Simple Harmonic Motion

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# Simple Harmonic Motion

## Simple Harmonic Motion

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##### Presentation Transcript

1. Simple Harmonic Motion • Simple harmonic motion (SHM) refers to a certain kind of oscillatory, or wave-like motion that describes the behavior of many physical phenomena: • a pendulum • a bob attached to a spring • low amplitude waves in air (sound), water, the ground • the electromagnetic field of laser light • vibration of a plucked guitar string • the electric current of most AC power supplies • …

2. SHM Position, Velocity, and Acceleration

3. Simple Harmonic Motion Periodic Motion: any motion of system which repeats itself at regular, equal intervals of time.

4. Simple Harmonic Motion

5. Simple Harmonic Motion • Equilibrium: the position at which no net force acts on the particle. • Displacement: The distance of the particle from its equilibrium position. Usually denoted as x(t) with x=0 as the equilibrium position. • Amplitude: the maximum value of the displacement with out regard to sign. Denoted as xmax or A.

6. The period and frequency of a wave • the periodT of a wave is the amount of time it takes to go through 1 cycle • the 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: • Since a cycle is 2p radians, the relationship between frequency and angular frequency is:

7. http://www.physics.uoguelph.ca/tutorials/shm/Q.shm.html

8. Here is a ball moving back and forth with simple harmonic motion (SHM): Its position x as a function of time t is: where A is the amplitude of motion : the distance from the centre of motion to either extreme T is the period of motion: the time for one complete cycle of the motion.

9. Restoring Force How does the restoring force act with respect to the displacement from the equilibrium position? F is proportional to -x Simple harmonic motion is the motion executed by a particle of mass m subject to a force that is proportional to the displacement of the particle but opposite in sign.

10. 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: • What is the angular frequency in this case? • Use Newton’s 2nd law, together with Hooke’s law, and the above description of the acceleration to find:

11. Springs and Simple Harmonic Motion

12. Equations of Motion Conservation of Energy allows a calculation of the velocity of the object at any position in its motion…

13. Conservation of Energy For A Spring in Horizontal Motion E =Kinetic + Elastic Potential E = ½ mv2 + ½ kx2 = Constant • At maximum displacement, velocity is zero and all energy is elastic potential, so total energy is equal to ½ kA2

14. To find the velocity @ any displacement… do conservation of Energy… @ some point at max displacement ½ mv2 + ½ kx2 = 1/2kA2 Solving for v This is wonderful but what about time??! v=dx/dt and separate variables! 

15. SHM Solution... • Drawing of A cos(t ) • A = amplitude of oscillation T = 2/ A     - A

16. SHM Solution... • Drawing of A cos(t + )      -

17. SHM Solution... • Drawing of A cos(t - /2)  = /2 A     - = A sin(t)!

18. 1989.M3 3.0 m/s, 0.63 s, 0.098 m, 0.41 m, 0.31 m • 1989M3. A 2‑kilogram block is dropped from a height of 0.45 m above an uncompressed spring, as shown above. The spring has an elastic constant of 200 N per meter and negligible mass. The block strikes the end of the spring and sticks to it. • a. Determine the speed of the block at the instant it hits the end of the spring. 3 m/s • b. Determine the period of the simple harmonic motion that ensues. 0.63s • Determine the distance that the spring is compressed at the instant the • speed of the block is maximum. 0.098 m • d. Determine the maximum compression of the spring. 0.41 m • e. Determine the amplitude of the simple harmonic motion. 0.31 m

19. SHM So Far • The most general solution is x = A cos(t + ) where A = amplitude  = angular frequency  = phase • For a mass on a spring • The frequency does not depend on the amplitude!!! • We will see that this is true of all simple harmonic motion!

20. Pendulums • When we were discussing the energy in a simple harmonic system, we talked about the ‘springiness’ of the system as storing the potential energy • But when we talk about a regular pendulum there is nothing ‘springy’ – so where is the potential energy stored?

21. The Simple Pendulum • As we have already seen, the potential energy in a simple pendulum is stored in raising the bob up against the gravitational force • The pendulum bob is clearly oscillating as it moves back and forth – but is it exhibiting SHM?

22. We can see that the restoring force is: F = -mg sinθ If we assume that the angle θ is small, then sinθ ~ θand our equation becomes F ~ -mgθ = - mgs/L = - (mg/L)s

23. F ~ -mgθ = - mgs/L = - (mg/L)sdividing both sides by m, a = -(g/L)s. Equate to a = -ω2 x, and ω = (g/L)1/2

24. The Physical Pendulum • Now suppose that the mass is not all concentrated in the bob? • In this case the equations are exactly the same, but the restoring force acts through the center of mass of the body (C in the diagram) which is a distance h from the pivot point

25. Going back to our definition of torque, we can see that the restoring force is producing a torque around the pivot point of: • where L is the moment arm of the applied force

26. The Simple Pendulum If we substitute τ = Iα, we get: • This doesn’t appear too promising until we make the following assumption – • that θ is small… • If θ is small we can use the approximation that sin θθ • (as long as we remember to express θ in radians)

27. Making the substitution we then get: The Physical Pendulum which we can then rearrange to get: which is the angular equivalent to a = -ω2 x • So, we can reasonably say that the motion of a pendulum is approximately SHMif the maximum angular amplitude is small

28. Making the substitution we then get: The Physical Pendulum which we can then rearrange to get: which is the angular equivalent to a = -ω2 x • So, we can reasonably say that the motion of a pendulum is approximately SHMif the maximum angular amplitude is small

29. The Physical Pendulum • So we go back to our previous equation for the period and replace L with h to get:

30. The Simple Pendulum The period of a pendulum is given by: where I is the moment of inertia of the pendulum • If all of the mass of the pendulum is concentrated in the bob, then I = mL2 and we get:

31. An AngularSimple Harmonic Oscillator • The figure shows an angular version of a simple harmonic oscillator • In this case the mass rotates around it’s center point and twists the suspending wire • This is called a torsional pendulum with torsion referring to the twisting motion

32. Torsional Oscillator • If the disk is rotated through an angle (in either direction) of θ, the restoring torque is given by the equation:

33. Sample Problem A uniform bar with mass m lies symmetrically across two rapidly rotating, fixed rollers, A and B, with distance L = 2.0 cm between the bar’s center of mass and each roller. The rollers slip against the bar with coefficient of friction µk = 0.40. Suppose the bar is displaced horizontally by a distance x and then released. What is the angular frequency of the resulting horizontal motion of the bar? ω = 14 rad/s

34. Sample Problem 1 • A 1 meter stick swings about a pivot point at one end at a distance h from its center of mass • What is the period of oscillation?

35. A penguin dives from a uniform board that is hinged at the left and attached to a spring at the right. The board has length L = 2.0 m and mass m = 12 kg; the spring constant k = 1300 N/m. When the penguin dives, it leaves the board and spring oscillating with a small amplitude. Assume that the board is stiff enough not to bend, and find the period T of the oscillations. T = 0.35 s