Download APB SHM pendulum Unit 5 packet 2010 KEY

AP B Physics:
Springs & Pendulum like to do….
Unit 5
(chap.10, sections 1-4)
1
Objectives:
Pendulums, Springs, Simple Harmonic Motion
1. Simple harmonic motion (dynamics and energy relationships)
a) Sketch or identify a graph of displacement as a function of time,
and determine from such a graph the amplitude, period, and frequency of the motion.
b) Write displacement in the form Atsinw or Atcosw to describe the motion.
d) State the relations between acceleration, velocity, and displacement, and identify locations
where these quantities are zero or achieve their greatest positive and negative values.
e) State and apply the relation between frequency and period.
g) State how the total energy of an oscillating system depends on the amplitude of the motion,
sketch a graph of kinetic or potential energy as a function of time for all points of motion
h) Calculate the kinetic and potential energies of an oscillating system as functions of time,
and prove that the sum of kinetic and potential energy is constant.
2. Mass on a spring
b) Apply the expression for the period of oscillation of a mass on a spring.
c) Analyze problems in which a mass hangs from a spring and oscillates vertically
.
d) Analyze problems in which a mass attached to a spring oscillates horizontally.
3. Pendulum and other oscillations
Students should be able to apply their knowledge of simple harmonic motion to the case of a
pendulum, so they can:
b) Apply the expression for the period of a simple pendulum.
c) State what approximation must be made in deriving the period.
Misconceptions
The period of oscillation depends on the amplitude.
The restoring force is constant at all points in the oscillation.
The heavier a pendulum bob, the shorter its period.
All pendulum motion is perfect simple harmonic motion, for any initial angle.
A pendulum accelerates through lowest point of its swing.
Amplitude of oscillations is measured peak-to-peak.
The acceleration is zero at the end points of the motion of a pendulum.
2
Cliff notes
1. Oscillating motion describes pendulums, springs, and sound waves. A restoring force
pulls back an object so that it repeatedly moves up and down or side to side.
2. SHM or simple harmonic motion is when the restoring force depends on displacement:
a. F=-kx
where the constant is “k” for springs or mg/l for pendulums
b. the force varies so the acceleration is not constant (use energy not tneoms!)
c. at maximum stretch: max position, max acceleration, minimum velocity
d. period of oscillations depends only on mass of object or length of pendulum
spring: T= 2π √ (m/k)
pendulum: T= 2π √ (l/g)
e. natural frequency of oscillation: ω = 2 π/T
π∗f
π = 2π∗
π∗
(ω
ω= radians/sec)
f. pendulums: restoring force = mg sin() = constant*x so SHM if < 15o
3. All springs: ∆x= stretch, F= -k∆
∆x ,
ω= √ (k/m),
Elastic energy= ½ k ∆x2
spring constant= “k” (but will be higher for shorter springs- stiffer)
Elastic energy= area under F vs. x curve (but F depends on x)
4. Horizontal springs: ∆x measure from equilibrium point before add mass
5. Vertical springs:
∆x measure from equilibrium point after add mass
3
What does the period of a simple pendulum depend on?
Pendulum Class Data
•
Fnet = T- mg (but note it’s not constant! depends on angle)
a = Fnet/m = also not constant so can’t use TNEOM’s (energy ok)
•
•
Only mg provides torque since tension attached to pivot point (lever arm =0)
mg*sin() is the restoring force which as small angles is proportional to angle
so is a simple harmonic motion (force varies with angle distance directly)
Which graph shows: Period (T) vs. length? _B_ T vs. mass _A___
A
B
C
more pendulums? See these web sites:
http://www.myphysicslab.com/pendulum2.html
http://phet.colorado.edu/sims/pendulum-lab/pendulum-lab_en.html
4
Pendulum: SHM at small angles
θ
L
Max acceleration
Max force
Max position
zero velocity
θ
r
L
Max acceleration
Max force
Max position
zero velocity
L
h
∆x ≈L* θ
Min acceleration
Min force
Min position
Max velocity
Bottom Line: Period depends only on the length of the string at small angles!
Torque due to gravity (tension always perpendicular to arc motion)
Period:
T= 2π √ (l/g)
Frequency:
f= 1/T
∆y :
h= L – Lcos(θ
θ)
radius:
r= L*sin(θ
θ)
velocity:
½ m V2 = mgh
∆x :
∆x= L* θ (in radians)
centripetal acceleration:
horizontal acceleration:
total energy:
gives the period at small angles
(frequency is measured in 1/sec or Hertz)
where h = ∆y
(only vertical drop matters)
(approximately horizontal for small angles)
a = mv2/r = Fnet/m = T- mgcos(θ
θ)/m
a = g sin(θ
θ)
θ
T
E= mgh + ½ mv2
θ
net force:
Fnet = T – mg (use vectors and free body diagrams!)
mg
Max values (derive from energy and free body diagrams!)
acceleration:
amax = gsin(θ
θ)… at highest point
velocity:
1/2m(Vmax )2 = mg ∆ymax … at lowest point
displacement:
ymax = L- Lcos(θ
θmax)
and
Xmax= L*(θ
θmax )… at highest point
5
SamplePendulum Problems
1. A pendulum follows simple harmonic motion only at small enough angles so that
sin(θ )≈ θ . Make a chart and estimate the largest angle that meets this criteria
*** if you use excel, use the formula: =SIN(RADIANS(A2)) A. close to 15
Angle
Sin()
Angle
Sin()
Angle
Sin()
Angle
Sin()
2. A grandfather clock swings back and forth in two second.
(mistake in original packet- period is 2 seconds- tic- toc)
a. What is the frequency in Hz of this papa clock? 0.5 Hz
f= 1/T = 1/2sec = 0.5 Hz
b. How long is the clock’s arm assuming it follows simple harmonic motion? 0.248m
T= 2π √ (l/g)
so l = g*T2/4π
π2= 9.8*4/4*pi*pi = 0.99 m
c. What would be the period if the mass of the clock arm was doubled? No change
d. How long should the arm be if we want it to work with the same period while
on Mars? (g= 4m/s2). l = g*T2/4π
π2= 4/9.8 of 0.99m = .40 m
6
3. A one meter long pendulum with a mass of 1kg is let go at θ= 70.
a. Draw a picture and determine the change in height. y= 0.007m
y= l-lcos(7) = 1-1cos(7) = .007
b. How much energy does it have when it is let go? 0.073J
PE= mgh = mgy = 1*9.8* .007 = .073
c. How fast is it traveling when the pendulum is at θ= 00? 0.38 m/s
1/2mv2 = mgh so v= √ (2*g*h)= √ (2*9.8*.007) = 0.38 m/s
d. what it its period? T= 2 sec
T= 2π √ (l/g) = T= 2π √ (1/9.8) = 2.007 sec
e. How much time did it take to go from θ= 70 to θ= 00? T= 0.5 sec
this is ¼ of a period so just ¼ * 2 = 0.5 sec
4. A heavy pendulum called a “charpy impact pendulum” is used to test the strength of
m = 10,0000 kg
materials by breaking them!
How much energy was needed to break the steel pieces above if a 2m long pendulum
had to be dropped from at least at θ = 100 to fracture and break the steel? E= 2978J
PE= mgh = 10,000 * 9.8 * (2- 2cos(10) = 2977.68
7
Angular Motion (see chap.8 for refresher)
(2πradians=360o)
∆θ = s/r
(radians= arc length/radius)
ω = ∆θ/t
(radians/sec of angular velocity)
α = ∆ω/t
(radians/sec2 of angular acceleration)
* angular frequency:
ω = 2π/T= 2 π f
s or x
r
θ
arc shown is about ½ the radius so it’s 0.5 radians!)
If compare to tangential motion:
x= r* θ
v= r*ω
a= r* α
ac= r *ω 2 ( Centripetal)
Simple Harmonic Motion:
*Harmonic force depends linearly on displacement: F= constant *∆
∆x
*Harmonic forces create sinusoidal motion such as in pendulums, springs, and elastic materials.
*velocity is 180 out of phase with displacement (at max velocity, displacement is min : x=0)
*acceleration is in phase with displacement (at max acceleration, displacement is max, x= “A” )
* displacement in angle :
∆x = A*cos(ωt))
V = −ωΑ * sin(ωt )
a = -ω2Α2 * cos(ωt))
θ = ωt
where ∆x max = Α
where v max = ω* Α
where a max = −ω2 ∗ Α2
Term
Springs (∆x = stretch)
Pendulum (small angles only)
Amplitude
A= max stretch from equilibrium
A ≈L* θ max
Force
F= -k* ∆x
F= T-mg≈ gsin(θ
θ)/m
Period
T= 2π √ (m/k)
T= 2π √ (l/g)
Energy
Emax = 1/2k(∆x)2
Emax = mg* (L-Lcos(θ
θ) )
8
Sinusoidal Motion due to harmonic forces
http://www.walter-fendt.de/ph14e/springpendulum.htm
http://www.myphysicslab.com/spring1.html
http://physics-animations.com/Physics/English/mech.htm
B
A
C
Fill out the plots below using the teacher demo or lame web site
C
B
A
B
C
B
A
B
C
B
A
B
C
B
A
B
force
B
acceleration
time
position
time
velocity
time
time
B
C
B
A
B
C
B
A
B
C
B
A
B
C
B
A
B
Summary:
At max position, what is else is maximum: _at C or A, max force and acceleration____?
At minimum position, only at B, velocity_____ is maximum.
If the force is minimum, then the acceleration is _min____________ (min or max?)
If the stretch is the most, the force is ___max____(the most or least?)
Work=Force times displacement, but force depends on displacement (depends or doesn’t
depend)
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Horizontal Spring motion!
A
1
Position
B
C
2
3
Displacement
Force
Total Energy
Position #1 :
(no stretch, max
speed, no force)
∆x =0
Position #2:
(little stretch, some
elastic and kinetic)
∆x = X2-X1
F=-k(X2-X1 )
E= 1/2k(X2-X1 )2 + 1/2mV22
Position #3:
(max stretch, most
force, no velocity)
∆x = X3-X1
F=-k(X3-X1 )
E= 1/2k(X3-X1 )2 + 0
•
•
•
•
•
•
•
(equilibrium) F=0
E= 0 + 1/2m(Vmax)2
Measure all distances from the equilibrium point where there is no stretching
No gravity involved, Friction is optional (motion similar, just won’t last as long)
Max stretch: most force, acceleration, displacement
No stretch: most velocity as mass oscillates back and forth
Find max speed using energy:
1/2m(Vmax)2 = ½ k (X3-X1 )2
Find spring constant using frequency:: ω = 2 π/T
π = 2π∗f
2π∗
Find max acceleration using amax = Fmax/ m = -k ∆xmax / m
10
Vertical Spring Motion! (JUST LIKE HORIZONTAL IF MEASURE X FROM “b”)
Force diagrams (g= +10m/s2)
A. No weight- spring not stretched
A
∆x= 0 , F= 0
B. Add weight but just sits there
Fs
F= -k(Xb-Xa) + mg=0
B
b
C
k= mg/(Xb-Xa)
mg
c
C. Pull weight down
F= -k (Xc-Xa) + mg
= -k(Xc-Xb) + -k (Xb-Xa) + mg………but note the second term is –mg from part B.
= -k(Xc-Xb) + -mg + mg
== -k(Xc-Xb) ……………………so just measure from “B” where it hangs naturally!
* TREAT LIKE A HORIZONTAL SPRING BUT MEASURE STRETCH FROM THE
EQUILIBRIUM POINT FOUND AFTER HANG ON MASS AT POINT B
SO…. F(x) = -k * (X - Xb)
•
Now Vertical Spring acts like a Horizontal spring!
( Fun by any stretch of the imagination)
11
Sample Spring Problems
5. A moving cart hits a spring, traveling at 5.0 m/s at the time of contact.
At the instant the cart is motionless, by how much is the spring compressed?
Initial
Ek Eg Ee
Energy Flow
Diagram
0
E k E g Ee
E int
0
Final
Initial
Final
1/2mv2= 1/2kx2 so x= v*sqrt (m/k)= 5*sqrt(8/50)= 2.0m
6..
A block is placed on a spring, compressing it 0.30m.
m = 500 g
v=0
Initial
Ek Eg Ee
Energy Flow
Diagram
Final
E k E g Ee
k = 100 N
m
²x = 0.30 m
Initial
0
Final
0
What height does the block reach when launched by the spring?
Mgh = 1/2k x2 so h= 1/2kx2/ mg = ½*100*(0.30) 2/ ( 0 .5kg *9.8) = .92m
7a... A spring whose spring constant is 850 N/m is compressed 0.40 m.
What is the maximum speed it can give to a 500. g ball?
1/2k x2 = 1/2mv2 so v= x* sqrt(k/m) = 0.40* sqrt(850/.5kg) = 16.5m/s
b. If spring were compressed twice as much, what would the velocity of the ball be?
Twice x means twice velocity so both are squared terms of energy
12
E int

N
8.. Suppose in the lab one group found that F = 1000.  ⋅ x Construct a graphical
m

representation of force vs. stretch. (Hint: make the maximum displacement 0.25 m. )
a.
Graphically determine the amount of
energy stored while stretching the spring described
above from x = 0 to x = 10. cm.
E=area = 2 boxes= 2* 50*.05= 5J
b.
Graphically determine the amount of
energy stored while stretching the spring described
above from x = 15 to x = 25 cm.
area = 8boxes= 8* 2.5= 20J
The graph at left was made from data
collected during an investigation of the
relationship between the amount two
different springs stretched, when different
forces were applied.
36.0
32.0
Data Set 1
Data Set 2
28.0
c.For each spring determine the spring
constant.
24.0
K= slope = 36/4.5 = 8.0 N/m
Force (N)
20.0
K= slope = 36/ 7 = 5.1 N/m
16.0
d.
12.0
For just the stiffest spring, find:
force to stretch the spring to 3.0 m.
8.00
F= kx = 8 N/m * 3m = 24 N
4.00
0.00
0.00
2.00
4.00
²x (m)
6.00
8.00
energy stored when stretched to 3.0 m.
E=1/2 kx2= ½ * 8 N/m * (3m)2 = 36 Nm
13
AP Problems
9. A pendulum consists of a small object of mass m fastened to the end of an inextensible
cord of length L. Initially, the pendulum is drawn aside through an angle of 60° with the
vertical and held by a horizontal string as shown in the diagram above. This string is burned
so that the pendulum is released to swing to and fro.
a.
draw a force diagram identifying forces acting on the object while it is held by the string.
b.
Determine the tension in the cord before the string is burned.
Wt = Tc*cos(60) so………Tc = Wt /cos(60)= 2Wt.= 2mg
c.
Show that the cord, strong enough to support the object before the string is burned, is
also strong enough to support the object as it passes through the bottom of its swing.
T –mg = mv2/r where r= L, v = sqrt (2gh) and h = L- Lcos(60) = L/2
V= sqrt(2gL/2) = sqrt(g/L) or v2 = g/L
plug in:
T = mg + m (g/L)/L = 2mg but this is equal to Tc (tension of cord so ok, won’t break)
d.
The motion of the pendulum after the string is burned is periodic.
Is it also simple harmonic? Why, or why not?
Not simple harmonic since the force = 2mg which is not –kx or ksin()
14
10. Suppose the object undergoes this motion because it is the bob of a
simple pendulum as shown above. If the object breaks loose from the
string at the instant the pendulum reaches its lowest point and hits the
ground at point P shown, what is the horizontal distance “d” it travels?
X= Vx* t
t= time to fall , get from tneom’s or energy since y= 0.5m= ½ *g*t2 or t=.319 sec
find Vx at bottom of swing using energy of pendulum (PE = KE)
Angle and length are not given (from another problem)
so pick L= 1m and angle=5 degrees:
1/2mVx2 = mgh where h = L- L*cos(θ) = 1- 1cos(5) = .0038
Vx = sqrt (2gh) = sqrt (2*9.8*.0038) = 0.27 m/s
Then x= Vx*t = 0.27 m/s * .319 sec = .087 m (yipeee)
15
11. A heavy ball swings at the end of a string as shown with negligible air resistance. Point P is
the lowest point reached by the ball in its motion. point Q is one of the two highest points.
a. Draw and label the vectors that could represent the velocity and acceleration of the ball at
points P & Q. If a vector is zero, state so. horizontal and vertical directions are dashed.
no horizontal acceleration, max speed
i. Point P
max acceleration, no speed
ii. Point Q
b. After several swings, the string breaks. The mass of the string and air resistance are
negligible. On the following diagrams, sketch the path of the ball if the break occurs when
the ball is at point P or Q. In each case, describe the motion of the ball after the break.
horizontal motion
i. Point P
free falls from test since v=0
ii. Point Q
16
12. A spring that can be assumed to be ideal hangs from a stand, as shown above.
a. You wish to determine experimentally the spring constant k of the spring.
i. What additional, commonly available equipment would you need?
force meter & meter stick (F=kx) or known mass, stopwatch (T= 2π √ (m/k)
ii. What measurements would you make?
force, stretching distance or time for a period
iii. How would k be determined from these measurements?
K= slope of F vs. x
or calculate from T= 2π √ (m/k)
b. Assume that the spring constant is 500 N/m. A 2.0-kg mass is attached to the lower end of
the spring and released from rest. Determine the frequency of oscillation of the mass.
f= 1/T but T=2π
2π √ (m/k) = 2π √ (2/500) = 0.4 sec ……..so T = 2.5 Hz
c. Suppose that the spring is now used in a spring scale that is limited to a maximum value of 25 N,
but you would like to weigh an object of mass M that weighs more than 25 N. You must use
commonly available equipment and the spring scale to determine the weight of the object
without breaking the scale.
i. Draw a clear diagram that shows one way that the equipment you choose could be used
with the spring scale to determine the weight of the object,
lots of ideas: levers, water (buoyance), counter weights, etc and equipment for these
ii. Explain how you would make the determination.
Include how avoid friction or unknowns
17
13. A 3.0 kg object subject to a restoring force F is undergoing simple
harmonic motion with a small amplitude. The potential energy U of the object
as a function of distance x from its equilibrium position is shown above. This
particular object has a total energy E: of 0.4 J.
(a) What is the object's potential energy when its displacement is +4 cm
from its equilibrium position?
read off graph, U or elastic potenial energy is about 0.05 J
(or use PE = 1/2kx2 a so 0.4J= 1/2k(0.1) 2 to gt k= 80 then do again U=
½*80*(.042) 2 = .06J (almost same)
(b) What is the farthest the object moves along the x axis in the positive
direction? Explain your reasoning.
10 cm since max energy is 0.4 J and will be all potential at ends (KE=0)
(c) Determine the object's kinetic energy when its displacement is 7 cm.
total is always 0.4J = potenial + kinetic
potential is about 0.18 from graph so kinetic = 0.40- 0.18 = 0.22J
(d) What is the object's speed at x = 0 ?
2
all kinetic energy = total energy = 0.4 J = 1/2mv = ½* 3kg * v
2
solve and get v= 0.5 m/s
18
14.
One end of a spring is attached to a solid wall while the other end just reaches to
the edge of a horizontal, frictionless tabletop, which is a distance h above the floor.
A block of mass M is placed against the end of the spring and pushed toward the
wall until the spring has been compressed a distance X, as shown above. The
block is released, follows the trajectory shown, and strikes the floor a horizontal
distance D from the edge of the table. Air resistance is negligible.
Determine expressions for the following quantities in terms of M, X, D, h, and g.
Note that these symbols do not include the spring constant.
a.
The time elapsed from the instant the block leaves the table to the instant it
strikes the floor
h= ½*g*t2 so t= √(2*h/g)
b.
The horizontal component of the velocity of the block just before it hits the floor
Vx = x/t = D/t = D/√(2*h/g)
c.
The work which is done on the block by the spring
Work = kinetic energy gained = 1/2mv2 = ½*m *D2/(2hg)= mD2g/4h
d.
The value of the spring constant “k”.
elastic = potential
½*k*x2 = ½ * m*v2
½*k*x2 = mD2g/4h
k= mD2g/(2h*x2)
19
15a. Label the graphs as either force, velocity, position, or acceleration
b. Which point (A,B,C) matches with equilibrium, max extension, and max compression?
B
A
B
C
B
A
B
C
C
B
A
B
C
B
A
B
C
B
A
B
force
?
acceleration
position
time
time
velocity
?
time
time
?
?
B
C
B
A
B
C
B
A
B
C
B
A
B
C
B
A
B
20
16. waves Unit I: Worksheet 2
A
The diagram to the right
shows a 0.100 kg block attached to a
Hookean spring. The block
experiences no net force when it is at
position C. When the block is to the
left of point C the spring pulls it to the
right. When the block is to the right of
point C, the spring pushes it to the
left.
B
C
5.0 cm
D
E
5.0 cm
10.0 cm
10.0 cm
0
0
The mass is pulled to the left from point C to point A and released. The block then oscillates
between positions A and E. Assume that the system consists of the block, spring, and the
horizontal surface on which the spring oscillates.
1. When there is no friction between the block and the surface, the force required to hold the
block at rest at position A is 5.00 N. Calculate the spring constant for the spring.
K= F/x = 5 N/ 0.1m = 50 N/m
2. Calculate the elastic energy at the following positions:
a. Position A
Ee= ½ k*x2 = ½ * 50 * .102 = 0.25 J
b. Position B.
c. Position C.
d. Position D.
e. Position E.
3.
Ee= ½ k*x2 = ½ * 50 * .052 = .0625 J
Ee= ½ k*x2 = ½ * 50 * 02 = 0
Ee= ½ k*x2 = ½ * 50 * .052 = .0625 J
Ee= ½ k*x2 = ½ * 50 * .12 = .25 J
Calculate the kinetic energy of the system at each of the following positions:
1. Position A
KE = 0
b. Position B
KE = total – Ee = 0.25- 0.625 = .1875J = ½ mv2 if want velocity
c. Position C
KE= .25 J
d. Position D
KE = 0.1875 J
e. Position E
KE = 0
4. Create a set of qualitative energy pie charts for this system starting from the instant of release
at point A and ending after the system has completed one full cycle and returns to position A
Position A
Position B
Position C
Position D
Position E
Position D
Position C
Position B
21
Position A
17. Waves Unit I: Worksheet 3
1. An ideal Hookean spring of spring constant 20.0 N/m is connected to a 0.500
kg block in the arrangement shown to the right. The (*) represents the
position of the center of the block when the spring is unstretched. (Positions
are not to scale.) From this position the experimenter slowly lowers the block
from (*) until it reaches point B where the system is at rest.
a. How far does the spring stretch when the 0.500 kg block is slowly
lowered to position B?
*
C
b. The block is then pulled to position A, 10.0 cm below position B. It is
10.0 cm
released and allowed to oscillate between positions A and C. In the
space below calculate the elastic energy, gravitational energy, and
kinetic energy of the system at positions A, B, and C.
B
10.0 cm
A
c. How fast is the block moving at the instant the center is even with position B?
d. Create quantitative energy bar graphs for the system at the positions indicated for one
complete cycle of the system starting when it is released at position A until it returns to
position A the first time.
Ek
Eg
Ee
Ediss
0
Ek
Eg
Ee
Ediss
0
Position A
Ek
Eg
Ee
Ediss
0
Position B
Ek
Eg
Ee
0
Position C
Ediss
Ek
Eg
Ee
Ediss
0
Position B
Position A
22