1. S H M a n d W a v e s B a s i c s

2. T h e O s c i l l a t o r When displaced from its vertical equilibrium position, this plastic ruler oscillates back and forth because of the restoring force opposing displacement. When the ruler is on the left, there is a force to the right and vice versa.

3. T h e O s c i l l a t o r a. The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. b. The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. c. The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. d. Now the ruler has momentum to the left. e. In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself

4. H o o k e ’ s L a w Hooke’s Law:Fs = -kx K is the spring constant and x is displacement Negative sign indicates that the force is opposite to the displacement from the equilibrium position. Motion that obeys Hooke’s Law is called simple harmonic motion.

5. H o o k e ’ s L a w a. A graph of absolute value of the restoring force versus displacement is displayed. The fact that the graph is a straight line means that the system obeys Hooke’s law. The slope of the graph is the force constant k. b. The data in the graph were generated by measuring the displacement of a spring from equilibrium while supporting various weights. The restoring force equals the weight supported, if the mass is stationary.

6. H o o k e ’ s L a w A graph of applied force versus distance for the deformation of a system that can be described by Hooke’s law is displayed. The work done on the system equals the area under the graph or the area under the graph or the area of the triangle.

7. B l o c k a n d S p r i n g S y s t e m A block and spring system disturbed from equilibrium will obey simple harmonic motion The time it takes to complete one period of oscillation is T. If the block is released from maximum displacement at t = 0, at what times (in terms of T) is the block at equilibrium? A. T B. T/4 C. T/2 D. 3T/4

8. C o n s e r v a t i o n o f E n e r g y The total potential energy stored in a spring can be expressed as PE = ½ kx 2 Total Mechanical Energy at any point in the motion of a mass and spring system will be the potential energy and the kinetic energy. Kinetic energy is maximum at equilibrium. Potential energy is zero at equilibrium. TME = PE + KE = ½ kx 2 + ½ mv 2 Since at maximum dispacement (+X or -X) the kinetic energy is zero, then the total mechanical energy stored in a mass-and-spring can be expressed by: TME = ½ k|X| 2 which means that the toal energy is proportional to maximum dispacement.

9. E x a m p l e P r o b l e m 1 A 0.50 kg object is attached to a spring of spring constant 20 N/m along a horizontal frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. a)What is the total energy of the system? b)What is the amplitude? c)At what location are the values for the potential and kinetic energies the same?

10. E x a m p l e P r o b l e m 2 An object is attached to a spring of spring constant 60 N/m along a horizontal, frictionless surface. The spring is initially stretched by a force of 5.0 N on the object and let go. It takes the object 0.50 s to get back to its equilibrium position after its release. a)What is the amplitude? b)What is the period? c)What is the frequency?

11. C o n c e p t C h e c k The maximum kinetic energy of a spring-mass system in SHM is equal to: a.A b.A 2 c.kA d.½ kA 2

12. C o n c e p t C h e c k If the amplitude of an object in SHM is doubled: a.how is the energy affected? a.how is the maximum speed affected?

13. D e s c r i b i n g S H M W i t h f o r m u l a s!

14. N e a t D i a g r a m s Simple harmonic motion is understood in relation to circular motion. The equations you will use can be thought of as the “shadows” of a point on an oscillating object.

15. N e a t D i a g r a m s

16. x = A cos  t

17. N e a t D i a g r a m s x = A sin  t

18. S H M E q u a t i o n s y = A sin (  t +  ) y is the vertical displacement (in meters) A is the amplitude (in meters)  is the angular frequency of motion (in rad/sec)  is the phase constant (in rad)  = 2  f = 2  /T The phase constant, , is determined by the initial displacement and velocity direction. It determines whether a sine or cosine function best describes the simple harmonic motion.

19. S i n e a n d C o s i n e

20. If at t = 0 displacement is zero and velocity is positive, then  = 0 and motion can be described by y = A sin  t. If at t = 0 displacement is +A and velocity is negative, then  = π /2 and motion can be described by y = A sin (  t + π /2) = A cos  t. If at t = 0 displacement is zero and velocity is negative, then  = π and motion can be described by y = A sin (  t + π ) = -A sin  t. If at t = 0 displacement is -A and velocity is positive, then  = 3 π /2 and motion can be described by y = A sin (  t + 3 π /s) = -A cos  t.

21. C o n c e p t C h e c k What would be the equation for the motion to the left? A) x = X sin  t. B) x = X cos  t. C) x = -X sin  t. D) x = -X cos  t.

22. C o n c e p t C h e c k In what phase is the oscillator to the left? A)  = 0 B)  = π /2 C)  = π D)  = 3 π /2

23. S H M E q u a t i o n s Q U I C K C A L C U L U S If x = A sin (  t +  ) Then dx/dt =  A cos (  t +  ) And d 2 x/dt 2 = -  2 A sin (  t +  ) Since A sin (  t +  ) = x we can substitute this back into the d 2 x/dt 2 equation. The second derivative of displacement, d 2 x/dt 2 is the acceleration. Therefore, a = -  2 x v =  A cos (  t +  ) a = -  2 A sin (  t +  ) = -  2 x

24. D i s p l a c e m e n t, V e l o c i t y a n d A c c e l e r a t I o n

25. A mass is lifted to +X displacement and then released. It enters simple harmonic motion since the restoring force of the spring serves to pull it back. Velocity is  / 2 out of phase with displacement. Acceleration is  out of phase with displacement.

26. V e r t i c a l M a s s a n d S p r i n g N e t F o r c e Does gravity matter when a mass and spring system is vertical rather than horizontal? F spring = - kx F spring = - k (x + ∆x g ) F spring = - kx – k∆x g If force mg is applied to a spring it will cause a displacement of Fg = -k∆x, so ∆x = mg / k and F spring = - kx – mg F net = F spring + mg F net = - kx - mg + mg F net = - kx

27. D a m p e n e d H a r m o n i c M o t i o n

28. S H M E q u a t i o n s The period of an object in simple harmonic motion depends only on the mass of the object an on the spring constant. Unlike a pendulum, gravity plays no role. The greater the mass, the longer the period and the smaller the frequency. A N D The greater the spring constant, the shorter the period and greater the frequency

29. A S i m p l e P e n d u l u m A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of −mg sin Ɵ toward the equilibrium position—that is, a restoring force.

30. A S i m p l e P e n d u l u m The period of the simple pendulum is independent of the mass of the bob and the amplitude of the oscillations. Where L is the length of the string and g is the acceleration due to gravity