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Oscillations

Oscillations. oscillations_02. Time variations that repeat themselves at regular intervals - periodic or cyclic behaviour Examples : Pendulum (simple); heart (more complicated) Terminology : Amplitude : max displacement from equilibrium position [m]

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Oscillations

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  1. Oscillations oscillations_02 Time variations that repeat themselvesat regular intervals - periodic or cyclic behaviour Examples:Pendulum (simple); heart (more complicated) Terminology: Amplitude:max displacement from equilibrium position [m] Period: time for one cycle of motion [s] Frequency: number of cycles per second [s-1 = hertz (Hz)] SHM How can you determine the mass of a single E-coli bacterium or a DNA molecule ? CP458 CP Ch 14

  2. Signal from ECG period T voltage  time  Period: time for one cycle of motion [s] Frequency: number of cycles per second [s-1 = Hz hertz] 1 kHz = 103 Hz 106 Hz = 1 MHz 1GHz = 109 Hz CP445

  3. Example: oscillating stars Brightness Time CP445

  4. oscillations_02: MINDMAP SUMMARY Reference frame (coordinate system, origin, equilibrium position), displacement (extension, compression), applied force, restoring force, gravitational force, net (resultant) force, Newton’s Second Law, Hooke’s Law, spring constant (spring stiffness), equilibrium, velocity, acceleration, work, kinetic energy, potential energy (reference point), gravitational potential energy, elastic potential energy, total energy, conservation of energy, ISEE, solve quadratic equations, SHM, period, frequency, angular frequency, amplitude, sine function (cos, sin), phase, phase angle, radian, SHM & circular motion

  5. Simple harmonic motion SHM x = 0 spring restoring force +X • object displaced, then released • objects oscillates about equilibrium position • motion is periodic • displacement is a sinusoidal function of time (harmonic) • T = period = duration of one cycle of motion • f = frequency = # cycles per second • restoring force always acts towards equilibrium position • amplitude – max displacement from equilibrium position CP447

  6. By viewing the animation You should have a better understanding of the following terms SHM – periodic motion Equilibrium position Displacement Amplitude Period Frequency

  7. origin 0 equilibrium position displacement x [m] velocity v [m.s ] - 1 - + x x max max acceleration a [m.s ] - 2 Force F [N] e Motion problems – need a frame of reference Vertical hung spring: gravity determines the equilibrium position – does not affect restoring force for displacements from equilibrium position – mass oscillates vertically with SHM Fe = - k y CP447

  8. What is the connection between circular motion and SHM ? What is the meaning of period T, frequency f, and angular frequency  ?

  9. w q = d / d t w p = 2 / T amplitude A A q T = 1 / f f = 1 / T w p = 2 f w p = 2 / T One cycle: period T [s] - Cycles in 1 s: frequency f [Hz] w Connection SHM – uniform circular motion 1 revolution = 2 radians = 360o Angles must be measured in radians [rad.s-1] Angular frequency CP453

  10. SHM & circular motion +1 0 -1 0 2 4 6 Displacement is sinusoidal function of time uniform circular motion radius A, angular frequency  x component is SHM x X CP453

  11. Simple Harmonic Motion Displacement is a sinusoidal function of time T amplitude displacement time T T By how much does phase change over one period? CP451

  12. displacement velocity = dx/dt acceleration = dv/dt Simple Harmonic Motion x k force m +X x = 0 angular frequency, frequency, period CP457

  13. Simple harmonic motion acceleration is  rad (180) out of phase with displacement CP457

  14. Describe the phase relationships between displacement, velocity and acceleration? What are the key points on these graphs (zeros and maximums)? CP459 3 4 5 6 7 8

  15. Problem solving strategy: I S E E Identity: What is the question asking (target variables) ? What type of problem, relevant concepts, approach ? Set up: Diagrams Equations Data (units) Physical principals Execute: Answer question Rearrange equations then substitute numbers Evaluate: Check your answer – look at limiting cases sensible ? units ? significant figures ? PRACTICE ONLY MAKES PERMANENT

  16. Problem 1 • If a body oscillates in SHM according to the equation • where each term is in SI units. What are • the amplitude? • the angular frequencies, frequency and period? • the initial phase at t = 0 ? • the displacement at t = 2.0 s ? use the ISEE method 9 10 11

  17. Solution 1 Identify / Setup SHM Execute (a) amplitude A = xmax = 5. 0 m (b) angular frequency  = 0.40 rad.s-1 frequency f =  / 2 = 0.40 / (2) Hz = 0.064 Hz period T = 1 / f = 1 / 0.064 s = 16 s (c) initial phase angle  = 0.10 rad (d) t = 2.0 s x = 5 cos[(0.4)(2) + 0.1] m = 3.1 m Execute OK

  18. Problem 2 An object is hung from a light vertical helical spring that subsequently stretches 20 mm. The body is then displaced and set into SHM. Determine the frequency at which it oscillates. use the ISEE method 12 13 14

  19. Solution 2 Identify / Setup SHM k = ? N.m-1 x = 20 mm f = ? Hz m x = 20 mm = 2010-3 m Execute Execute OK

  20. What are all the values at times t = T/4, T/2, 3T/4, T ? Problem 3 -A 0 +A

  21. Problem 4 A spring is hanging from a support without any object attached to it and its length is 500 mm. An object of mass 250 g is attached to the end of the spring. The length of the spring is now 850 mm. (a) What is the spring constant? The spring is pulled down 120 mm and then released from rest. (b) What is the displacement amplitude? (c) What are the natural frequency of oscillation and period of motion? (d) Describe the motion on the object attached to the end of the spring. Another object of mass 250 g is attached to the end of the spring. (e) Assuming the spring is in its new equilibrium position, what is the length of the spring? (f) If the object is set vibrating, what is the ratio of the periods of oscillation for the two situations? use the ISEE method

  22. Solution 4 Identify / Setup L0 = 500 mm = 0.500 m L1 = 850 mm = 0.850 m m1 = 250 g = 0.250 kg ymax = 120 mm = 0.120 m k = ? N.m-1 f1 = ? Hz T1 = ? s m2 = 0.500 kg L2 = ? m T2 / T1 = ? L0 L1 L2 m1 equilibrium position y = 0 ymax F = k (L1 – L0) m2 m1a = 0 F = FG FG = m g

  23. Execute (a) Object at end of spring – stationary F = FG k (L1 - L0) = m g k = m g /(L1 – L0) k = (0.250)(9.8) / (0.850 – 0.500) N.m-1 k = 7.00 N.m-1 (b) (c) (d) Object vibrates up and down with SHM about the equilibrium position with a displacement amplitude A = ymax = 0.120 m Evaluate (e) Again F = m g = k y y = m2 g / km2 = 0.500 kg k = 7 N.m-1 y = (0.5)(9.8) / 7 m = 0.700 m L2 = L0 + y = (0.500 + 0.700) m = 1.20 m (f)

  24. Problem 5 A 100 g block is placed on top of a 200 g block. The coefficient of static friction between the blocks is 0.20. The lower block is now moved back and forth horizontally in SHM with an amplitude of 60 mm. (a) Keeping the amplitude constant, what is the highest frequency for which the upper block will not slip relative to the lower block? Suppose the lower block is moved vertically in SHM rather than horizontally. The frequency is held constant at 2.0 Hz while the amplitude is gradually increased. (b) Determine the amplitude at which the upper block will no longer maintain contact with the lower block. use the ISEE method

  25. Solution 5 Identify / Setup FN 1 m1 = 0.1 kg m2 = 0.2 kg  = 0.20 A2 = 60 mm = 0.06 m max freq f = ? Hz SHM amax= A 2 = A(2  f)2 = 4 2 f 2 A 2 m1a1y = 0 Ff = N = m g FN = m g FG FN 1 2 m1 FG

  26. Execute (a) max frictional force between blocks Ff =  m1g max acceleration of block 1 a1max = Ff / m1 =  g max acceleration of block 2 a2max = a1max =  g SHM a2max = 4 2f2A  g = 4 2f2A f = [ g / (4 2A)] = [(0.20)(9.8)/{(4)(2)(0.06)}] Hz = 0.91 Hz

  27. Execute (b) max acceleration of block 1 (free fall) a1max = g max acceleration of block 2 a2max = a1max = g SHM a2max = 4 2f2A g = 4  2f 2A f = 2 Hz A = g / (4 2f 2) = (9.8) / {(4)(2)(22)} m = 0.062 m Evaluate

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