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10.2 Simple Harmonic Motion and the Reference Circle

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Presentation on theme: "10.2 Simple Harmonic Motion and the Reference Circle"— Presentation transcript:

1 10.2 Simple Harmonic Motion and the Reference Circle
DISPLACEMENT - Reference circle circulating the ball whose shadow is being projected on the film.

2 10.2 Simple Harmonic Motion and the Reference Circle
A – radius of the circle 𝒙 – displacement of ball from equilibrium position. The displacement, x, of the shadow is just the projection of the radius, A onto the x-axis on the screen Theta is angular displacement: angular speed X time

3 10.2 Simple Harmonic Motion and the Reference Circle
amplitude A: the maximum displacement period T: the time required to complete one cycle frequency f: the number of cycles per second (measured in Hz)

4 10.2 Simple Harmonic Motion and the Reference Circle
DISPLACEMENT - Reference circle circulating the ball whose shadow is being projected on the film.

5 Velocity Using the reference circle model to determine the velocity of an object: 𝑣𝑥 – 𝑥 component of 𝑣𝑡 = −𝑣𝑡𝑠𝑖𝑛𝜃 Negative sign indicates direction of 𝑣𝑥 Where: 𝜃=𝜔𝑡 𝑣𝑡 = 𝑟𝜔 (since r = A) then: 𝑣𝑡 = 𝐴𝜔 Therefore the velocity in SHM is given by: (ω in rad/s)

6 Oscillating Spring/mass Systems
Maximum velocity is experienced at the equilibrium point, where Felastic = 0. At maximum displacement, the velocity is zero. Use bouncing spring and mass to teach following. (0) Where is KE going to be greatest? Why does mass oscillate? Where is the force the greatest? © 2002 HRW

7 10.2 Simple Harmonic Motion and the Reference Circle
VELOCITY

8 Acceleration Using the reference circle model to determine the acceleration of an object: The centripetal acceleration of the ball points towards the center, 𝑎𝑐 𝑎𝑥 – 𝑥 component of 𝑎𝑐 = −𝑎𝑐𝑐𝑜𝑠𝜃 Negative sign indicates direction of 𝑎𝑥 Where: 𝑎𝑐=𝑟𝜔2 (since r = A) then: 𝑎𝑐 = 𝐴𝜔2 Therefore the acceleration in SHM is given by: (ω in rad/s)

9 10.2 Simple Harmonic Motion and the Reference Circle
ACCELERATION

10 Consider the graph shown for the position of a ball attached to a spring as it oscillates in simple harmonic motion. At which of the following times is the ball at its equilibrium position? a) 0 s only b) 2 s only c) 4 s only d) at 0 s and 8 s e) at 0 s, 4 s, and 8 s

11 Which one of the following statements concerning the total mechanical energy of a harmonic oscillator at a particular point in its motion is true? a) The mechanical energy depends on the acceleration at that point. b) The mechanical energy depends on the velocity at that point. c) The mechanical energy depends on the position of that point. d) The mechanical energy does not vary during the motion. e) The mechanical energy is equal to zero joules if the point is the equilibrium point.

12 10.3 Energy and Simple Harmonic Motion
A stretched or compressed spring has elastic potential energy and therefore it can do work.

13 10.3 Energy and Simple Harmonic Motion
From the work equation: 𝑊 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 = 𝐹𝑐𝑜𝑠𝜃 𝑑 From the diagram below: 𝑑= 𝑥 𝑜 − 𝑥 𝑓 The magnitude of the spring force given by 𝐹 𝑥 =−𝑘𝑥 is not constant and changes from 𝑘𝑥 𝑜 to 𝑘𝑥 𝑓 therefore the magnitude of the average force is: 𝐹 𝑥 = 1 2 ( 𝑘𝑥 𝑜 + 𝑘𝑥 𝑓 ) The magnitude of the force if not going to be constant, it changes from kxo  kxf The average force becomes ½(kxo+kxf)

14 10.3 Energy and Simple Harmonic Motion
The elastic potential energy is the energy that a spring has by virtue of being stretched or compressed. For an ideal spring, the elastic potential energy is SI unit: Joule (J)

15 Example 5: Adding a Mass to a Simple Harmonic Oscillator
10.3 Energy and Simple Harmonic Motion Example 5: Adding a Mass to a Simple Harmonic Oscillator A 0.20-kg ball is attached to a vertical spring. The spring constant is 28 N/m. When released from rest, how far does the ball fall before being brought to a momentary stop by the spring? G: m=0.20 kg, k=28N/m, hf=0m, vo=0m/s,vf=0m/s U: ho=? E: S: Spring extension: yf = ho


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