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PHYS 20 LESSONS Unit 6: Simple Harmonic Motion Mechanical Waves Lesson 5: Pendulum Motion as SHM

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Reading Segment #1: Kinematics and Dynamics of Pendulum Motion To prepare for this section, please read: Unit 6: p. 13

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Kinematics and Dynamics of Pendulum Motion You will soon discover that pendulum motion can be equivalent to SHM. However, before you can see this, you need to understand the nature of the forces and acceleration during pendulum motion.

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Pendulum Motion Applet To analyze the velocity, acceleration, and forces during pendulum motion, click on the following link: Instructions: Click on mass and move pendulum to an angle Click on “Play” and it will oscillate Click on Button 1 to see the forces Click on Button 2 to see the velocity and acceleration anaxis/apply/physicalpendulum/applet.html

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Consider moving a pendulum to a deviation angle and then releasing it. rest

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FTFT F g = mg rest First, we draw the forces on the mass: The force of gravity always acts downward. Tension force acts along the string and towards the pivot (i.e. away from the mass).

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FTFT F g = mg rest y x Next, we establish the y-axis along the string. Notice that F g is the diagonal force here.

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FTFT F g = mg rest F g y F g x Next, we draw the x- and y-components of F g.

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FTFT F g = mg rest F g y F g x We can determine each component as follows:

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FTFT F g = mg mg cos mg sin rest The two “vertical” forces are equal but opposite in this position, and so there is no acceleration along the y-axis.

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FTFT F g = mg mg cos mg sin rest a The net force is mg sin , and so the pendulum’s acceleration is entirely along the x-axis. That is, the acceleration is tangential.

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FTFT F g = mg mg cos mg sin (F s ) rest a mg sin acts in a direction opposite (approximately) to the displacement x, much like a spring force. It is trying to return the mass to the equilibrium position (i.e. a restoring force). x

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FTFT F g = mg mg cos mg sin (F s ) xrest av The pendulum speeds up until it reaches equilibrium position, where it has achieved maximum speed. Equilibrium: Max speed

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FTFT v FgFg Again, we draw the forces on the mass: The force of gravity always acts downward. Tension force acts along the string and towards the pivot (i.e. away from the mass). FTFT F g = mg mg cos mg sin (F s ) xrest a

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FTFT v FgFg At this moment, the mass is in uniform circular motion. The acceleration is entirely vertical and towards the centre, and so it is centripetal. acac FTFT F g = mg mg cos mg sin (F s ) xrest a

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FTFT F g = mg mg cos xrest a acac FTFT v FgFg mg sin (F s ) It is very similar when it moves back up. mg sin acts to slow down the mass until it comes to rest at its maximum displacement. FTFT F g = mg mg cos mg sin (F s ) xrest a

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Reading Segment #2: Pendulum Motion as SHM To prepare for this section, please read: Unit 6: pp

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Pendulum Motion as SHM Consider a pendulum deviated at an angle . x F s = mg sin L Again, we consider mg sin as being equivalent to the restoring force F s.

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x F s = mg sin L For the triangle shown:

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x F s = mg sin L Combining this with we discover that or

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x F s = mg sin L Since m, g, and L are constant for the pendulum, it follows that Thus, the “spring force” has a direct relationship with the displacement x.

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x F s = mg sin L However, technically speaking, for the pendulum to be in SHM, the restoring force must be directly proportional to the displacement along the arc. This is not actually true, so a pendulum does not undergo SHM. a

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< 15 x a F s = mg sin L However, if the angle of deflection is less than 15 , then we can safely assume that the displacement is equal to the arclength. Under this condition, F s x and the pendulum undergoes SHM. a

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Period of Pendulum Motion As we have seen, for small amplitudes ( < 15 ), we can consider mg sin as a restoring force, much like F s. x F s = mg sin L As a result, since this is equivalent to a mass-spring system, we can use Hooke’s Law (F s = k x) to create a formula.

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x F s = mg sin L Starting with the formula we derived earlier: We now substitute Hooke’s Law F s = k x. This leaves us with

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x F s = mg sin L Thus, or,

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If we substitute this into the period formula for a mass-spring system, we discover the following: but, or So,

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Equation: The period T of a pendulum is calculated using the formula SI Units: s where L is the length of the pendulum (in m) g is the magnitude of the acceleration due to gravity (in m/s 2 )

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Note: L is measured from the axis of rotation to the centre of the mass. For this formula to be used, < 15 L axis

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Example A pendulum oscillates at a frequency of 0.42 Hz. Determine its length. Try this example on your own first. Then, check out the solution.

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= 2.38 s Find period:

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Find length:

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= 1.4 m

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Practice Problems Try these problems in the Physics 20 Workbook: Unit 6 p. 66 #1 - 4

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Reading Segment #3: Energetics of Pendulum Motion To prepare for this section, please read: Unit 6: p. 15

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Energetics of Pendulum Motion Again, consider moving a pendulum to a deviation angle and then releasing it. rest

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Pendulum Motion Applet Analysis of Pendulum Motion To analyze the energy of pendulum motion, click on the following link: Note: Click on “show” if you wish to see the acceleration vector and its components.

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rest = A At the position of maximum displacement, the angle of deviation represents the amplitude of oscillation. i.e. max = A

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rest A h max Equilibrium position At maximum displacement, the mass is at a maximum height (relative to the equilibrium position). Thus, it has maximum gravitational potential energy. Max Ep g Ref h: h = 0

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rest A h max Equilibrium position The mass is also at rest, and so it has no kinetic energy. Max Ep g E k = 0

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rest A h max Equilibrium position The total mechanical energy at this position is Max Ep g E k = 0 The mass only has gravitational potential energy

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rest v max A h max Equilibrium position Max Ep g E k = 0 As the pendulum falls, it speeds up. When it reaches the equilibrium position, the mass has attained its maximum speed.

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rest v max A h max Equilibrium position Max Ep g E k = 0 Since the speed is at a maximum, the mass has maximum kinetic energy. Max E k

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rest v max A h max Equilibrium position Max Ep g E k = 0 However, since the mass has reached the equilibrium position, it has no gravitational potential energy. Ref h: h = 0 Max E k Ep g = 0

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rest v max A h max Equilibrium position Max Ep g E k = 0 Max E k Ep g = 0 The total mechanical energy at the equilibrium position is The mass only has kinetic energy

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rest v max A h max A Equilibrium position Max Ep g E k = 0 Max E k Ep g = 0 Max Ep g E k = 0 rest It is similar when it moves up to maximum displacement. The mass slows down until it comes to rest at its maximum height. At this position, it has no kinetic energy and maximum gravitational potential energy. Maximum displacement

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Energy Formulas of Pendulum Motion To derive an energy formula for pendulum motion, we compare maximum displacement with equilibrium position. rest v max A h max Equilibrium position Max Ep g E k = 0 Max E k Ep g = 0

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If this is an ideal pendulum, then there is no energy lost due to friction or air resistance. rest v max A h max Equilibrium position Max Ep g E k = 0 Max E k Ep g = 0 Thus, the total mechanical energy must remain constant. i.e. E m (top) = E m (bottom)

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rest v max A h max Equilibrium position Max Ep g E k = 0 Max E k Ep g = 0 E m (top) = E m (bottom)

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rest v max A h max Equilibrium position Max Ep g E k = 0 Max E k Ep g = 0 E m (top) = E m (bottom)

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rest v max A h max Equilibrium position Max Ep g E k = 0 Max E k Ep g = 0 E m (top) = E m (bottom) This is not on the formula sheet, and so you must derive this each time.

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Example On a different planet, an 80 cm long pendulum oscillates with a period of 2.9 seconds. If this pendulum reaches a maximum height of 1.5 cm (above its equilibrium position), then determine its maximum speed. Assume no air resistance or friction. Try this example on your own first. Then, check out the solution.

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Find the acceleration due to gravity:

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= m/s 2

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Find the maximum speed: E m (top) = E m (bottom) rest v max A h max Equilibrium position Max Ep g E k = 0 Max E k Ep g = 0

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E m (top) = E m (bottom)

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= 0.34 m/s

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Practice Problems Try these problems in the Physics 20 Workbook: Unit 6 p. 6 #11

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