Calculus Of One Real Variable By Pheng Kim Ving

6.1.7 
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1. The Simple Harmonic Motion 
Consider a mass m suspended vertically
by an elastic spring attached to a beam. See Fig. 1.1. It has stretched the
spring
for some distance. It hangs unmoving in its position of rest, which is its
equilibrium position. Let's abbreviate equilibrium
position as EP.
Let the yaxis be vertical, with origin at
the same height as the EP of the mass, and directed upward. So
the EP of the mass is y = 0. Now
let's pull the mass vertically downward to some distance below its EP, without
exceeding
the elastic limit of the spring, and then release it. It moves up and down,
between points above and below its EP. Its
position on the yaxis oscillates between
positive and negative quantities. It experiences a restoring force that tends
to
restore it back to its EP. This force is exerted by the spring. The same motion
can be generated by instead raising the
mass vertically upward to some distance above its EP and then releasing it.
When the mass is displaced to some distance y_{0} from its EP (displacement = y_{0}) and before it's released, it's acted upon
by the restoring force. While it's moving up and down, at each position y of it (displacement = y)
it's also acted upon by
Fig. 1.1
Mass m suspended
by a spring is at its equilibrium position y = 0.

the restoring force. By Hooke's law, this force is
proportional to the displacement. Motion governed by such a restoring
force is called simple harmonic motion. Note that the
displacement of the mass from its EP is its yposition.
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2. Equations 
We now are going to establish the equations of the simple harmonic
motion, which relate its position to time. First let's
think of what functions we should expect to be involved in the equations. Refer
to Fig. 1.1. Suppose the mass m
is moving
up and down between points above and below its EP. When m
moves up from point below to point above, it speeds up at
successive points between point below and EP, attains maximum speed at EP,
slows down at successive points between EP
and point above, stops at point above, moves down towards point below, speeds
up at successive points between point
above and EP, attains maximum speed at EP, slows down at successive points
between EP and point below, stops at point
below, moves up towards point above, etc, repeating the same type of motion.
Now look at the unit circle in Fig. 2.1. Suppose point P moves counterclockwise on the circle at a
constant speed. Then
clearly the perpendicular projection M of P on the vaxis, which
is the sine axis, moves between 1 and 1 in the same
fashion as does the mass m between
points above and below described above, with point 0 corresponding to EP, point
1
to point above, and point 1 to point below. Since the position of M is a value of the sine function, we expect that
trigonometric functions are involved in the equations of the simple harmonic
motion. Note that considering instead the
motion of the perpendicular projection of P on the uaxis, which is the cosine axis, will yield the
same comparison.
Fig. 2.1 Point M moves in same fashion as mass m. 
The spring constant is the constant k > 0 such that F = ky, where F is the restoring force exerted by the spring on the mass and y is the position of the mass. 
Assuming that there are no air resistance and no friction, F is the only force acting on the mass. By
Newton's law of
motion, we have F = ma,
where a is the acceleration of the mass m produced by the force F.
Now a = d^{2}y/dt^{2} = y'',
as
by Section
5.8 Part 3. Consequently F = my''.
Hence ky = my'', or my'' + ky
= 0. Dividing by m yields y'' + (k/m)y =
A differential equation is an equation that involves
the derivatives of some orders greater than or equal to 1 of a function.
Recall that a function is considered as its own derivative of order 0: f ^{(0)} = f, or y ^{(0)} = y.
The differential equation of the simple harmonic motion is:

Eq. [2.1] is an example of a 2nd order homogeneous linear
differential equation (
2nd order because the order of the
highestorder derivative is 2, not because the exponent of w is
2), which is usually encountered in an introductory
differential equations course. Here we admit the solution without proof. Problem & Solution 1 provides a way of proving
it.
Consider the simple differential equation y' 2x = 0. We
have y' = 2x. So y = x^{2}, or y = x^{2} + 3, or y = x^{2} 100.38, since if
we differentiate each of these functions then we get y'
= 2x, so that y'
2x = 0. Each of these functions
satisfies and thus
solves the differential equation, and as a consequence is called a solution
or a particular solution of the differential equation.
The function y = x^{2} + C, where C is an arbitrary constant, also is a solution
and represents all the solutions, and hence is
called the general solution of the differential equation.
The general
solution of Eq. [2.1] is the equation of the general simple harmonic motion
and is: where A and B are arbitrary constants. 
This is a relationship between position y
and time t. The position y
is a function of time t. We can
check the correctness
of this general solution as follows:
Indeed this general solution is correct. As expected,
trigonometric functions are involved in the equation of the simple
harmonic motion.
The equation of
the particular simple harmonic motion with initial position y_{0} and initial velocity v_{0} is: 
Eq. [2.4] is a relationship between position y and time t. The position y is a function of time t.
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3. Sinusoidal Motion 
We wish to write the equation:

Let's find the period of the motion, denoted T, for time, since the period is a length
of intervals of the independent
variable and here the independent variable is time t,
and thus the period here is a time period. Recall from
Section
6.1.5 Part 2 that a function f(x) is periodic with period p > 0 if f(x + p) = f(x) for all x in its domain. We have:
Hence:
The period of the motion is: 
Now let's find the amplitude of
the motion. Since the value of the sine function oscillates between 1 and 1,
Eq. [3.1]
shows that y oscillates between R and R.
Consequently:
The amplitude of the motion is: 
Note from Eq. [3.2] that if v_{0} = 0, which is the case where the mass is released, not thrown or flicked, then R = y_{0}.
Fig. 3.1 Graph Of Simple Harmonic Motion. 
The motion will go on and on forever if indeed there are no
air resistance and no friction as assumed. However in
ordinary circumstances these two forces do exist. The motion will eventually be
stopped by them, and the mass will be
back at its EP.
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4. Frequency 
Let the unit of time be the second. Suppose the period of a
simple harmonic motion is 1 second. This means that the
motion repeats itself, ie completes 1 cycle, every 1 second. So the number of
cycles per second is 1. Now suppose the
period is 1/2 second, which means that the motion completes 1 cycle every 1/2
second, and thus the number of cycles per
second is 2 = 1/(1/2). Similarly, if the period is 2 seconds, then the number
of cycles per second is 1/2. Generally, given
that the period is T seconds, the
number of cycles per second is 1/T.

The position in cm of a mass in simple harmonic motion is
given by y = 3 cos 2t
after t
sec. Find:
a. The amplitude of the motion.
b. The period.
Problems & Solutions 
2. The position
in cm of a mass in simple harmonic motion is given by y
= 4 cos 3t after t sec. Find:
a. The amplitude of the
motion.
b. The period.
y = (5/2) sin 2t,
where y is measured in cm and t in sec.
4. A 100gm
mass is suspended from a spring. A force of 3 x 10^{4} dynes (1 dyne = 1 gm cm/sec^{2}) is required
to displace
the mass from its equilibrium
position by 1/3 cm. The mass is pulled down 2 cm below equilibrium, and then
released.
Find:
5. An elastic spring has a constant k = 9 x 10^{4} dynes/cm.
a. What mass should be suspended from the
spring to provide a system whose frequency is 10 Hz?
b. Find the position of such a mass from its equilibrium position t sec after
it's pulled down 1 cm from equilibrium and
flicked upward with a speed of
2 cm/sec.
c. Find the amplitude of the motion of the mass.
Solution
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