1 Oscillations oscillations_02 CP Ch 14 How can you determine the mass of a single E-coli bacterium...
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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
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
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
Slide 3
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
Slide 4
3 Example: oscillating stars Brightness Time CP445
Slide 5
4 oscillations_02: MINDMAP SUMMARY Reference frame (coordinate
system, origin, equilibrium position), displacement (extension,
compression), applied force, restoring force, gravitational force,
net (resultant) force, Newtons Second Law, Hookes 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
Slide 6
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
Slide 7
6 By viewing the animation You should have a better
understanding of the following terms SHM periodic motion
Equilibrium position Displacement Amplitude Period Frequency
Slide 8
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
Slide 9
8 What is the connection between circular motion and SHM ? What
is the meaning of period T, frequency f, and angular frequency
?
Slide 10
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
Slide 11
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
Slide 12
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
Slide 13
12 Simple Harmonic Motion CP457 x = 0 x m k +X displacement
velocity = dx/dt acceleration = dv/dt force angular frequency,
frequency, period
Slide 14
13 acceleration is rad (180 ) out of phase with displacement
CP457 Simple harmonic motion
Slide 15
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
Slide 16
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
Slide 17
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 9 10 11
Slide 18
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 = 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
Slide 19
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
Slide 20
19 Solution 2 Identify / Setup SHM Execute OK x = 20 mm f = ?
Hz k = ? N.m -1 m x = 20 mm = 20 10 -3 m
Slide 21
20 txVaKE PE 0A0 - 2 A 0 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
Slide 22
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
Slide 23
22 Solution 4 Identify / Setup L 0 = 500 mm = 0.500 m L 1 = 850
mm = 0.850 m m 1 = 250 g = 0.250 kg y max = 120 mm = 0.120 m k = ?
N.m -1 f 1 = ? Hz T 1 = ? s m 2 = 0.500 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
Slide 24
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 = 0.120 m (e) Again F = m g = k y y = m 2 g / k
m 2 = 0.500 kg k = 7 N.m -1 y = (0.5)(9.8) / 7 m = 0.700 m L 2 = L
0 + y = (0.500 + 0.700) m = 1.20 m (f) Evaluate
Slide 25
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
Slide 26
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
Slide 27
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
Slide 28
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 = 0.062 m Evaluate