1. . Physics 207, Lecture 19, Nov. 5 Goals: Chapter 14 Chapter 15
Understand and use energy conservation in oscillatory systems.
Understand the basic ideas of damping and resonance.
Chapter 15
Understand pressure in liquids and gases
Use Archimedes’ principle to understand buoyancy
Understand the equation of continuity
Use an ideal-fluid model to study fluid flow.
Investigate the elastic deformation of solids and liquids
Assignment
HW8, Due Wednesday, Nov. 12th
Monday: Read all of Chapter 15 . 1

2. SHM So Far The most general solution is x(t) = A cos(t + )
where A = amplitude
 = (angular) frequency = 2p f = 2p/T
 = phase constant
For SHM without friction,
The frequency does not depend on the amplitude !
This is true of all simple harmonic motion!
The oscillation occurs around the equilibrium point where the force is zero!
Energy is a constant, it transfers between potential and kinetic
Velocity: v(t) = -A sin(t + )
Acceleration: a(t) = -2A cos(t + )
Simple Pendulum:

3. The shaker cart
You stand inside a small cart attached to a heavy-duty spring, the spring is compressed and released, and you shake back and forth, attempting to maintain your balance. Note that there is also a sandbag in the cart with you.
At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground.
What effect does jettisoning the sandbag at the equilibrium position have on the amplitude of your oscillation?
A. It increases the amplitude.
B. It decreases the amplitude.
C. It has no effect on the amplitude.
Hint: At equilibrium, both the cart and the bag
are moving at their maximum speed. By
dropping the bag at this point, energy
(specifically the kinetic energy of the bag) is
lost from the spring-cart system. Thus, both the
elastic potential energy at maximum displacement
and the kinetic energy at equilibrium must decrease

4. The shaker cart
Instead of dropping the sandbag as you pass through equilibrium, you decide to drop the sandbag when the cart is at its maximum distance from equilibrium.
What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation?
A. It increases the amplitude.
B. It decreases the amplitude.
C. It has no effect on the amplitude.
Hint: Dropping the bag at maximum
distance from equilibrium, both the cart
and the bag are at rest. By dropping the
bag at this point, no energy is lost from
the spring-cart system. Therefore, both the
elastic potential energy at maximum displacement
and the kinetic energy at equilibrium must remain constant.

5. The shaker cart
What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the maximum speed of the cart?
A. It increases the maximum speed.
B. It decreases the maximum speed.
C. It has no effect on the maximum speed.
Hint: Dropping the bag at maximum distance
from equilibrium, both the cart and the bag
are at rest. By dropping the bag at this
point, no energy is lost from the spring-cart
system. Therefore, both the elastic
potential energy at maximum displacement
and the kinetic energy at equilibrium must
remain constant.

6. What about Vertical Springs?
For a vertical spring, if y is measured from the equilibrium position
Recall: force of the spring is the negative derivative of this function:
This will be just like the horizontal case: -ky = ma = j k
y = 0
F= -ky m
Which has solution y(t) = A cos( t + )
where

7. Exercise Simple Harmonic Motion
A mass oscillates up & down on a spring. It’s position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration ?
Remember: velocity is slope and acceleration is the curvature
y(t) = A cos( t + )
v(t) = -A  sin( t + )
a(t) = -A 2 cos( t + ) t
y(t)
(a)
(b)
(c)

8. Example A mass m = 2 kg on a spring oscillates with amplitude
A = 10 cm. At t = 0 its speed is at a maximum, and is v=+2 m/s
What is the angular frequency of oscillation  ?
What is the spring constant k ?
General relationships E = K + U = constant, w = (k/m)½
So at maximum speed U=0 and ½ mv2 = E = ½ kA2
thus k = mv2/A2 = 2 x (2) 2/(0.1)2 = 800 N/m, w = 20 rad / sec k x m

9. Home Exercise Initial Conditions
A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0.
Which of the following describe its velocity and acceleration as a function of time (upwards is positive y direction)?
(A) v(t) = - vmax sin( wt ) a(t) = -amax cos( wt )
(B) v(t) = vmax sin( wt ) a(t) = amax cos( wt ) k y
t = 0 m d
(C) v(t) = vmax cos( wt ) a(t) = -amax cos(wt )
(both vmax and amax are positive numbers)

10. Exercise Simple Harmonic Motion
You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1.
Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2.
Which of the following is true recalling that w = (g / L)½ T1 T2
(A) T1 = T2
(B) T1 > T2
(C) T1 < T2

11. Energy in SHM
For both the spring and the pendulum, we can derive the SHM solution using energy conservation.
The total energy (K + U) of a system undergoing SMH will always be constant!
This is not surprising since there are only conservative forces present, hence energy is conserved. -A A x U K E

12. SHM and quadratic potentials
SHM will occur whenever the potential is quadratic.
For small oscillations this will be true:
For example, the potential between H atoms in an H2 molecule looks something like this: -A A x U K E U x

13. SHM and quadratic potentialsx
Curvature reflects the spring constant
or modulus (i.e., stress vs. strain or
force vs. displacement)
Measuring modular proteins with an AFM
See:

14. What about Friction? A velocity dependent drag force
We can guess at a new solution.
and now w02 ≡ k / m
Look
With,

15. What about Friction? if
What does this function look like?

16. Variations in the damping
Small damping time constant (b/2m)
Low friction coefficient, b << 2m
Moderate damping time constant (b/2m)
Moderate friction coefficient (b < 2m)

17. Damped Simple Harmonic Motion
A downward shift in the angular frequency
There are three mathematically distinct regimes
underdamped
critically damped
overdamped

18. Physical properties of a globular protein (mass 100 kDa)
Mass x kg
Density x 103 kg / m3
Volume 120 nm3
Radius 3 nm
Drag Coefficient 60 pN-sec / m
Deformation of protein in a viscous fluid

19. Driven SHM with Resistance
Apply a sinusoidal force, F0 cos (wt), and now consider what A and b do,
b small
b middling
b large w 
w  w0

20. Microcantilever resonance-based DNA detection with nanoparticle probes
Change the mass of the cantilever and change the resonant frequency and the mechanical response.
Su et al., APPL. PHYS. LETT. 82: 3562 (2003)

21. Stick - Slip Friction
How can a constant motion produce resonant vibrations?
Examples:
Strings, e.g. violin
Singing / Whistling
Tacoma Narrows Bridge …

22. Dramatic example of resonance
In 1940, a steady wind set up a torsional vibration in the Tacoma Narrows Bridge 

23. Dramatic example of resonance
Eventually it collapsed 

24. Exercise Resonant Motion
Consider the following set of pendulums all attached to the same string D A B C
If I start bob D swinging which of the others will have the largest swing amplitude ?
(A) (B) (C)

25. Exercise
Damped oscillations:
A can of coke is attached to a spring and is displaced by hand (m = 0.25 kg & k = 25.0 N/m)
The coke can is released, and it starts oscillating with an amplitude of A = 0.3 m.
How damped is the system?
Underdamped (multiple oscillations with an exponential decay in amplitude)
Critically damped (simple decaying motion with at most one overshoot of the system's resting position)
Overdamped (simple exponentially decaying motion, without any oscillations)