waves4.1 Oscillationsx-2.00.02.0ETEPEK 2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesEXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring.(c) Determine the spring constant k of the spring.SOLUTION: Recall EP = (1/2)kx2.Note that EK = 0 at x = xMAX = 2.0 cm. ThusEK + EP = ET = CONST ET = 0 + (1/ 2)kxMAX2 so that 4.0 = (1/ 2)k 0.0202 k = 20000 Nm-1.

Topic 4: Waves4.1 Oscillationsx-2.00.02.0 2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesEXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring.(d) Determine the acceleration of the mass at x = 1.0 cm.SOLUTION:From Hookes law, F = -kx we get F = -20000(0.01) = -200 N.From F = ma we get -200 = 0.125a a = -1600 ms-2.

Topic 4: Waves4.1 Oscillationsx-2.00.02.0 2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: A 4.0-kg mass is placed on a springs end and displaced 2.0 m to the right.The spring force F vs. its displacement x from equilibrium is shown in the graph.(a) How do you know that the mass is undergoing SHM?SOLUTION:In SHM, a -x. Since F = ma, then F -x also.The graph shows that F -x. Thus we have SHM.x-2.00.02.0

2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: A 4.0-kg mass is placed on a springs end and displaced 2.0 m to the right.The spring force F vs. its displacement x from equilibrium is shown in the graph.(b) Find the spring constant of the spring.SOLUTION: Use Hookes law: F = -kx.Pick any F and any x. Use k = -F / x.Thus k = -(-5.0 N) / 1.0 m = 5.0 Nm-1.x-2.00.02.0

F = -5.0 Nx = 1.0 m 2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: A 4.0-kg mass is placed on a springs end and displaced 2.0 m to the right.The spring force F vs. its displacement x from equilibrium is shown in the graph.(c) Find the total energy of the system.SOLUTION: Use ET = (1/2)kxMAX2. Then ET = (1/2)kxMAX2 = (1/2) 5.0 2.02 = 10. J.x-2.00.02.0

2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: A 4.0-kg mass is placed on a springs end and displaced 2.0 m to the right.The spring force F vs. its displacement x from equilibrium is shown in the graph.(d) Find the maximum speed of the mass.SOLUTION: Use ET = (1/2)mvMAX2. 10. = (1/2) 4.0 vMAX2 vMAX = 2.2 ms-1.x-2.00.02.0

2006 By Timothy K. LundSketching and interpreting graphs of simple harmonic motion examplesTopic 4: Waves4.1 OscillationsEXAMPLE: A 4.0-kg mass is placed on a springs end and displaced 2.0 m to the right.The spring force F vs. its displacement x from equilibrium is shown in the graph.(e) Find the speed of the mass when its displacement is 1.0 m.SOLUTION: Use ET = (1/2)mv 2 + (1/2)kx 2. Then 10. = (1/2)(4)v 2 + (1/2)(5)12 v = 1.9 ms-1.x-2.00.02.0

2006 By Timothy K. Lund