2. Chapter 14 Springs

3. A TRAMPOLINE exerts a restoring force on the jumper that is directly proportional to the average force required to displace the mat. Such restoring forces provide the driving forces necessary for objects that oscillate with simple harmonic motion.

4. Simple Harmonic Motion In periodic motion, a body repeats a certain motion indefinitely, always returning to its starting point after a constant time interval and then starting a new cycle. Every system that has this force exhibits a simple harmonic motion (SHM) and is called a simple harmonic oscillator.

5. Objectives: After finishing this unit, you should be able to: Write and apply Hooke’s Law for objects moving with simple harmonic motion. Describe the motion of pendulums and calculate the length required to produce a given frequency. Write and apply formulas for finding the frequency f, period T, velocity v, or acceleration a in terms of displacement x or time t.

6. Periodic Motion Simple periodic motion Simple periodic motion is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time. Amplitude A Period (seconds,s) Period, T, is the time for one complete oscillation. (seconds,s) Frequency Hertz (s -1 ) Frequency, f, is the number of complete oscillations per second. Hertz (s -1 )

7. Example 1: The suspended mass makes 30 complete oscillations in 15 s. What is the period and frequency of the motion? xF Period: T = 0.500 s Frequency: f = 2.00 Hz

8. GRAPH OF SHM The graph shown below depicts the up and down oscillation of the mass at the end of a spring. One complete cycle is from a to b, or from c to d. The time taken for one cycle is T, the period.

9. PERIOD AND FREQUENCY The period T of a body of mass m attached to a spring of force constant k UNITS: T in seconds f in Hz (s -1 )

10. A mass m attached to a spring executes SHM when the spring is pulled out and released. The spring’s PE becomes K as the mass begins to move, and the K of the mass becomes PE again as its momentum causes the spring to overshoot the equilibrium position and become compressed.

12. State the conditions necessary for simple harmonic motion. A spring wants to stay at its equilibrium or resting position. However, if a distorting force pulls down on the spring (when hanging an object from the spring, the distorting force is the weight of the object), the spring stretches to a point below the equilibrium position. The spring then creates a restoring force, which tries to bring the spring back to the equilibrium position.

13. Cont’d The distorting force and the restoring force are equal in magnitude and opposite in direction. F NET and the acceleration are always directed toward the equilibrium position.

14. Cont’d Applet showing the forces, displacement, and velocity of an object oscillating on a spring. https://ngsir.netfirms.com/englishhtm/Spring SHM.htm

15. Displacement Velocity Acceleration

16. Cont’d at equilibrium: – speed or velocity is at a maximum – displacement (x) is zero – acceleration is zero – F NET is zero (magnitude of Restoring Force = magnitude of Distorting Force which is the weight) – object continues to move due to inertia

17. Cont’d at endpoints: – speed or velocity is zero – displacement (x) is at a maximum equal to the amplitude – acceleration is at a maximum – restoring force is at a maximum F= -kx (hook’s law) – F NET is at a maximum

18. Hooke’s Law When a spring is stretched, there is a restoring force that is proportional to the displacement. F = -kx The spring constant k is a property of the spring given by: k = FxFx F x m

19. Work Done in Stretching a Spring F x m Work done ON the spring is positive; work BY spring is negative. To stretch spring from x 1 to x 2, work is: F (x) = kx x1x1 x2x2 F

20. Example 2: a) A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant? F 20 cm m The stretching force is the weight (W = mg) of the 4-kg mass: F = (4 kg)(9.8 m/s 2 ) = 39.2 N Now, from Hooke’s law, the force constant k of the spring is: k = = FxFx 39.2 N 0.2 m k = 196 N/m

21. Example 2: b) The mass m is now stretched a distance of 8 cm and held. What is the potential energy? (k = 196 N/m) F 8 cm m U = 0.627 J The potential energy is equal to the work done in stretching the spring: 0 PE

22. Displacement in SHM m x = 0x = +Ax = -A x Displacement is positive when the position is to the right of the equilibrium position (x = 0) and negative when located to the left. The maximum displacement is called the amplitude A.

23. Velocity in SHM m x = 0 x = +A x = -A v (+) Velocity is positive when moving to the right and negative when moving to the left. It is zero at the end points and a maximum at the midpoint in either direction (+ or -). v (-)

24. Acceleration in SHM m x = 0x = +Ax = -A Acceleration is in the direction of the restoring force. (a is positive when x is negative, and negative when x is positive.) Acceleration is a maximum at the end points and it is zero at the center of oscillation. +x -a -x +a

25. Example 3: A 2-kg mass hangs at the end of a spring whose constant is k = 400 N/m. The mass is displaced a distance of 12 cm and released. What is the acceleration at the instant the displacement is x = +7 cm? m +x+x a = -14.0 m/s 2 a +7 cm -14.0 m/s 2 Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s 2 (upward) independent of motion direction.

26. Example 4: What is the maximum acceleration for the 2-kg mass in the previous problem? (A = 12 cm, k = 400 N/m) m +x+x The maximum acceleration occurs when the restoring force is a maximum; i.e., when the stretch or compression of the spring is largest. F = ma = -kx x max =  A a max = ± 24.0 m/s 2 Maximum Acceleration:

27. 6.3 As shown in the figure, a long, light piece of spring steel is clamped at its lower end and a 2.0-kg ball is fastened to its top end. A horizontal force of 8.0 N is required to displace the ball 20 cm to one side as shown. Assume the system to undergo SHM when released. Find: a. The force constant of the spring F = 8 N x = 0.2 m m = 2 kg = 40 N/m b. Find the period with which the ball will vibrate back and forth. = 1.4 s SHM

28. 6.3 In a laboratory experiment a student is given a stopwatch, a wooden bob, and a piece of cord. He is then asked to determine the acceleration of gravity. If he constructs a simple pendulum of length 1 m and measures the period to be 2 s, what value will he obtain for g? L = 1 m T = 2 s = 9.86 m/s 2 SHM

29. Conservation of Energy The total mechanical energy (PE + KE) of a vibrating system is constant; i.e., it is the same at any point in the oscillating path. m x = 0x = +Ax = -A x v a For any two points A and B, we may write: ½mv A 2 + ½kx A 2 = ½mv B 2 + ½kx B 2

30. The Period and Frequency as a Function of a and x. For any body undergoing simple harmonic motion: The frequency and the period can be found if the displacement and acceleration are known. Note that the signs of a and x will always be opposite.

31. Period and Frequency as a Function of Mass and Spring Constant. For a vibrating body with an elastic restoring force: Recall that F = ma = -kx Recall that F = ma = -kx: The frequency f and the period T can be found if the spring constant k and mass m of the vibrating body are known. Use consistent SI units.

32. Example 5: The frictionless system shown below has a 2-kg mass attached to a spring (k = 400 N/m). The mass is displaced a distance of 20 cm to the right and released. What is the frequency of the motion? m x = 0 x = +0.2 m x v a x = -0.2 m f = 2.25 Hz

33. 6.2 A 200-g mass vibrates horizontally without friction at the end of a horizontal spring for which k = 7.0 N/m. The mass is displaced 5.0 cm from equilibrium and released. Find: a. Maximum speed m = 0.2 kg k = 7 N/m x o = 0.05 m v max is at x = 0 then: v = 0.295 m/s COE

34. b. Speed when it is 3.0 cm from equilibrium. x = 0.03 m v = 0.236 m/s

35. c. What is the acceleration in each of these cases? F = ma = - kx a. x = 0 therefore a = 0 b. x = 0.03 m therefore = - 1.05 m/s 2

36. Chapter 10 The Pendulum A Foucault pendulum consists of a tall pendulum free to oscillate in any vertical plane. The direction along which the pendulum swings rotates with time because of Earth's daily rotationpendulumoscillate

37. The Simple Pendulum The period of a simple pendulum is given by: mg L For small angles 

38. For small displacements a pendulum obeys SHM. Its period is: Simple Pendulum The period and frequency DO NOT depend on the mass.

39. For the motion shown in the figure, find: a. Amplitude b. Period c. Frequency a. Amplitude: maximum displacement from equilibrium A = +- 0.75 cm b. T = time for one complete cycle T = 0.2 s c. f = 1/T = 1/0.2 = 5 Hz

40. Example 6. What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)? L L = 0.993 m

41. Chapter 10 Simple Harmonic Motion 1. A) 84,810 N/mB) 288.3 N 2. 666.6 N/m 3. A) 30 N/mB) 9 N 4. 35,092 N/m 5. A) 0.45 mB) 3.31 Hz C) 1.49 m/s 6. 1524 N/m4.29 kg 7. A) 46.94 JB) 55.9 m/s 8. 1.5 m/s 9. 0.556 m/s 10. 30 rad/s

42. DAMPED HARMONIC MOTION A system undergoing SHM will exhibit damping. Damping is the loss of mechanical energy as the amplitude of motion gradually decreases. In the mechanical systems studied in the previous sections, the losses are generally due to air resistance and internal friction and the energy is transformed into heat. For the amplitude of the motion to remain constant, it is necessary to add enough energy each second to offset the energy losses due to damping.

43. In many instances damping is a desired effect. For example, shock absorbers in a car remove unwanted vibration.

44. FORCED VIBRATIONS: RESONANCE An object subjected to an external oscillatory force tends to vibrate. The vibrations that result are called forced vibrations. These vibrations have the same frequency as the external force and not the natural frequency of the object. If the external forced vibrations have the same frequency as the natural frequency of the object, the amplitude of vibration increases and the object exhibits resonance. The natural frequency (or frequencies) at which resonance occurs is called the resonant frequency. EXAMPLES OF RESONANCE

46. 18. A person bounces up and down on a trampoline, while always staying in contact with it. The motion is SHM an it takes 1.90 s to complete one cycle. The height of each bounce is 45.0cm. Determine (a) the amplitude (b) the angular frequency (c) What is the maximum speed attained by the person?

47. Example 9: A man enters a tall tower, needing to know its height h. He notes that a long pendulum extends from the roof almost to the ground and that its period is 15.5 s. (a) How tall is the tower? a) 59.7 m

48. (b) If this pendulum is taken to the Moon, where the free-fall acceleration is 1.67 m/s 2, what is the period of the pendulum there? b) 37.6 s