2. Simple Harmonic Motion Wenny Maulina

3. Simple harmonic motion  Simple harmonic motion (SHM) Solution: What is SHM? A simple harmonic motion is the motion of an oscillating system which satisfies the following condition: 1.Motion is about an equilibrium position at which point no net force acts on the system. 2.The restoring force is proportional to and oppositely directed to the displacement. 3.Motion is periodic. t=0 t=-  Acos  f=  By Dr. Dan Russell, Kettering University      

4. 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

5. Oscillations of a Spring Hooke’s Law states F s =  kx F s is the restoring force. It is always directed toward the equilibrium position. Therefore, it is always opposite the displacement from equilibrium. k is the force (spring) constant. x is the displacement.

6. Oscillations of a Spring  In a, the block is displaced to the right of x = 0. The position is positive. The restoring force is directed to the left (negative).  In b, the block is at the equilibrium position. x = 0 The spring is neither stretched nor compressed. The force is 0.  In c, the block is displaced to the left of x = 0. The position is negative. The restoring force is directed to the right (positive).

7. 6 Example A 0.42-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the spring stretches by 2.1 cm relative to its length. When the block is released, it moves with an acceleration of 9.0 m/s 2. What is the spring constant of the spring?

8. 7 2.1cm kx = ma

9. Displacement in SHM 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. m x = 0x = +Ax = -A x

10. 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. v (-)

11. 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

12. 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.

13. Simple harmonic motion  Displacement, velocity and acceleration in SHM Displacement Velocity Acceleration

14. The diaphragm of a loudspeaker moves back and forth in simple harmonic motion to create sound. The frequency of the motion is f = 1.0 kHz and the amplitude is A = 0.20 mm. (a)What is the maximum speed of the diaphragm? (b)Where in the motion does this maximum speed occur? Example

15. (b) The speed of the diaphragm is zero when the diaphragm momentarily comes to rest at either end of its motion: x = +A and x = –A. Its maximum speed occurs midway between these two positions, or at x = 0 m. (a)

16. A loudspeaker diaphragm is vibrating at a frequency of f = 1.0 kHz, and the amplitude of the motion is A = 0.20 mm. (a)What is the maximum acceleration of the diaphragm, and (b)where does this maximum acceleration occur? Example

17. (b) the maximum acceleration occurs at x = +A and x = –A (a)

18. The drawing shows plots of the displacement x versus the time t for three objects undergoing simple harmonic motion. Which object, I, II, or III, has the greatest maximum velocity? Example

19. Oscillations of a Spring F = - k x k x m x

20. Displacement : x = L θ Returning force : F = - mg sin θ Newton Law : F = m d 2 θ/dt 2 Simple Pendulum

21. LC Circuit Solutions : I(t) = A cos ωt

22. A spring stretches 0.150 m when a 0.300-kg mass is gently attached to it. The spring is then set up horizontally with the 0.300-kg mass resting on a frictionless table. The mass is pushed so that the spring is compressed 0.100 m from the equilibrium point, and released from rest. Determine: (a) the spring stiffness constant k and angular frequency ω ; (b) the amplitude of the horizontal oscillation A ; (c) the magnitude of the maximum velocity v max ; (d) the magnitude of the maximum acceleration a max of the mass; (e) the period T and frequency f ; (f) the displacement x as a function of time; and (g) the velocity at t = 0.150 s. Exercise

23. Solution:

24. What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)? L Exercise

25. 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 Exercise