CHAPTER 20
### Chapter 20 - Applications of the Sine Function - Oscillatory Motion

Having gone through a long and detailed study of the Sine function the time arrives for us to study its applications.
There are many applications of the Sine function, however, the
most common are relate the dynamics of vibrating or oscillating systems. By
oscillating system we mean any object whose motion can be characterized as a
continuous back and forth never ending motion. Such motion occurs when the
resistance to motion of a body in motion is directly proportional to the
distance covered and acts in a direction that is opposite to the direction of
movement.

The typical and simplest example of
this is just a simple spring with a body or block of mass m attached to it.

If we assume that the block with
mass m rests on a friction less surface then the block is free to oscillate
with any horizontal force. According to Hooke's law the force required to
stretch a spring is directly proportional to the length of the spring. Each
spring therefore has a constant, known as the spring constant k which tells us
the force required to stretch the spring a given distance.

For example if the spring constant,
k, of a rather stiff spring such as those used in cars as opposed to a slinky,
were 1800 newtons per meter or k= 1800 N/m then to stretch the spring one meter
would require a force of 1800 Newtons and to further stretch it three meters
would require a force three times as great or 5400 N. This is all in accord
with Hooke’s Law, which states that the more you stretch it the greater the
force required.

Consider now the block above of mass
m at equilibrium position. By equilibrium position we mean the object is at
rest with no unbalanced force acting on it. This tells us that the spring is
being neither compressed nor stretched but is just lying there attached to the
block. If we stretch the block a distance x from its equilibrium position and
let go of it, then it will oscillate back and forth. Our goal is to
mathematically describe this motion.

The amplitude of oscillation is 2x
and the maximum force exerted by the spring on the block is kx. Keep in mind
however that the force is zero at the equilibrium position , since the spring
is un-stretched there and at this point the blocks entire energy is kinetic
while at the endpoints of its oscillation it possesses entirely potential
energy stored in the spring and its velocity is zero.

Since the spring exerts a force kx
when stretched, the work done stretch a spring a distance x is given by:

The work required to stretch a
spring a distance x is equal to
. When any mass m is stretched a distance x then let go of it
will be pulled back with a force that varies with the distance stretched.

The total energy of the system
varies between kinetic and potential energy depending on the mass’s position.
The kinetic energy of the block in motion is always equal to
, where v is its velocity at a particular point. At that
point x the potential energy will be
or the energy used to
stretch the spring to that position. The sum of the potential and kinetic
energy of the block at **any** point is
a constant that never changes. It corresponds to the net energy of the
oscillating system:

At x=0, the
velocity is a maximum and when v=0, x is a maximum and the block is at its
endpoint.

. Let us first solve the equation for v:

Multiplying both
sides by 2 gives:

Now substituting
velocity for
and separating
variable yields:

Integrating both
sides:

Recall that E
stood for the total energy of the system. At its endpoints the velocity of the
block is zero. The total energy is then equal to
where
equals half the
amplitude of oscillation.

Substituting
back into our integral gives us:

We can factor
out a
from the denominator
to get:

We can take out
the constant
from the integral to get:

We can now
factor out
from
and then remove the
constant A from the integral to get:

This integral
now has the familiar solution that we have studied:

where
and

has the
solution:

Now that we have solved it, how do
we interpret it or make sense of this solution? Basically this solution tells
us that the time required for the mass to a distance x from the equilibrium
position. We can therefore easily calculate the period of the oscillation, the
time required for the mass to complete one cycle. The period would be four
times the time required for the mass to move from x=0 to x=
. Substituting this into our solution gives:

This answer tells us that the period
of an oscillating mass on a spring is independent of the amplitude of the
oscillation. For this reason springs are used in watches. Despite
air-resistance and friction between the spring and other parts that would cause
the Amplitude to gradually decrease over time, the period of vibration would
remain unchanged.

Just as we are able to find time as
a function of x, we can find x as a function of time using the Sine function or
the inverse Sine Function. We saw that:

Multiplying both
sides by
gives us:

Taking the Sine
of both sides just switching around the dependent and independent variable:

( Many physics
books replace
with
w, where w=2p
f and f, frequency=
1/t, t= period =
. The reason being w
stands for radians/sec and
hence keeps the units in order and simplifies calculations.)

Now that we have the position x as a
function of time we can differentiate the function one to find the velocity and
twice to find the acceleration.

This expression
in
with its original
reference, x or position since
. This now gives us:

Since Force is
by definition mass* acceleration, if we multiply the function for acceleration
by m we get:

This is what we
originally started with, Hooke’s Law. The force in a spring is proportional to
the distance it is stretched and acts in the opposite direction of the motion
to oppose the motion.

Before ending you might be wondering
what happened to the constants? I decided not to put them in as to avoid any unnecessary
variables or numbers that would impede your understanding of the important
concepts being studied. The solutions with the constants included should here
forth read:

Summary

In this chapter we saw how
in a spring the force required to stretch the spring is directly proportional
to the length stretched. Recalling that
force is defined as mass* acceleration or F= ma, where acceleration is
the second derivative of the position function of an object in motion:

This can be
rewritten as

The only
function whose second derivative is proportional to negative of the original
function is the Sine function, since:

Therefore this chapter was in
essence devoted to the understanding of how the Sine function and its
derivative apply to applications where the second derivative is directly
proportional to the negative of the original function. This is why the Sine
function is restricted to Oscillatory motion, since a force -kx, will always
act so as to oppose the motion of the body in any direction. From the
Conservation of Energy the body will then rebound back in the opposite
direction once whatever force acting on it brings its velocity to zero because
all the energy will now have been transferred from the mass to the system. In
such a way this back and forth process goes on forever until interrupted by
other forces such as friction.