Lecture 43: Damped and Driven Oscillations
m d2x dx +b + kx = 0 dt 2 dt
Final examination and evaluations
6:30 - 9:30 PM, Monday 17th December Last names A-H in Anderson Hall 370 I-Z in Moos Tower 2650 Make-up exam 8.00-11:00 AM Tuesday 18th Dec, in Anderson Hall 350 If you have a clash of finals you can take the Make-up but you MUST make arrangements with the office first Final will cover everything we have done on the course Format 5 long problems, you get credit for your 4 best, worth 25 points each (100 total) 10 multiple choice problems, 5 points each (50 total) You can bring in one sheet of normal sized paper with notes on both sides Please tell us your opinions of the course at http://eval.umn.edu/
k b2 m 4m 2
b = 2m
Physics 1301: Lecture 42, Pg 1
Physics 1301: Lecture 42, Pg 2
Problem: Vertical Spring
A mass m = 102 g is hung from a vertical spring. The equilibrium position is at y = 0. The mass is then pulled down a distance d = 10 cm from equilibrium and released at t = 0. The measured period of oscillation is T = 0.8 s. What is the spring constant k? Write down the equations for the position, velocity, and acceleration of the mass as functions of time. What is the maximum velocity? What is the maximum acceleration? m t=0 k y 0 -d So:
Problem: Vertical Spring...
What is k ?
k m
k = 2m
k y 0
2 = 7 .8 5 s 1 T
2
k = ( 7.85 s 1 ) 0.102kg = 6.29
N m
-d m t=0 m=0.102kg T=0.8s
UIUC
Physics 1301: Lecture 42, Pg 3
Physics 1301: Lecture 42, Pg 4
Problem: Vertical Spring...
What are the equations of motion? Solution of SHM is y=ymaxcos( t+) At t = 0, y = -d = -ymax v=0 -d=dcos -1=cos So = cos(t+)=-cos(t) m t=0 y(t) = -d cos(t) v(t) = d sin(t) a(t) = 2d cos(t) k y 0 -d
Problem: Vertical Spring...
y(t) = -d cos(t) v(t) = d sin(t) a(t) = 2d cos(t)
t k
ymax = d = .1m vmax = d = (7.85 s-1)(.1m) = 0.78 m/s amax = 2d = (7.85 s-1)2(.1m) = 6.2 m/s2 m t=0
y 0 -d
Physics 1301: Lecture 42, Pg 5
Physics 1301: Lecture 42, Pg 6
Damped Harmonic Motion
We have only been considering perfect motion, no friction, no energy loss. SIMPLE harmonic motion In real life the oscillations die away. How can we treat this? In many cases the energy loss is due to drag forces which are proportional to velocity -kx dx
Damped Harmonic Motion
m d2x dx +b + kx = 0 dt 2 dt
We want a function that oscillates and dies away with time Try
x = Ae t sin( t + )
F = bv = b
For a drag force plus a restoring force the 2nd law says
dt
The extra term added to the sin term is an exponential m
bv
At t=0 At t=
dx d2x F = kx b =m 2 dt dt
e 0 = 1 e = 0
Large
Small
We can solve this equation
Physics 1301: Lecture 42, Pg 7 Physics 1301: Lecture 42, Pg 8
Damped Harmonic motion
To solve
Damped Harmonic Motion
The differential equation will be satisfied if the coefficients of the sin and cosine terms separately equal 0 For the cosine terms
d x dx +b + kx = 0 dt 2 dt
Guess the solution Differentiate
x = Ae t sin( t + )
mA mA + Ab = 0 b = 2m
mA 2 mA 2 bA + kA = 0 mb 2 b2 m 2 +k=0 4m 2 2m 2 k b 2 = m 4m 2
Physics 1301: Lecture 42, Pg 10
dx = A e t sin ( t + ) + A e t cos ( t + ) dt
For the sin terms
d2x = A 2 e t sin ( t + ) A e t cos ( t + ) dt 2 -A e t cos ( t + ) A 2 e t sin ( t + )
Physics 1301: Lecture 42, Pg 9
Damped Harmonic Motion
The solution is
Damped Harmonic Motion
Increasing b, more damping, increases the speed with which the oscillation decreases Increasing b also reduces
x = Ae t sin( t + )
The angular frequency is modified
bt 2m
k b2 b2 = 02 4m 2 m 4m 2
0 is the natural frequency if
there was no damping And the amplitude is reduced by the exponential factor
k b2 = m 4m 2
When
b = 4km
=0
bt 2m
=e
t 2
m is the mean life of the oscillation b
Physics 1301: Lecture 42, Pg 11
The oscillation stops, the system is critically damped
Physics 1301: Lecture 42, Pg 12
Damped Harmonic Motion
Damping is extremely important in designing many objects One wants some systems to respond to forces but not to oscillate for long Shock absorbers, Dials and Pointers They need appropriate damping Other systems one doesnt want to oscillate at all Bridges, Buildings They need to be critically damped Electrical circuits also oscillate and can be described by this formalism Coming your way in 1302
ICQ: Damped Harmonic Motion
The solution for the differential equation for a damped oscillator when the damping is small is
x = Ae t sin( t + )
What do you think determines the phase constant ? a) The spring constant k b) The mass m of the system c) The damping coefficient b of the system d) The initial velocity and displacement of the system
Physics 1301: Lecture 42, Pg 13
Physics 1301: Lecture 42, Pg 14
ICQ: Damped Harmonic Motion
The solution for the differential equation for a damped oscillator when the damping is small is
Driven Harmonic Motion
Quite often we want something to keep oscillating Garden swing Clock pendulum. Because there is always energy loss in a system they are naturally damped To keep it oscillating we need to supply an oscillating driving force, for example
x = Ae t sin( t + )
What do you think determines the phase constant ? a) The spring constant k b) The mass m of the system c) The damping coefficient b of the system d) The initial velocity and displacement of the system
F ( t ) = F0 sin t
The equation we have to solve now is
defines the initial conditions of the oscillation at t=0 and
thus depends on the displacement and velocity at that time, just as for Simple Harmonic Motion x = A sin
0
Physics 1301: Lecture 42, Pg 15
kx b
dx d2x + F0 sin t = m 2 dt dt
Physics 1301: Lecture 42, Pg 16
Driven Harmonic Motion
kx b dx d2x + F 0 sin t = m dt dt 2
A=
Resonance
The amplitude of oscillation is given by
F0 m 2 ( 2 02 ) + b 2 2
2
This has a solution (substitute it yourself to check)
x = A s in ( t + )
Where
A=
F0
2 m 2 ( 2 0 ) + b 2 2 2
If there was no damping (b=0) the amplitude would be infinite at =0. This cannot happen in practice but still there can be large increases in the amplitude of oscillation when comes close to 0. Designers must be very careful to avoid resonances, the natural frequency must be different from any possibly occurring driving frequencies
Physics 1301: Lecture 42, Pg 18
The system oscillates at the frequency (the driven frequency) The natural frequency is 0 =
k m
Physics 1301: Lecture 42, Pg 17
ICQ: Swing
To keep a child on a swing moving with a large amplitude, you should a) Push with as large a force as possible b) Push with a periodic force as often as possible c) Push with a periodic force, with a period that depends on the weight of the child d) Push with a periodic force, with a period that depends on the length of the ropes on the swing e) Push with a force equal to the weight of the child
Physics 1301: Lecture 42, Pg 19
ICQ: Swing
To keep a child on a swing moving with a large amplitude, you should Push with as large a force as possible Push with a periodic force as often as possible Push with a periodic force, with a period that depends on the weight of the child Push with a periodic force, with a period that depends on the length of the ropes on the swing Push with a force equal to the weight of the child L
a) b) c) d)
T The natural frequency is that of a simple pendulum Resonance occurs when the driving frequency (your pushes) equals the natural L T = 2 frequency g
Physics 1301: Lecture 42, Pg 20
e)
Transport Tunnel
A straight tunnel is dug from Minneapolis through the center of the Earth and out the other side. A physics 1301 student jumps into the hole at noon. What happens to her? What time does she get back to Minneapolis? FG
Transport Tunnel...
FG (R ) = GmM R R2
where MR is the mass inside radius R R RE MR
FG ( R ) M R2 = R E FG (RE ) R 2 M E
but M R R (assuming the earth has constant density)
3
2 FG ( R ) R 3 RE R = = 3 FG (RE ) R 2 RE RE
Physics 1301: Lecture 42, Pg 21
Physics 1301: Lecture 42, Pg 22
Transport Tunnel...
FG (R ) = R RE
Transport Tunnel...
k=
mg RE
The SHM has a frequency
= k g = m RE
FG (RE )
FG MR
R RE
FG ( RE ) = mg
R FG = mg = kR RE
FG MR
R RE plug in g = 9.81 m/s2 and RE = 6.38 x 106 m get = .00124 s-1 2 and so T = = 5067 s 84 min
Equation for a SHM! Like a mass on k = mg RE a spring with
Physics 1301: Lecture 42, Pg 23
Physics 1301: Lecture 42, Pg 24
Transport Tunnel...
So she will make a Simple Harmonic motion whose amplitude is the radius of the earth and whose period is 84 minutes She gets back to Minneapolis 84 minutes later, at 1:24 p.m. 42 minutes later her feet will poke out the other side of the earth
Transport Tunnel...
Strange but true: The period of oscillation does not require that the tunnel be straight through the middle!! Any straight tunnel gives the same answer, as long as it is frictionless and the density of the Earth is constant.
Physics 1301: Lecture 42, Pg 25
Physics 1301: Lecture 42, Pg 26
Transport Tunnel...
Another strange but true fact: An object orbiting the earth near the surface will have a period of the same length as that of the transport tunnel. Acceleration towards center = g g = 2R
End of the Syllabus
That is the end of the syllabus Try problems Chapter 13 # 68, 82 We will spend the last two lectures going back over what we have done and reminding ourselves of the type of problem that may come up in the final and how to solve them.
g R
9.81 = 2 6.38(10)6 m = .00124 s-1 2 so T = = 5067 s = 84 min
Physics 1301: Lecture 42, Pg 27 Physics 1301: Lecture 42, Pg 28
