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Oscillations!
(Today: Springs)
Extra Practice: 5.34, 5.35, C13.1, C13.3, C13.11, 13.1, 13.3, 13.5,
13.9, 13.11, 13.17, 13.19, 13.21, 13.23, 13.25, 13.27, 13.31
Test #3 is this Wednesday!
• April 12, 7-10pm, Eiesland G24 as usual.
MAKE-UP EXAMS: If you haven’t gotten a
confirmation email, you’re not scheduled for a
make-up.
• Practice test, practice key, and equation sheet
are on my website.
• Today’s homework due Thursday.
No homework due Sunday (no class Friday!).
http://sarahspolaor.faculty.wvu.edu/home/physics-101
Commencing regularly-scheduled homework after Sunday.
The rest of this semester.
wave concepts
sound
springs
string vibrations/
music
ultrasonic waves
pendula
Springs will be used to introduce harmonic/oscillatory motion that will be used to discuss waves, including sound waves and music. This is the last general topic we’ll be
considering. Today we’ll talk about springs. The reason we’re so excited about springs is because 1) there are a lot of common uses for them, 2) they’ll help you review
some previous concepts both for your knowledge and the final, and 3) they are a great way to set up your thinking about oscillations and waves, which again is what the
rest of this course will focus on.
Familiar concepts…
• Forces/free body diagrams!
Fspring
• Newton’s laws!
Fg
ΣF = ma = 0
• Work and energy conservation!
PEg = mgh
All converted to
spring potential
Lest it seem like we’re always learning new physics, I want you to notice throughout this lecture that we’ll be applying the same concepts
- Forces/FBD’s.
- Force equilibrium.
- Potential energy and energy conservation.
Let’s think about springs.
Fs = -k x
spring constant [Units: N/m]
Fpull
Fspring
“Hooke’s Law!”
Let’s start by doing some basic observations.
[See light board notes]
This spring is relaxed. We like to say, it’s at equilibrium. At its equilibrium position, we put our axis there and call it x=0. As I extend this spring beyond that x=0 position and I feel the spring resisting my pull
with more and more force as I stretch it or compress it. So the amount of FORCE it has is larger and larger with displacement from x. In fact, for an “IDEAL SPRING,” if you experimentally measure the
amount of force it takes to compress the spring, it changes constantly with x. Or in other words, the force is equal to a constant times x. This constant is what we call the spring constant, using symbol k. We
call this “restoring force” because it’s a force that wants to push the spring back to its equilibrium position (let go of spring so it goes back to x=0). NOTE THERE’S A NEGATIVE SIGN! So whatever way x is
going, the Force you feel is opposite in direction. But no matter if you’re pushing or pulling, the MAGNITUDE of the force is always proportional to kx!
Fs = -k x
spring constant [Units: N/m]
Low k
(soft spring)
A word about spring constant…
[HANG WEIGHTS ON THE SPRINGS to show low and high k!]
High k
(stiff spring)
Familiar problem #1:
Newton’s second (and third) laws
A 2.5 kg object is hanging vertically on a spring, which is
stretched by the weight by a distance d. What is the magnitude
of the force the spring is applying to the object?
d
A.
B.
C.
D.
6.3 N
16 N
25 N
Not enough information
Q107
Now we are going to try measuring a spring constant by hanging a weight on a string. Let’s practice this by applying and reviewing a familiar concept, Newton’s second
and third laws.
ANSWER: C
Familiar problem #1:
Newton’s second (and third) laws
A 2.5 kg object is hanging vertically on a spring, which is
stretched by the weight by a distance d. What is the magnitude
of the force the spring is applying to the object?
b) If the string stretches by 2.76 cm due to this
mass, what is the spring constant?
d
A.
B.
C.
D.
0.55 N/m
77 N/m
888 N/m
5 x 105 N/m
Q108
c) What is the spring force if you stretch the
spring to 8 cm?
NOTE! Both of these questions are good test fodder.
[See light board notes for solutions]
Springs as harmonic oscillators.
Let’s keep thinking about spring dynamics for a moment, and I want you to consider me tying a mass to one end of the spring, and pulling it out. If I pull it out and let go, the mass will release and speed by the
equilibrium point. Will it be going faster when I drop it from position 1 (a bit) or position 2 (a lot)? You’d probably guess Position 2, right? Why is this? You might intuitively say it’s because F = -kx and since it
tugs with greater force here than here, it goes faster in the middle. In this case, your intuition for once didn’t fail you! THAT’S TRUE! Now I want you to consider that if you had a spring with no internal friction,
and no friction along the ground. If I let go, the spring would be pushed by its restoring force to accelerate beyond equilibrium to a compressed state, only to be pushed back to its original position. IF THERE
WERE NO FRICTION, this spring would go on and on in this way forever. It’s this type of motion that we refer to as “simple harmonic motion.” And hence, refer to springs doing this as “simple harmonic
oscillators”. They are perhaps the most straight forward example of this type of motion. We can even get springs doing this vertically, and see that it’s oscillating with a fixed period, and a fixed AMPLITUDE, or
the most extreme points of the motion, defined by the place where I let go of the mass. In this graphic you can see it’s defined as amplitude, A. We’ll get back to this later.
Spring (Elastic) Potential Energy
When stretched or
compressed, a spring
gains potential energy!
But first, I wanted to introduce the concept of spring potential energy. Any time you compress or stretch the spring you give it potential energy, because it has the potential to become unsprung.
Recall: Work and
Gravitational Potential Energy.
Work:
Energy associated with a force and its action.
W = F||Δx
Wg = -ΔPEg
PEg = mgh
PEg = 0 J
If an object is
released from height
h, gravity has the
potential to do an
amount of work
equal to mgh.
Remember we spoke about GRAVITATIONAL POTENTIAL ENERGY a few months ago. Work is defined as the force in the direction of movement times the amount it’s
moved by that force. For gravity, we also had the condition that the negative of the change in potential energy is equal to the work done by gravity. In other words, if you
load this box up with potential energy by lifting it, it’s capable of converting that potential to a lot of kinetic energy when we drop it and gravity starts acting.
How Do We Find Spring
Potential Energy?
F0 = 0 N
(PEs = 0 N)
Fs = -kx (Hooke’s Law)
W = F||Δx (Work done by spring)
Ws = -ΔPEs (Work done by spring)
Let’s apply this same principle to springs, starting with these three things we know. We start with the spring on its own; it’s not active so it has no spring force and hence no
potential to do any work. Now let’s load up the spring with potential by squishing it, and while I’m doing that, the spring is doing some work. How much work is it doing?
[See LIGHT BOARD notes]
Springs clearly make a nice review topic!
Conservation of energy applies to springs, too!
PEs = ½ kx2
(KE + PEg + PEs)f = (KE + PEg + PEs)i
(if no friction)
Wnc = (KE + PEg + PEs)f - (KE + PEg + PEs)i
(if friction or other non-conservative
forces are present)
Energy lost to non-conservative forces
Think about giving a spring a greater and greater force as you stretch it further and further: it gets more potential energy.
Where does an oscillating spring
provide a maximum kinetic energy?
A.
B.
C.
D.
E. Both B and D
Q109
Let’s go back to this graphic and think about the oscillating spring. In this oscillation, you see a literal view of constant conversion between kinetic and potential energies.
Note that GPE is zero here because it’s on a frictionless, horizontal surface. I want you to tell me… AT WHAT POINT IN THE CYCLE of this spring does it give the block a
maximum kinetic energy?
ANSWER: E
Kinetic and Potential
A weight is suspended by a spring. The spring is then
stretched until the weight is just above some eggs, and then it
is released. The weight springs upward and then comes back
down due to gravity. Ignoring air resistance, what happens?
A. the weight bounces back well
above the eggs
Q110
B. the weight bounces back just
before it reaches the eggs
C. it smashes the eggs!
There’s a fixed reservoir of energy available for this weight to move, and I put it all in potential spring energy at the bottom of the motion.
ANSWER: B.
In the initial situation, with the spring stretched until the bottom of the weight is just above the eggs, all of the energy of the system is potential. Upon release, some of this
energy is converted to kinetic energy in the moving spring and weight. When the weight returns to its position just above the eggs, however, all of the energy in the system
must again be potential. Having no kinetic energy, the weight stops at the point from which it was initially released. The eggs are safe!
I wanted to remind you of this graphic to say that we’re in part excited about springs because they are a nice way to introduce waves.
Total energy in an oscillator
Total energy = KE + PE
x=A
Total energy in
oscillator
= ½ kA2
At max
displacement,
energy is all
in PEs!
= ½ kx2
at x = A
If you think back to our horizontal oscillator, boinging around infinitely, you know that the total energy in the system is conserved and is swapping over and over again
between potential and kinetic energy. Now, we just saw that at this particular maximum displacement point, ALL THE ENERGY IS IN SPRING POTENTIAL.
Velocity as a Function of Position
v=0
max -v
1
2
KE
2
PEs
2
Total E
mv + 12 kx = 12 kA2
• We can derive the velocity of the
object at any position
v=0
max +v
– Speed is a maximum at x = 0
– Speed is zero at x = ±A
– The ± indicates the object can
be traveling in either direction
We’ll be discussing things as a function of time next Monday, but I wanted to show you that using the total energy of that oscillator, we can solve for velocity and get a
function for how fast something will be going at any point in an oscillation!
A block whose mass is 0.5 kg is retracted against a
release spring with spring constant 75 N/m. If the
spring is initially stretched A=0.25 m, what is the
velocity of the block when the spring returns to its
equilibrium position? (assume a flat, frictionless
release surface)
[See light board notes for solution]