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: Waves 4.1 – Oscillations Nature 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.

Understandings: 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 examples Topic 4: Waves 4.1 – Oscillations

International-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: Waves 4.1 – Oscillations

Utilization: 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) Aims: 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: Waves 4.1 – Oscillations

EXAMPLE: Oscillations can be driven externally, like a pendulum in a gravitational field. EXAMPLE: Oscillations can be driven internally, like a mass on a spring. Oscillations Oscillations are vibrations which repeat themselves. Topic 4: Waves 4.1 – Oscillations x v = 0 v = v max v = 0 v = v max v = 0 FYI In all oscillations, v = 0 at the extremes… and v = v max in the middle of the motion.

Topic 4: Waves 4.1 – Oscillations EXAMPLE: Oscillations can be very rapid vibrations such as in a plucked guitar string or a tuning fork. Oscillations Oscillations are vibrations which repeat themselves.

Time period, amplitude and displacement Consider a mass on a spring that is displaced 4 meters to the right and then released. We call the maximum displacement x 0 the amplitude. In this example x 0 = 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: Waves 4.1 – Oscillations x0x0 x equilibrium

EXAMPLE: 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 = Hz Time period and frequency The 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: Waves 4.1 – Oscillations f = 1 / T relation between T and f T = 1 / f

Phase difference We 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: Waves 4.1 – Oscillations Start stretched Start compressed The two motions are half a cycle out of phase. x x

PRACTICE: 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 difference Topic 4: Waves 4.1 – Oscillations Start stretched and then release Start unstretched with a push left x x

PRACTICE: 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: Waves 4.1 – Oscillations Start stretched and then release Start unstretched with a push right x x Phase difference

EXAMPLE: 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 Hooke’s law, show that the acceleration a of a mass-spring system is related to the spring’s displacement x by the proportion a  -x. SOLUTION: Recall Hooke’s law: F = -kx (see Topic 2-3). From Newton’s second law F = ma we then have ma = -kx or a = -(k / m) x. Thus, a  -x. Conditions for simple harmonic motion Topic 4: Waves 4.1 – Oscillations x

Conditions for simple harmonic motion Topic 4: Waves 4.1 – Oscillations EXAMPLE: 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) = ms -2. (c) a = -25x  -42 = - 25 x  x = -42 / -25 = m. x

PRACTICE: 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: Waves 4.1 – Oscillations a  -x definition of SHM x SHM: A type of vibratory motion in which acceleration of a body is directly proportional to its displacement and the acceleration is always directed towards the equilibrium (mean) position.

Conditions for simple harmonic motion The minus sign in Hooke’s 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), but that’s not just any kind of force, its specific! Topic 4: Waves 4.1 – Oscillations a  -x definition of SHM F x 0 F x x F and x oppose each other.

Conditions 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: Waves 4.1 – Oscillations x t

Qualitatively 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 E K and E P occurs. Topic 4: Waves 4.1 – Oscillations

Qualitatively 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 E K and E P occurs. Note that the sum of E K and E P is constant. Topic 4: Waves 4.1 – Oscillations x E K + E P = E T = CONST relation between E K and E P FYI * If friction and drag are both zero E T = CONST.

Qualitatively 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: Waves 4.1 – Oscillations Energy time x E K + E P = E T = CONST relation between E K and E P

EXAMPLE: 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 = 6.0  s. Then f = 1 / T = 1 / = 170 Hz. Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: 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 x MAX = 2.0 mm = 2.0  m.

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: 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.

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: 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: ● The period is 6 ms, the difference is 1 ms We see that it is T / 6 (= 360  / 6 = 60  = 2  / 6 rad).

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: The displacement x vs. time t for a system undergoing SHM is shown here. T= 3.0 sec. Sketch in red the velocity vs. time graph. SOLUTION: At the extreme compression/extension, v = 0. At x = 0, v = v MAX. The slope determines sign of v MAX. x-black v-red (different y-scale) t (+) ( -)(+)( -) (+) x v = 0 v = v MAX

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: 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-black v-red (different y-scale) t x v = 0 v = v MAX a-blue (different y-scale)

Sketching and interpreting graphs of simple harmonic motion examples EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 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)mv MAX 2 so that 4.0 = (1/ 2)(.125)v MAX 2  v MAX = 8.0 ms -1. Topic 4: Waves 4.1 – Oscillations x

Sketching and interpreting graphs of simple harmonic motion examples EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a kg mass on a spring. (b) Sketch E P and determine the total energy of the system. SOLUTION: Since E K + E P = E T = CONST, and since E P = 0 when E K = E K,MAX, it must be that E T = E K,MAX = 4.0 J. Thus the E P graph will be the “inverted” E K graph. Topic 4: Waves 4.1 – Oscillations x ETET EPEP EKEK

Sketching and interpreting graphs of simple harmonic motion examples EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a kg mass on a spring. (c) Determine the spring constant k of the spring. SOLUTION: Recall E P(spring) = (1/2)kx 2. Note that E K = 0 at x = x MAX = 2.0 cm. Thus E K + E P = E T = CONST  E T = 0 + (1/ 2)kx MAX 2 so that 4.0 = (1/ 2)k  k = Nm -1. Topic 4: Waves 4.1 – Oscillations x

Sketching and interpreting graphs of simple harmonic motion examples EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a kg mass on a spring. (c) Determine the acceleration of the mass at x = 1.0 cm. SOLUTION: From Hooke’s law, F = -kx we get F = (0.01) = -200 N. From F = ma we get -200 = 0.125a  a = ms -2. Topic 4: Waves 4.1 – Oscillations x

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: A 4.0-kg mass is placed on a spring’s 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

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: A 4.0-kg mass is placed on a spring’s 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 Hooke’s law: F = -kx. Pick any F and any x. Then use k = -F / x. Thus k = -(-5.0 N) / 1.0 m = 5.0 Nm -1. x F = -5.0 N x = 1.0 m

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: A 4.0-kg mass is placed on a spring’s 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 total energy of the system. SOLUTION: Use E T = (1/2)kx MAX 2. Then E T = (1/2)kx MAX 2 = (1/2)  5.0  = 10. J. x

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: A 4.0-kg mass is placed on a spring’s 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 maximum speed of the mass. SOLUTION: Use E T = (1/2)mv MAX = (1/2)  4.0  v MAX 2 v MAX = 2.2 ms -1. x

Sketching and interpreting graphs of simple harmonic motion examples Topic 4: Waves 4.1 – Oscillations EXAMPLE: A 4.0-kg mass is placed on a spring’s 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 speed of the mass when its displacement is 1.0 m. SOLUTION: Use E T = (1/2)mv 2 + (1/2)kx 2. Then 10. = (1/2)(4)v 2 + (1/2)(2)1 2 v = 2.1 ms -1. x