Simple Harmonic Motion

DIFFERENTIAL EQUATIONS
SIMPLE HARMONIC MOTIONS

WHAT IS A DIFFERENTIAL EQUATION?
• Differential Equations – Real world relations
involving rates at which one variable, say y, may
change with respect to another variable, t. Most
often, these relations take the form of equations
containing y and certain of the derivatives y', y'' ,
. . . , y(n) of y with respect to t. The resulting
equations are then referred to as differential
equations

CLASSIFICATION OF DIFFERENTIAL
EQUATIONS
Differential equations are classified according to
various criteria for a sensible treatment of
theory, solution methods, and solution behavior:
• ordinary differential equations and partial
differential equations
• single (or scalar) equations and systems of
equations
• linear equations and nonlinear equations
• first order equations and higher order equations

APPLICATONS OF DIFFERENTIAL
EQUATIONS
• Differential equations find their application in
various fields.Some of them are:
• Simple pendulum

Any vibrating system where the
restoring force is proportional to the
negative of the displacement is in
simple harmonic motion (SHM), and is
often called a simple harmonic
oscillator (SHO).
Simple Harmonic Motion

Substituting F = kx into Newton’s
second law gives the equation of
motion:
with solutions of the form:
F = -kx = m
dv
dt
= m
d
dt
dx
dt
æ
è
ç
ö
ø
÷ = m
d2
x
dt2
d2
x
dt2
+
k
m
x = 0

Substituting, we verify that this solution does
indeed satisfy the equation of motion, with:
The constants A and φ
will be determined by
initial conditions; A is
the amplitude, and φ
gives the phase of the
motion at t = 0.

The velocity can be found by differentiating the
displacement:
These figures illustrate the effect of φ:

Because then

Determine the period and frequency of a
car whose mass is 1400 kg and whose
shock absorbers have a spring constant
of 6.5 x 104 N/m after hitting a bump.
Assume the shock absorbers are poor,
so the car really oscillates up and down.

The velocity and
acceleration for simple
harmonic motion can
be found by
differentiating the
displacement:

A vibrating floor.
A large motor in a factory causes the
floor to vibrate at a frequency of 10 Hz.
The amplitude of the floor’s motion
near the motor is about 3.0 mm.
Estimate the maximum acceleration of
the floor near the motor.

The cone of a loudspeaker oscillates in SHM at a
frequency of 262 Hz (“middle C”). The amplitude
at the center of the cone is A = 1.5 x 10-4 m, and at
t = 0, x = A. (a) What equation describes the
motion of the center of the cone? (b) What are the
velocity and acceleration as a function of time? (c)
What is the position of the cone at t = 1.00 ms (=
1.00 x 10-3 s)?

Spring calculations.
A spring stretches 0.150 m when a 0.300-kg mass is
gently attached to it. The spring is then set up
horizontally with the 0.300-kg mass resting on a
frictionless table. The mass is pushed so that the spring
is compressed 0.100 m from the equilibrium point, and
released from rest. Determine: (a) the spring stiffness
constant k and angular frequency ω; (b) the amplitude of
the horizontal oscillation A; (c) the magnitude of the
maximum velocity vmax; (d) the magnitude of the
maximum acceleration amax of the mass; (e) the period T
and frequency f; (f) the displacement x as a function of
time; and (g) the velocity at t = 0.150 s.

Spring is started with a push.
Suppose the spring of the former example
(where ω = 8.08 s-1) is compressed 0.100 m
from equilibrium (x0 = -0.100 m) but is given
a shove to create a velocity in the +x
direction of v0 = 0.400 m/s. Determine (a) the
phase angle φ, (b) the amplitude A, and (c)
the displacement x as a function of time,
x(t).

We already know that the potential energy of a
spring is given by:
The total mechanical energy is then:
The total mechanical energy will be
conserved, as we are assuming the system
is frictionless.
Energy in the SHO

If the mass is at the
limits of its motion, the
energy is all potential.
If the mass is at the
equilibrium point, the
energy is all kinetic.
We know what the
potential energy is at the
turning points:
Energy in the SHO

The total energy is, therefore,
And we can write:
This can be solved for the velocity as a
function of position:
where
Energy in the SHO

This graph shows the potential energy
function of a spring. The total energy is
constant.
Energy in the SHO

Suppose this spring is
stretched twice as far (to
x = 2A).What happens to
(a) the energy of the
system, (b) the maximum
velocity of the oscillating
mass, (c) the maximum
acceleration of the mass?
Energy in the SHO

Simple Harmonic Motion

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