1 Oscillations oscillations_02 CP Ch 14 How can you determine the mass of a single E-coli bacterium or a DNA molecule ? CP458 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] Period: time for one cycle of motion [s] Frequency: number of cycles per second [s -1 = hertz (Hz)] SHM

2 Signal from ECG period T Period: time for one cycle of motion [s] Frequency: number of cycles per second [s -1 = Hz hertz] CP445 1 kHz = 10 3 Hz 10 6 Hz = 1 MHz 1GHz = 10 9 Hz time  voltage 

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 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 spring restoring force CP447 x = 0 +X

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

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

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

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

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

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

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

14 Describe the phase relationships between displacement, velocity and acceleration? What are the key points on these graphs (zeros and maximums)? CP

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

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

17 Solution 1 Identify / Setup SHM Execute (a) amplitude A = x max = 5. 0 m (b) angular frequency  = 0.40 rad.s -1 frequency f =  / 2  = 0.40 / (2  ) Hz = Hz period T = 1 / f = 1 / 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

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

20 txVaKE PE 0A0 -  2 A 0½ k A 2 T / 4 T / 2 3T / 4 T What are all the values at times t = T/4, T/2, 3T/4, T ? 0 +A+A -A-A Problem 3

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 L 0 = 500 mm = m L 1 = 850 mm = m m 1 = 250 g = kg y max = 120 mm = m k = ? N.m -1 f 1 = ? Hz T 1 = ? s m 2 = kg L 2 = ? m T 2 / T 1 = ? L0L0 m1m1 y max equilibrium position y = 0 F = k (L 1 – L 0 ) F G = m g m 1 a = 0 F = F G L1L1 L2L2 m2m2

23 Execute (a) Object at end of spring – stationary F = F G  k (L 1 - L 0 ) = m g k = m g /(L 1 – L 0 ) 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 = y max = m (e) Again F = m g = k y  y = m 2 g / k m 2 = kg k = 7 N.m -1 y = (0.5)(9.8) / 7 m = m L 2 = L 0 + y = ( ) m = 1.20 m (f) Evaluate

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 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 2 FNFN FGFG m 1 a 1y = 0 F N = m g F f =  N =  m g m 1 = 0.1 kg m 2 = 0.2 kg  = 0.20 A 2 = 60 mm = 0.06 m max freq f = ? Hz SHM a max = A  2 = A(2  f) 2 = 4  2 f 2 A 1 2 FNFN FGFG m1m1 1

26 Execute (a) max frictional force between blocks F f =  m 1 g max acceleration of block 1 a 1max = F f / m 1 =  g max acceleration of block 2 a 2max = a 1max =  g SHM a 2max = 4  2 f 2 A  g = 4  2 f 2 A f =  [  g / (4  2 A)] =  [(0.20)(9.8)/{(4)(  2 )(0.06)}] Hz = 0.91 Hz

27 Execute (b) max acceleration of block 1 (free fall) a 1max = g max acceleration of block 2 a 2max = a 1max = g SHM a 2max = 4  2 f 2 A g = 4  2 f 2 A f = 2 Hz A = g / (4  2 f 2 ) = (9.8) / {(4)(  2 )(2 2 )} m = m Evaluate