Calculus Of One Real Variable – By Pheng Kim Ving
Chapter 6: The Trigonometric Functions And Their Inverses – Section 6.1.7: The Simple Harmonic Function

 

6.1.7
The Simple Harmonic Motion

 

 

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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 y-axis 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 y-axis 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 y0 from its EP (displacement = y0)  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 y-position.

 

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2. Equations

 

Expecting Trigonometric Functions

 

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 v-axis, 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 u-axis, which is the cosine axis, will yield the same comparison.

 

Fig. 2.1

 

Point M moves in same fashion as mass m.

 

Equation Of General Simple Harmonic Motion

 

 

 

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 = d2y/dt2 = 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
highest-order 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 = x2, or y = x2 + 3, or y = x2 – 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 = x2 + 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 y0 and initial velocity v0 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

 

Sinusoidal Motion

 

We wish to write the equation:

 

 

 

 

 

Periodicity

 

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:

 

 

 

Amplitude

 

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 v0 = 0, which is the case where the mass is released, not thrown or flicked, then R = y0.

 

Graph

 

 

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.

 

 

 

 

Example 4.1

 

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.

 

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Problems & Solutions

 

 

 

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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.

 

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 y = – (5/2) sin 2t,

 

where y is measured in cm and t in sec.

 

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4.  A 100-gm mass is suspended from a spring. A force of 3 x 104 dynes (1 dyne = 1 gm cm/sec2) 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:

 

 

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5.  An elastic spring has a constant k = 9 x 104 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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