F ∆t = ∆p
21
1
Replacing F with − F we obtain
21
12
− F ∆t = ∆p (26-4)
12
1
Now add equation 26-3 ( F ∆t = ∆p ) and equation 26-4 together. The result is:
12
2
F ∆t − F ∆t = ∆p + ∆p
12
12
1
2
0 = ∆p + ∆p
1
2
On the right is the total change in momentum for the pair of particles p
∆
= p
∆ + ∆p so
TOTAL
1
2
what we have found is that
0 = ∆p
TOTAL
which can be written as
∆p
= 0 (26-5)
TOTAL
Recapping: If the net external force acting on a pair of particles is zero, the total momentum of
the pair of particles does not change. Add a third particle to the mix and any momentum change
that it might experience because of forces exerted on it by the original two particles would be
canceled by the momentum changes experienced by the other two particles as a result of the
interaction forces exerted on them by the third particle. We can extend this to any number of
particles, and since objects are made of particles, the concept applies to objects. That is, if,
during some time interval, the net external force exerted on a system of objects is zero, then the
momentum of that system of objects will not change.
As you should recall from Chapter 4, the concept is referred to as conservation of momentum for
the special case in which there is no net transfer of momentum to the system from the
surroundings, and you apply it in the case of some physical process such as a collision, by
picking a before instant and an after instant, drawing a sketch of the situation at each instant, and
writing the fact that, the momentum in the before picture has to be equal to the momentum in the
after picture, in equation form: p = p′ . When you read this chapter, you should again consider
yourself responsible for solving any of the problems, and answering any of the questions, that
you were responsible for back in Chapter 4.
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Chapter 27 Oscillations: Introduction, Mass on a Spring
27 Oscillations: Introduction, Mass on a Spring
If a simple harmonic oscillation problem does not involve the time, you should
probably be using conservation of energy to solve it. A common “tactical error”
in problems involving oscillations is to manipulate the equations giving the
position and velocity as a function of time, x = x
2
cos(
f
π )t and
max
v = v
−
2
sin( f
π )t rather than applying the principle of conservation of energy.
max
This turns an easy five-minute problem into a difficult fifteen-minute problem.
When something goes back and forth we say it vibrates or oscillates. In many cases oscillations
involve an object whose position as a function of time is well characterized by the sine or cosine
function of the product of a constant and elapsed time. Such motion is referred to as sinusoidal
oscillation. It is also referred to as simple harmonic motion.
Math Aside: The Cosine Function
By now, you have had a great deal of experience with the cosine function of an angle as the ratio
of the adjacent to the hypotenuse of a right triangle. This definition covers angles from 0 radians
π
to
radians (0° to 90°). In applying the cosine function to simple harmonic motion, we use the
2
extended definition which covers all angles. The extended definition of the cosine of the angle θ
is that the cosine of an angle is the x component of a unit vector, the tail of which is on the origin
of an x-y coordinate system; a unit vector that originally pointed in the +x direction but has since
been rotated counterclockwise from our viewpoint, through the angle θ , about the origin.
Here we show that the extended definition is consistent with the “adjacent over hypotenuse”
π
definition, for angles between 0 radians and
radians.
2
For such angles, we have:
y
u
u
θ
y
u
x
x
in which, u, being the magnitude of a unit vector, is of course equal to 1, the pure number 1 with
no units. Now, according to the ordinary definition of the cosine of θ as the adjacent over the
hypotenuse:
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Chapter 27 Oscillations: Introduction, Mass on a Spring
ux
cosθ =
u
Solving this for u we see that
x
u = u cosθ
x
Recalling that u = 1, this means that
u = cosθ
x
Recalling that our extended definition of cosθ is, that it is the x component of the unit vector uˆ
when uˆ makes an angle θ with the x-axis, this last equation is just saying that, for the case at
π
hand (θ between 0 and
radians) our extended definition of cosθ is equivalent to our ordinary
2
definition.
π
3π
At angles between
and
radians (90° and 270°) we see that u takes on negative values
2
2
x
(when the x component vector is pointing in the negative x direction, the x component value is,
by definition, negative). According to our extended definition, cosθ takes on negative values at
such angles as well.
y
u
u
y
θ
u
x
x
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Chapter 27 Oscillations: Introduction, Mass on a Spring
With our extended definition, valid for any angle θ, a graph of the cosθ vs. θ appears as:
cos θ
1
0
π
π
2
π
3
π
4
θ [radians]
-1
Some Calculus Relations Involving the Cosine
The derivative of the cosine of θ, with respect to θ :
d cosθ = −sinθ
dθ
The derivative of the sine of θ, with respect to θ :
d sinθ = cosθ
dθ
Some Jargon Involving The Sine And Cosine Functions
When you express, define, or evaluate the function of something, that something is called the
argument of the function. For instance, suppose the function is the square root function and the
expression in question is 3x . The expression is the square root of 3x, so, in that expression, 3x
is the argument of the square root function. Now when you take the cosine of something, that
something is called the argument of the cosine, but in the case of the sine and cosine functions,
we give it another name as well, namely, the phase. So, when you write cosθ, the variable θ is
the argument of the cosine function, but it is also referred to as the phase of the cosine function.
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Chapter 27 Oscillations: Introduction, Mass on a Spring
In order for an expression involving the cosine function to be at all meaningful, the phase of the
cosine must have units of angle (for instance, radians or degrees).
A Block Attached to the End of a Spring
Consider a block of mass m on a frictionless horizontal surface. The block is attached, by means
of an ideal massless horizontal spring having force constant k, to a wall. A person has pulled the
block out, directly away from the wall, and released it from rest. The block oscillates back and
forth (toward and away from the wall), on the end of the spring. We would like to find equations
that give the block’s position, velocity, and acceleration as functions of time. We start by
applying Newton’s 2nd Law to the block. Before drawing the free body diagram we draw a
sketch to help identify our one-dimensional coordinate system. We will call the horizontal
position of the point at which the spring is attached, the position x of the block. The origin of our
coordinate system will be the position at which the spring is neither stretched nor compressed.
When the position x is positive, the spring is stretched and exerts a force, on the block, in the
x
− direction. When the position of x is negative, the spring is compressed and exerts a force, on
the block, in the +x direction.
k
m
+x
x=0
Equilibrium Position
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Chapter 27 Oscillations: Introduction, Mass on a Spring
Now we draw the free body diagram of the block:
FN
a
F
x
S = k
m
F
+x
= mg
g
and apply Newton’s 2nd Law:
1
a
=
→
∑F
∑
m
→
1
a =
(− kx)
m
k
a = −
x
m
This equation, relating the acceleration of the block to its position x, can be considered to be an
equation relating the position of the block to time if we substitute for a using:
dv
a =
dt
and
dx
v =
dt
so
d dx
a =
dt dt
which is usually written
2
d x
a =
(27-1)
2
dt
and read “d squared x by dt squared” or “the second derivative of x with respect to t .”
k
Substituting this expression for a into a = −
x (the result we derived from Newton’s 2nd Law
m
above) yields
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Chapter 27 Oscillations: Introduction, Mass on a Spring
d 2 x
k
= −
x (27-2)
dt 2
m
We know in advance that the position of the block depends on time. That is to say, x is a
function of time. This equation, equation 27-2, tells us that if you take the second derivative of x
with respect to time you get x itself, times a negative constant (−k/m).
We can find the an expression for x in terms of t that solves 27-2 by the method of “guess and
check.” Grossly, we’re looking for a function whose second derivative is essentially the negative
of itself. Two functions meet this criterion, the sine and the cosine. Either will work. We
arbitrarily choose to use the cosine function. We include some constants in our trial solution
(our guess) to be determined during the “check” part of our procedure. Here’s our trial solution:
2π rad
x = x
cos
max
t
T
Here’s how we have arrived at this trial solution: Having established that x, depends on the
cosine of a multiple of the time variable, we let the units be our guide. We need the time t to be
part of the argument of the cosine, but we can’t take the cosine of something unless that
2π rad
something has units of angle. The constant
, with the constant T having units of time
T
(we’ll use seconds), makes it so that the argument of the cosine has units of radians. It is,
2π rad
however, more than just the units that motivates us to choose the ratio
as the constant.
T
To make the argument of the cosine have units of radians, all we need is a constant with units of
2π rad
radians per second. So why write it as
? Here’s the explanation: The block goes back
T
and forth. That is, it repeats its motion over and over again as time goes by. Starting with the
block at its maximum distance from the wall, the block moves toward the wall, reaches its
closest point of approach to the wall and then comes back out to its maximum distance from the
wall. At that point, it’s right back where it started from. We define the constant value of time T
to be the amount of time that it takes for one iteration of the motion.
Now consider the cosine function. We chose it because its second derivative is the negative of
itself, but it is looking better and better as a function that gives the position of the block as a
function of time because it too repeats itself as its phase (the argument of the cosine) continually
increases. Suppose the phase starts out as 0 at time 0. The cosine of 0 radians is 1, the biggest
the cosine ever gets. We can make this correspond to the block being at its maximum distance
from the wall. As the phase increases, the cosine gets smaller, then goes negative, eventually
reaching the value −1 when the phase is π radians. This could correspond to the block being
closest to the wall. Then, as the phase continues to increase, the cosine increases until, when the
phase is 2π, the cosine is back up to 1 corresponding to the block being right back where it
started from. From here, as the phase of the cosine continues to increase from 2π to 4π, the
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Chapter 27 Oscillations: Introduction, Mass on a Spring
cosine again takes on all the values that it took on from 0 to 2π. The same thing happens again
as the phase increases from 4π to 6π, from 8π to 10π, etc.
2π rad
Getting back to that constant
that we “guessed” should be in the phase of the cosine in
T
our trial solution for x(t):
2π rad
x = x
cos
max
t
T
With T being defined as the time it takes for the block to go back and forth once, look what
happens to the phase of the cosine as the stopwatch reading continually increases. Starting from
2π rad
0, as t increases from 0 to T, the phase of the cosine,
t , increases from 0 to 2π radians.
T
So, just as the block, from time 0 to time T, goes though one cycle of its motion, the cosine, from
time 0 to time T, goes through one cycle of its pattern. As the stopwatch reading increases from
T to 2T, the phase of the cosine increases from 2π rad to 4π rad. The block undergoes the second
cycle of its motion and the cosine function used to determine the position of the block goes
through the second cycle of its pattern. The idea holds true for any time t —as the stopwatch
reading continues to increase, the cosine function keeps repeating its cycle in exact
synchronization with the block, as it must if its value is to accurately represent the position of the
2π rad
block as a function of time. Again, it is no coincidence. We chose the constant
in the
T
phase of the cosine so that things would work out this way.
A few words on jargon are in order before we move on. The time T that it takes for the block to
complete one full cycle of its motion is referred to as the period of the oscillations of the block.
2π rad
Now how about that other constant, the “x
x =
max” in our educated guess
x
cos
?
max
t
T
Again, the units were our guide. When you take the cosine of an angle, you get a pure number, a
value with no units. So, we need the xmax there to give our function units of distance (we’ll use
meters). We can further relate xmax to the motion of the block. The biggest the cosine of the
phase can ever get is 1, thus, the biggest xmax times the cosine of the phase can ever get is xmax.
2π rad
So, in the expression x = x
cos
, with x being the position of the block at any time t,
max
t
T
xmax must be the maximum position of the block, the position of the block, relative to its
equilibrium position, when it is as far from the wall as it ever gets.
2π rad
Okay, we’ve given a lot of reasons why x = x
cos
should well describe the motion
max
t
T
of the block, but unless it is consistent with Newton’s 2nd Law, that is, unless it satisfies equation
27-2:
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Chapter 27 Oscillations: Introduction, Mass on a Spring
d 2x
k
= −
x
dt 2
m
which we derived from Newton’s 2nd Law, it is no good. So, let’s plug it into equation 27-2 and
2
d x
see if it works. First, let’s take the second derivative
of our trial solution with respect to t
2
dt
(so we can plug it and x itself directly into equation 27-2):
Given
2π rad
x = x
cos
,
max
t
T
the first derivative is
dx
2π rad π
=
2 rad
xmax − sin
t
dt
T
T
dx
2π rad
2π rad
= −
x
sin
max
t
dt
T
T
The second derivative is then
d 2x
2π rad
2π rad π
= −
2 rad
x
cos
2
max
t
dt
T
T
T
2
d 2x
2π rad
2π rad
= −
x
cos
2
max
t
dt
T
T
2π rad
Now we are ready to substitute this and x itself, x = x
cos
, into the differential
max
t
T
d 2x
k
equation
= −
x (equation 27-2) stemming from Newton’s 2nd Law of Motion. The
dt 2
m
substitution yields:
2
2π rad
2π rad
k
2π rad
−
x
cos
max
t = −
x
cos
max
t
T
T
m
T
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Chapter 27 Oscillations: Introduction, Mass on a Spring
which we copy here for your convenience.
2
2π rad
2π rad
k
2π rad
−
x
cos
max
t = −
x
cos
max
t
T
T
m
T