2. Simple Harmonic Motion

3. 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 )

4. 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

5. Simple Harmonic Motion, SHM Simple harmonic motion is periodic motion in the absence of friction and produced by a restoring force that is directly proportional to the displacement and oppositely directed. A restoring force, F, acts in the direction opposite the displacement of the oscillating body. F = -kx A restoring force, F, acts in the direction opposite the displacement of the oscillating body. F = -kx xF

6. 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

7. The period and frequency of a wave the period T of a wave is the amount of time it takes to go through 1 cyclethe period T of a wave is the amount of time it takes to go through 1 cycle the frequency f is the number of cycles per secondthe frequency f is the number of cycles per second –the unit of a cycle-per-second is commonly referred to as a hertz (Hz), after Heinrich Hertz (1847-1894), who discovered radio waves. frequency and period are related as follows:frequency and period are related as follows: Since a cycle is  radians, the relationship between frequency and angular frequency is:Since a cycle is  radians, the relationship between frequency and angular frequency is:

8. The reference circle compares the circular motion of an object with its horizontal projection.  2  f The Reference Circle x = Horizontal displacement. A = Amplitude (x max ).  = Reference angle.

9. Velocity in SHM The velocity (v) of an oscillating body at any instant is the horizontal component of its tangential velocity (v T ). v T =  R =  A;  2  f v = -v T sin  ;  =  t v = -  A sin  t v = -2  f A sin 2  f  t

10. The acceleration (a) of an oscillating body at any instant is the horizontal component of its centripetal acceleration ( a c ). Acceleration Reference Circle a = -a c cos  = -a c cos(  t) R = A a = -    cos(  t)

11. The Period and Frequency as a Function of a and x. For any body undergoing simple harmonic motion: Since a = -4   f 2 x and T = 1/f 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.

12. Springs and SHM Attach an object of mass m to the end of a spring, pull it out to a distance A, and let it go from rest. The object will then undergo simple harmonic motion:Attach an object of mass m to the end of a spring, pull it out to a distance A, and let it go from rest. The object will then undergo simple harmonic motion: –Use Newton’s 2nd law, together with Hooke’s law, and the above description of the acceleration to find:

13. 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.

14. Example 6: 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

15. Example 6 (Cont.): Suppose the 2-kg mass of the previous problem is displaced 20 cm and released (k = 400 N/m). What is the maximum acceleration? (f = 2.25 Hz) m x = 0 x = +0.2 m x v a x = -0.2 m Acceleration is a maximum when x =  A a =  40 m/s 2

16. Example 6: The 2-kg mass of the previous example is displaced initially at x = 20 cm and released. What is the velocity 2.69 s after release? (Recall that f = 2.25 Hz.) m x = 0 x = +0.2 m x v a x = -0.2 m v = -0.916 m/s v = -2  f A sin 2  f  t (Note:  in rads) The minus sign means it is moving to the left.

17. Example 7: At what time will the 2-kg mass be located 12 cm to the left of x = 0? (A = 20 cm, f = 2.25 Hz) m x = 0 x = +0.2 m x v a x = -0.2 m t = 0.157 s -0.12 m

18. To Be Continued…

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

20. Example 2: 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FxxFFxx  0.2 m k = 196 N/m

21. Example 2(cont.: 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

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.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 -).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 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. Acceleration vs. Displacement m x = 0x = +Ax = -A x v a Given the spring constant, the displacement, and the mass, the acceleration can be found from: or Note: Acceleration is always opposite to displacement.

26. 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 Note: When the displacement is +7 cm (downward), the acceleration is -14.0 m/s 2 (upward) independent of motion direction.

27. 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:

28. Conservation of Energy The total mechanical energy (U + K) 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

29. Energy of a Vibrating System: m x = 0x = +Ax = -A x v a At any other point: U + K = ½mv 2 + ½kx 2At any other point: U + K = ½mv 2 + ½kx 2 U + K = ½kA 2 x =  A and v = 0. At points A and B, the velocity is zero and the acceleration is a maximum. The total energy is:At points A and B, the velocity is zero and the acceleration is a maximum. The total energy is: A B

30. Velocity as Function of Position. m x = 0x = +Ax = -A x v a v max when x = 0:

31. Example 5: A 2-kg mass hangs at the end of a spring whose constant is k = 800 N/m. The mass is displaced a distance of 10 cm and released. What is the velocity at the instant the displacement is x = +6 cm? m +x+x ½mv 2 + ½kx 2 = ½kA 2 v = ±1.60 m/s

32. Example 5 (Cont.): What is the maximum velocity for the previous problem? (A = 10 cm, k = 800 N/m, m = 2 kg.) m +x+x ½mv 2 + ½kx 2 = ½kA 2 v = ± 2.00 m/s 0 The velocity is maximum when x = 0:

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

34. Example 8. Example 8. 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

35. To Be Continued…