Download F = 0 x = 0 F =

PHYSICS 151 – Notes for Online Lecture 4.1
Periodicity
Periodic means that something repeats itself. For example, every twenty-four hours, the Earth makes a
complete rotation. Heartbeats are an example of periodic behavior. If you look at heartbeats on an
electrocardiogram, they make a regular pattern. The pattern that the heart obeys is rather complicated.
In this section, we’re going to be dealing with a specific type of periodic motion called simple
harmonic motion
Harmonic means that the motion can be described using sines and cosines.
Simple means that the motion can be described using a single frequency.
F=0
x=0
A mass on a spring
(horizontal or vertical) is a
good example of simple
harmonic motion (or SHM
for short). The motion of
the spring is repeated over
and over. Let’s start with
a horizontal spring, resting
on a frictionless table.
We pick a reference point
on the mass –for example,
the center of the mass.
The position of the center
of the mass when the
spring is unstretched is
called the ‘equilibrium
point’ (x = 0). Now I pull
x = x1
the mass an arbitrary
distance x to the right.
The spring exerts a force in the direction opposite the displacement (to the left in this case). The force
is given by Hooke’s Law:
F = -k x1
F = − kx
where k is the spring constant and has units of N/m. If I pull the spring to the right, the spring exerts a
force to the left. Alternately, I can push the spring in a distance x. Now the spring exerts a force
toward the right. Remember that Hooke’s law only works when the displacements are small. If you
make a very large displacement, Hooke’s law doesn’t apply anymore and none of what I’m about to
tell you will apply either.
A special characteristic of simple harmonic motion is that the acceleration is directly proportional to
the displacement. We can start with Newton's second law F = ma, and then insert Hooke's law for the
force on the spring.
F
m
− kx
a=
m
k
a=− x
m
a=
Any system in which the acceleration is proportional to the displacement will exhibit simple harmonic
motion. This can be tested experimentally. Plot F vs. x on a graph and take the slope of the resulting
straight line. If you do this and you don’t get a straight line, it means that the spring can’t be described
by Hooke’s law.
Simple Harmonic Motion Vocabulary
d isp la c e m e n t
tim e
When I pull the mass on a
spring and release it, the
mass exhibits a periodic
motion – the position of
the
spring
constantly
repeats itself. If I were to
plot the displacement of
the mass as a function of
time, it would look
something like this:
If you want to find out
where the mass is at any
point in time, you follow
the x-axis out to the time
you're interested in and they move up to the curve to see where the mass's position is.
We can define a number of characteristics of simple harmonic motion.
For example, the amplitude is the maximum displacement of the mass. The symbol for amplitude is
xo. This is a distance, so the units should be meters.
The time it takes for the mass to make one complete cycle – that is, to go from stretched to compressed
and back again – is called the period, which we represent by ‘T’. Remind yourself that the "picture"
of the wave is a picture of the mass as a function of time. It's not a snapshot of the wave itself.
The frequency is the number of cycles that are completed in one second. The frequency is given by
1
f=
T
If the mass takes 3.0 s to complete a cycle, the frequency is 1/3.0 = 0.33 (1/s). We have a special name
for the unit of frequency, which is the Hertz (Hz).
Hz =
1
s
Quantity
Symbol
Definition
Units
Period
T
time for one cycle
Frequency
f
number of cycles per second
Amplitude
xo
maximum displacement
s
1/s = Hz
m
d isp la cem en t
a m p litu d e
tim e
p erio d
Describing SHM using sines or cosines
The graph of the wave we have been diagramming can be expressed as a sine or cosine wave. In
general, we can write any SHM as a sine wave or a cosine wave. You know from trig that the sine and
cosine waves have periods of 2 π . The argument of the trig function has to be multiplied by a factor
such that the period of your wave is a multiple of 2 π . The scale factor turns out to be t/T. When the
time is equal to one period, you want your wave to be back where it started. At t = T, the argument is
equal to 2 π .
x( t ) = x o cos(2 π ?)
FG
H
x( t ) = x o cos 2 π
t
T
IJ
K
If you take a cosine wave and shift it by one quarter of a cycle (90 degrees or π/2 radians), you find
that the result is a sine wave.
⎛ 2π t π ⎞
⎛ 2π t ⎞
− ⎟ = xo sin ⎜
x(t ) = xo cos ⎜
⎟
2⎠
⎝ T
⎝ T ⎠
How do you know which is which? The answer is that you have to figure out how the wave starts. For
example, at t =0, the sine function will always be zero, regardless of the value of omega. The wave
below in blue must be a sine wave because it starts at zero.
displacement
Cosine wave
time
Sine wave
We can also find the velocity and the acceleration of the mass as a function of time. If
FG 2πt IJ
HTK
x( t ) = x o cos
then
v( t ) = − v o sin
and
FG 2πt IJ
HTK
FG 2πt IJ
HTK
a ( t ) = − a o cos
Note that our constraint that x and a must be proportional to each other is satisfied by these
expressions.
Ex. 1: The motion of an oscillator of mass 0.2 kg is given by:
b
x( t ) = (0.50 m) cos 2.09 t
g
where x is in m and t is in s
a) Find the amplitude
b) Find the period
c) Find the frequency of oscillation
d) Find the position of the mass at t = 0 s, 0.75 s,1.5 s, 3.0 s and 6.0 s
FG
H
We first have to put this in the same form -as x( t ) = x o cos 2π
t
T
IJ
K
This give us
t ⎞
⎛
x(t ) = ( 0.5m ) cos ⎜ 2π
⎟
⎝ 3.0 ⎠
a) xo = 0.50 m
b) The argument in the cosine function is
2πt
. The period must therefore be 3.0 s.
T
c) f = 1/T = 1/3.0 s = 0.33 Hz
Time
2 π t/T
FG
H
cos 2 π
t
3.0
IJ
K
X(t)
0
0
1
0.5 m
0.75 s
π /2
0
0
1.5 s
π
-1
-0.5 m
3.0
2π
1
0.5 m
Ex. 2: A 0.50-kg mass at the end of a horizontal spring has position 0 when t = 0. The amplitude is
0.15 m and the cycle starts by moving to the right first. The mass makes 2.0 complete oscillations each
second. What is the equation for the position as a function of time?
Solution:
The function will be either a sine or a cosine. How do we know which to pick? We’re told that the
position at t = 0 is x = 0. Compare the cosine and sin functions.
function
cos 2π
FG
H
t
T
FG
H
t
T
sin 2π
t=0
value
IJ
K
cos(0)
1
IJ
K
sin(0)
0
So anytime that the mass starts from x =0, you will have a sin function. If the mass starts from its
amplitude value, x0, you need to have a cosine function. Since we’re starting from 0, we need to use a
sin function.
x( t ) = x o sin
FG 2πt IJ
HTK
We are told that the system completes two oscillations every second. This is the frequency, f
f = 2 1/s
The period, T, is given by T = 1/f = 0.5 s
The amplitude is given to us as xo = 0.15 m. Putting these in our equation, we have:
b
. mg sinb4 πt g
g FGH 20π.5t IJK = b015
x( t ) = 015
. m sin
Why were we told that the oscillations started toward the right? So that we would know whether we
needed a positive or a negative sign out front. When the mass starts at zero, it can go either positive or
negative in displacement. If we take to the right as positive, the equation will not need a negative sign.
If the mass were going to the left, we would have a negative sign out front.
The vertical spring
What if the spring you have is hung vertically instead of horizontally? Does what we just discovered
still hold?
Ex. 3: A spring of spring constant k = 25 N/m has a mass of 0.5 kg hung from it. How far does
the spring stretch when the mass is placed on it?
When the mass is on the spring, it pulls the spring down, but then it just hangs there. We can draw a
free-body diagram for the mass. The acceleration is zero, and the only forces acting are gravity down
and the force of the spring up.
x=0
x = xeq
ΣF = 0
mg − kx eq = 0
m
g
k
(0.5kg)
=
9.8 sm2
25 N / m
x eq =
x eq
. m = 0.20 m
x eq = 0196
This is where the effect of gravity come in- it shifts the equilibrium position of the spring. Once this
has been accounted for - by taking the potential energy to be zero when the mass is at xo. Gravity has
no effect on the SH motion at all. Let’s look at the spring when it’s displaced a distance x. Draw the
free-body diagram.
F = k(x-xo)
x=0
F = mg
x = xeq
x = x + xeq
The net force is
F = k ( x + x eq ) − mg
We found in part a that xeq = mg/k.
FG mg IJ − mg
H kK
F mg IJ − mg
F = kG x +
H kK
F= k x+
F = kx + mg − mg
F = kx
The only force causing the SHM is the spring!
So analyzing SHM in the vertical and the horizontal directions is the same, except that the equilibrium
position shift must be accounted for.
Conservation of Energy for SHM
As the spring is stretched or compressed, energy is converted from the motion of the mass and spring
to energy stored in the coils and back again. The elastic potential energy due to a spring (and other
stretchy things like rubber bands) is:
PE el =
1 2
kx
2
where, unlike gravitational potential energy, we take the zero to be the equilibrium position of the
spring (i.e. x = 0 corresponds to the point at which there is zero potential energy).
We can write the total mechanical energy for a
spring as:
F=0
E = KE + PE
1
1
E = mv 2 + kx 2
2
2
xo
F = -k xo
At the maximum displacement (x = xo), the
mass is momentarily standing still. The total
energy is then:
v=0
1
1
mv 2 + kx 2
2
2
1
E = kx 20
2
E=
F=0
v =vo
All potential energy!
xo
F = k xo
When x = 0, the mass has a velocity vo, which
is the maximum velocity that the mass can
have. The total energy is then:
v=0
F=0
v = vo
1
1
mv 2 + kx 2
2
2
1
E = mv 2o
2
E=
All kinetic energy
Because the total energy is constant at every place along the motion,
1
1
mv o 2 = kx o2
2
2
k
v o 2 = x o2
m
k
vo =
xo
m
There is one other relationship that we will need to use (which can be derived by considering SHM is
the projection of circular motion).
T = 2π
m
k
Let's review the definitions and relationships we have
xo = amplitude = maximum displacement - occurs when v = 0 (A is also used for amplitude)
vo = maximum velocity – occurs when x = 0
T = period (s)
f = frequency (Hz = 1/s)
k = spring constant (N/m)
Relationships:
m
T = 2π
k
T=
1
f
1
2
mv o2 = 21 kx o2
v 2o =
f=
k 2
xo
m
1
2π
k
m
Ex. 4: The motion of an oscillator of mass 0.2 kg is given by:
b
x( t ) = (0.50 m) cos 2.09 t
g
where x is in m and t is in s. Note that this is the same equation as Example 23-1.
e) Find the spring constant
f) Find the total energy
g) Find the maximum velocity
FG
H
We first have to put this in the same form as x( t ) = x o cos 2π
t
T
IJ
K
This gives us
t ⎞
⎛
x(t ) = ( 0.5m ) cos ⎜ 2π
⎟
⎝ 3.0 ⎠
a) spring constant: We notice first that the period is 3.0 s, so
m
k
T = 2π
m
k
2 m
k = ( 2π ) 2
T
2 ( 0.2 kg )
k = ( 2π )
2
( 3.0 s )
T 2 = ( 2π )
2
k = 0.88 kg
= 0.88
s2
N
m
b) Total energy
E = 21 kx 2o
E=
1
2
c0.88 hb0.50mg
N
m
2
. x 10 −1 J
E = 11
c) Maximum velocity: The maximum velocity occurs when x = 0, so the energy is entirely kinetic
E = 21 mv o2
2E
= vo
m
2(11
. x 10 −1 J )
= vo
0.2 kg
105
. ms = v o
11
. ms = v o
You Try
It!
A 0.50-kg mass at the end of a horizontal spring is pulled back to a distance of 0.15 m. At t
= 0, the mass is released and makes 3.0 complete oscillations each second. Find:
a) the velocity when the mass passes the equilibrium point
b) the velocity when the mass is 0.10 m from equilibrium
c) the total mechanical energy of the system
Known: A = 0.15 m m = 0.50 kg f = 3.0 Hz
a) The quantity we are looking for is vo. In examining the equations for velocity and position, we
found that
vo =
k
xo
m
Unfortunately, we don't know k, but we can find k from
m
k
T = 2π
1
m
= 2π
f
k
1
=
2 πf
m
k
FG 1 IJ = m
H 2 πf K k
k = b 2 πf g m
2
2
Now put this in our expression for vo
vo =
k
xo
m
d 2 πf i m x
2
vo =
m
o
v o = 2 πf 2 x o
v o = 2 πfx o
v o = 2 π(3.0 Hz)(015
. m)
v o = 2.8 ms
1
2
mv 2 + 21 kx 2 = 21 mv o2
k 2
x = v o2
m
k
v 2 = v o2 − x 2
m
v2 +
b 2 πf g m x
−
2
2
Using conservation of energy: v =
v o2
v =
v 2o
2
v=
v=
2
m
b gx
v − b 2 πf g x
. mg
c2.8 h − c2π(3.0 h b010
− 2 πf
2
o
m 2
s
2
2
2
2
1 2
s
v = 2.1 ms
Stop to see if this makes sense. The velocity must be less than vo, which it is.
c) Total energy
2
E = 21 mv o2
E=
1
2
b0.5kggc2.8 h
m 2
s
E = 2.0J
Pendula
A simple pendulum
consists of an object
suspended from a
string. The motion
of the pendulum
swinging back and
forth can also be
described by simple
harmonic
motion
(under
certain
conditions).
L
θ
T
mg
θ mg cos(θ)
mg sin(θ)
let’s first draw a free-body diagram for the bob on the pendulum. We can decompose the weight, mg,
into a component along the direction of the string and a component perpendicular to the string.
The component of the weight in the direction of the string will be:
mg cos(θ)
The component of the weight perpendicular to the string is
-mg sin(θ)
r
Write ΣF = 0 . We have to write one equation for each direction.
Along the string
Perpendicular to the string
In the direction along the string, there is no There is, however, acceleration in the direction
acceleration, so taking toward the pivot as perpendicular to the string due to the
positive,
unbalanced force:
r
ΣFalong the string = 0
r
ΣFperpendicular to the string = ma
T − mg cos(θ) = 0
F = − mg sin(θ)
T = mg cos(θ)
One of the things we learned in the last lecture, is that object that undergo simple harmonic motion
following something that looks like Hooke’s law, with the force proportional to the displacement. This
equation doesn’t look like Hooke’s law.
θ
L
θ
L
θ
L
If θ is very small, we can make an approximation. If we let s be the displacement of the pendulum
s
bob, s will be an arc. If θ is small, we can write that sin θ =
L
We can then write
F = − mg sin(θ)
F = − mg
F=−
so that the ‘spring constant’ for this problem is
s
L
mg
s
L
mg
s
L
mg
F=−
s
L
F = −" k" s
F=−
FG IJ
H K
where the effective spring constant is given by
k=
mg
L
We can use the same relationships we derived for the mass on a spring to find the similar quantities for
the pendulum
Spring
f=
T=
1
2π
Pendulum
k
m
1
m
= 2π
f
k
f=
T=
1
2π
g
L
1
L
= 2π
f
g
Note that, for the pendulum, all of these results are independent of the mass of the pendulum.
Energy Concerns for the Pendulum
Take h = 0 to be the lowest part of the pendulum swing. The height at any point is then
h = L-Lcos(θ)
At the top of the swing,
the velocity is zero, so
there is no kinetic energy
and the total energy is
entirely potential
L
Θ
L
L cos(θ)
E = PE = mgh
E = mgL(1-cos(θ))
At the bottom of the
swing,
the
potential
energy is zero and all of
the energy is kinetic, so
h = 0 here
L-L cos(θ)
where vo is the maximum velocity.
E = 21 mv 2o
From conservation of energy, the total energy at these two points must be equal.
b
E = 21 mv o2 = mgL 1 − cos(θ)
v o2
b
= 2gL 1 − cos(θ)
b
g
v o = 2gL 1 − cos(θ)
g
g
Ex. 5: A pendulum of length 0.50 m makes 21.0 oscillations in 30.0 seconds. What is the value of g?
frequency =
number of oscillations 21.0
=
= 0.7 Hz
time
30.0
f=
g
L
1
2π
b2πg f = Lg
b2πg f = Lg
b2 π g f L = g
b2πg c0.7 h (0.5m) = g
2
2
2
2
2
2
2
1 2
s
9.67 sm2 = g
Ex. 6: A pendulum with a frequency of 6 oscillations per second is taken to the moon, where gravity is
1/6th the gravity of earth. What will the pendulum’s frequency be on the moon?
Solution: The pendulum length is constant.
f=
1
2π
g
L
b g
b 2 πf g
g
L
g
=
L
g
f 2π =
2
L=
b2πf g
L=
1 g
4π 2 f 2
2
Earth
L=
Moon
1 g Earth
4 π 2 f Earth 2
L=
b g
1 g Moon
4 π 2 f Moon 2
b
g
These two must be equal, so
L=
g Earth
b
g
g Moon
bf g bf g
bf g = bf g g
bf g = bf g gg
2
Earth
g Earth
=
1 g Earth
1 g Moon
=
2
2
4 π f Earth
4 π 2 f Moon 2
2
Moon
2
Moon
2
Earth
2
Moon
Moon
2
Moon
Earth
f Moon = f Earth
f Moon = f Earth
Earth
g Moon
g Earth
1
6
g Earth
g Earth
f Moon = f Earth
1
6
f Moon = 6 Hz
1
= 2.5 Hz
6
b
g