Simple Harmonic Motion

1. Simple Harmonic Motion AP Physics B

2. Simple Harmonic Motion Back and forth motion that is caused by a force that is directly proportional to the displacement. The displacement centers around an equilibrium position.

3. Springs – Hooke’s Law One of the simplest type of simple harmonic motion is called Hooke's Law. This is primarily in reference to SPRINGS. The negative sign only tells us that “F” is what is called a RESTORING FORCE, in that it works in the OPPOSITE direction of the displacement.

4. Hooke’s Law Common formulas which are set equal to Hooke's law are N.S.L. and weight

5. Example A load of 50 N attached to a spring hanging vertically stretches the spring 5.0 cm. The spring is now placed horizontally on a table and stretched 11.0 cm. What force is required to stretch the spring this amount? 110 N 1000 N/m

6. Hooke’s Law from a Graphical Point of View Suppose we had the following data: k =120 N/m

7. We have seen F vs. x Before!!!! Work or ENERGY = FDx Since WORK or ENERGY is the AREA, we must get some type of energy when we compress or elongate the spring. This energy is the AREA under the line! Area = ELASTIC POTENTIAL ENERGY Since we STORE energy when the spring is compressed and elongated it classifies itself as a “type” of POTENTIAL ENERGY, Us. In this case, it is called ELASTIC POTENTIAL ENERGY.

8. Elastic Potential Energy The graph of F vs.x for a spring that is IDEAL in nature will always produce a line with a positive linear slope. Thus the area under the line will always be represented as a triangle. NOTE: Keep in mind that this can be applied to WORK or can be conserved with any other type of energy.

10. Conservation of Energy in Springs

11. Example A slingshot consists of a light leather cup, containing a stone, that is pulled back against 2 rubber bands. It takes a force of 30 N to stretch the bands 1.0 cm (a) What is the potential energy stored in the bands when a 50.0 g stone is placed in the cup and pulled back 0.20 m from the equilibrium position? (b) With what speed does it leave the slingshot? 3000 N/m 60 J 49 m/s

12. Springs are like Waves and Circles The amplitude, A, of a wave is the same as the displacement ,x, of a spring. Both are in meters. CREST Equilibrium Line Period, T, is the time for one revolution or in the case of springs the time for ONE COMPLETE oscillation (One crest and trough). Oscillations could also be called vibrations and cycles. In the wave above we have 1.75 cycles or waves or vibrations or oscillations. Trough Ts=sec/cycle. Let’s assume that the wave crosses the equilibrium line in one second intervals. T =3.5 seconds/1.75 cycles. T = 2 sec.

13. Frequency The FREQUENCY of a wave is the inverse of the PERIOD. That means that the frequency is the #cycles per sec. The commonly used unit is HERTZ(HZ).

14. SHM and Uniform Circular Motion Springs and Waves behave very similar to objects that move in circles. The radius of the circle is symbolic of the displacement, x, of a spring or the amplitude, A, of a wave.

15. SHM and Uniform Circular Motion • The radius of a circle is symbolic of the amplitude of a wave. • Energy is conserved as the elastic potential energy in a spring can be converted into kinetic energy. Once again the displacement of a spring is symbolic of the amplitude of a wave • Since BOTH algebraic expressions have the ratio of the Amplitude to the velocity we can set them equal to each other. • This derives the PERIOD of a SPRING.

16. Example A 200 g mass is attached to a spring and executes simple harmonic motion with a period of 0.25 s If the total energy of the system is 2.0 J, find the (a) force constant of the spring (b) the amplitude of the motion 126.3 N/m 0.18 m

17. The kinetic energy of a vibrating object = ½ mv2. The maximum kinetic energy = ½ mvmax2 = ½ m2A2 (since vmax = A). This makes clear that the energy of an oscillator is proportional to the square of its amplitude, A.

18. Pendulums

19. Pendulums Pendulums, like springs, oscillate back and forth exhibiting simple harmonic behavior. A shadow projector would show a pendulum moving in synchronization with a circle. Here, the angular amplitude is equal to the radius of a circle.

20. Pendulums Consider the FBD for a pendulum. Here we have the weight and tension. Even though the weight isn’t at an angle let’s draw an axis along the tension. q mgcosq q mgsinq

21. Pendulums What is x? It is the amplitude! In the picture to the left, it represents the chord from where it was released to the bottom of the swing (equilibrium position).

22. Example A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 9.40 s. What is the height of the tower? L = Height = 21.93 m

23. Section 11.1 to 11.3 AP Physics Periodic Motion

24. Periodic motion Motion that repeats in a regular cycle is known as periodic motion. Examples include a mass attached to a spring moving up and down, the motion of a pendulum (a weight on a string or rod), and a vibrating string such as the strings on a musical instrument. There will be one point in the motion where the object is in equilibrium, and the net force is zero. When the object that is in motion moves away from this equilibrium position a restoring force will exist that will try to pull the object back to equilibrium.

25. Simple harmonic motion If the restoring force, FS, is directly proportional to the displacement from equilibrium, the motion produced is called simple harmonic motion (SHM). Two important properties describe simple harmonic motion, period and amplitude. Period, T, measured in seconds, is the time needed to complete one full cycle. It is the seconds per cycle. Amplitude, A, measured in meters, is the maximumdisplacement of the object from the equilibrium position.

26. Frequency Frequency units Frequency (f) — the number of cycles that occur in one second. Many physics texts use nu: ν. Since period is seconds per wave, and frequency is waves per second, they are reciprocal values: 1 f = T Frequency measures cycles per second, but “cycles” is a count, and is therefore unitless. The standard units are therefore 1/s or s–1. Another acceptable SI name for the unit is hertz (Hz). 1 Hz = 1/s

27. Mass on a spring We saw the relationship between the force applied and the displacement in Section 6.4. To review: The displacement, x, is directly proportional to the applied force, FP. FP∝ x A constant of proportionality lets us write the equation: FP = kx k is the stiffnessconstant.

28. As you push or pull on the spring, the spring supplies an equal reaction force, FS. Recall that FS , tries to pull the mass back to equilibrium, and is called the restoringforce. Hooke’s Law FS = –kx This equation for an elastic object deformed from equilibrium is known as Hooke’slaw. FS Fg

29. As the equation and the previous image show, to create more displacement, more force must be supplied, and the restoring force gets larger as well. F = ½(0 + kx) W = Fx = (½kx)x Therefore, the force is a non-constant, but linearly increasing force. The work done will be the area under the F–x graph. The work equals the average force times the displacement. W = ½kx2 Therefore W PEelastic = US = ½kx2 PE in the textbook US on the AP test sheet

30. Mechanical energy of an oscillator The total mechanical energy of the oscillator is: Emechanical = K + US Emechanical = ½mv2 + ½kx2 x is the position at any point in the motion, and v is the velocity at that point. If the total energy is calculated from the maximum displacement, we can find the total energy in terms of amplitude, A. Velocity at A is zero, so: Emechanical = ½kA2 The total mechanical energy of a simple harmonic oscillator is proportional to the square of the amplitude.

31. Therefore, we can write a single equation for mechanical energy: ½mv2 + ½kx2 = ½kA2 Solving this for v2, we get: k k x2 (A2 – x2) 1 – v2 = = A2 m m A2 At equilibrium, the system reaches its maximum speed, vmax. Therefore, K at this point equals E, since US = 0. k and ½mvmax2 = ½kA2 vmax2 A2 = m Substituting this into the v2 equation gives: ½ x2 ± v = vmax 1 – A2

32. The last equation is exactly the same as the speed of a revolving object as seen from the side! The maximum speed at x0 is the same as the speed of the object revolving in a circle. The radius of the circle equals A. x0 circum. 2πA vmax = = period T Solving for period: 2πA T = vmax Substitute in the following value for vmax from the previous frame: k vmax A = m

33. SHM in a weighted spring m T = 2π k So, the period of a mass spring system is given by: FS Notice that period does not depend on A! It canceled. –A FS x0 a +A Fg Fg

34. The position, x, of the oscillator at any point is given by: x = A cosθ x0 A θ x θ Since ω = θ = ωt x t x = A cos ωt Recall from section 8-1 that: ω = 2πf x = A cos 2πft Since T is the reciprocal of f: 2πt x = A cos T Note: motion will repeat after t = T.

35. Graphing x as a function of t The SHM function equation can be graphed for position as a function of time: sinusoidal function

36. Sinusoidal functions Note that we usually start the spring away from the equilibrium position at A or –A. This gives us the cosine equation: x = A cos ωt 2πt x = A cos T Timing from equilibrium gives a sine equation. Both are sinusoidal. x = A sin ωt 2πt 2T x = A sin T