SIMPLE HARMONIC MOTION

• Simple harmonic motion is an example of

oscillatory motion. SHM is caused by
restoring forces. It is defined as the motion
of an object whose acceleration is
proportional to the distance from a fixed
point and is always directed towards that
point.
• i.e. a= -ω2x …… 1 where ω is the
constant of proportionality, a is
acceleration and x is the distance.
• For instance, when a body is
displaced from its equilibrium
position, it experiences a restoring
force which causes it to undergo
back and forth motion about the
equilibrium position. Such is an
illustration of simple harmonic
motion.
• Figure 1
• The figure below is an illustration of SHM.
Let us consider a point C, moving round a
circle of radius, r, with uniform speed, w. As
C moves round the circle once, the foot of
the perpendicular D moves from B through
O to A and back to B. The velocity of D
along AB. It has its greatest velocity at O,
the center motion of D but a zero velocity
at A and B. This to and fro motion along the
fixed diameter is defined as SHM.
• NOTE: The negative sign in equation 1
indicates that although the
acceleration is larger at larger
displacements, it is always in opposite
direction to the displacement, i.e
towards O (equilibrium point).
• In practice, many mechanical oscillations
are nearly simple harmonic, especially at
small amplitudes, or are combinations of
such oscillations. Any system which obeys
Hooke’s law will exhibit this type of
motion when vibrating.
• It is defined as an oscillatory motion in
which the acceleration a is directly
proportional to the displacement from
the equilibrium position and is in
• Another example of simple harmonic
oscillation is the oscillatory motion of a
simple pendulum, which consists of a
mass suspended from a string. In effect,
equation 1 also means that the
acceleration of a periodic motion at
each instant is proportional to the
negative of the displacement at that
instant.
 Characteristics of SHM

• Repetitive motion through a central

equilibrium point.
• Symmetry of maximum displacement.
• Period of each cycle is constant.
• Force causing the motion is directed
toward the equilibrium point (minus sign).
• F directly proportional to the
displacement from equilibrium.
 Equations of SHM
Suppose a point P moves round a circle of radius r
and center O with uniform angular velocity ω, its
speed V round the circumference will be constant
and equal to ωr (fig 2), As P revolves, N, the foot of
the perpendicular from P on to the diameter AOB,
moves from A to O to B and returns through O to A
as P completes each revolution. Let P and N be in
the positions shown at time t after leaving A, with
radius OP making angle θ with OA and distance ON
being x. we will then show that N describes SHM
about O.
1. Acceleration: The motion of N is directed to P, the
acceleration of P is ω2r (or v2/r) along PO and so the
component of this parallel to AB is ω2r Cosθ
Hence, a ₌ - ω2r Cos θ where x = r Cos θ therefore;
a = -ω2x
The negative sign indicates that a is always directed
towards O.
2. Period: The period T of N is the time taken for N to
make one complete oscillation from A to B and back
again.
T = (circumference of auxiliary circle)/(speed
of P)
T =2πr/V = 2π/ω (since V = ωr)
Velocity: The velocity of N is the displacement of P’s
velocity parallel to AB which is
= -V Sin θ
= - ωr Sin θ since Sin θ is positive when 0˂
θ˂1800 therefore, V = -ωr Sin ωt (θ₌ωt)
The variation of the velocity of N with displacement χ
= - ωr Sin θ
= ± ωr cos2θ ( sin2θ + cos2θ = 1)
= ± ωr
=±ω-
Hence the velocity of N is ± ωr ( a maximum) when χ =
±r
4. Displacement: χ = r Cos θ = r Cos ωt
 Applications of SHM
1. The helical spring

Figure 3
Consider the figure above, a helical spring with a mass m
attached to one end while the other end is fixed. When
the mass is pulled down through distance x from its
equilibrium position to a new position, the downward
force F required to stretch the spring is directly
proportional to the displacement x, i.e F= kx …..2

where k is the spring constant. Equation 2 is referred to as

being exerted on the mass by the spring. If the mass is
now released, the net force on the mas, or the restoring
force, is equal to kx. By Newton’s second law,
-kx= ma…… 3
where a is the resulting acceleration of the mass. The
negative sign before the kx term in eqn 3 indicates
that the restoring force and the displacement are in
opposite directions, so from eqn 3
a= - k/m x……..4 .
Comparing eqn 4 and 1, it can be concluded that the
motion of the mass is simple harmonic with ω2=.
Note that the acceleration a can also be written as
d2x/dt2 , whereby equation 4 can be integrated to
yield
x= A sin (ωt + ⱷo ) ……5
where x is the displacement of the mass at any time t and
ⱷo is a constant which is frequently referred to as the
initial phase angle of the simple harmonic motion. The
parameter ω is called the angular frequency of the motion.
The velocity V and acceleration of motion can be found by
differentiating eqn. 5 with respect to time to give
v= ωA cos (ωt + ⱷo ) …….6 and

a= -ω2A sin (ωt + ⱷo)….7

dividing eqn. 7 by 5 gives a/x = -ω2x which is in agreement
with eqn. 1.
The magnitude of the maximum velocity and acceleration
are Vmax= ωA and amax= --ω2A, from eqns. 6 and 7.
The period T of oscillation is equal to 2π/ω, or since
ω2= k/m for the helical spring,
T = 2π√m/k …………..8
The frequency of vibration, or the number of
vibrations per unit time , is
F= 1/T = ω/2π……………9
The unit of F is cycles per second or hertz. If the
mass is pulled beyond the equilibrium position and
then released, the mass executes a simple harmonic
motion. The period of the motion is
T = 2π√e/g ……10 where e is
the elongation produced by mass m.
2. The simple pendulum
This is another application of simple harmonic motion.
Suppose a bob of mass m suspended from a light
inextensible string of length L, is displaced through an
angle ϴ from the vertical. The restoring force on it is

F= -mg sin ϴ

If θ is small sin θ ₌₌θ=ϰ/L, and applying Newton’s second

law to the displaced bob gives –mg ϰ/L = ma, or a=
-g/lϰ…..11
Comparing equatns. 10 to 1 gives ω2 = g/L for the simple
pendulum and the period is

2π/ω, or T=1/2π√g/L ……12.

Correspondingly, the frequency of oscillation is

f=1/(2π )√g/L ……13 for a simple pendulum.

Eqn 11 suggests a simple method of experimentally
determining the acceleration due to gravity. A bob of
any convenient mass is suspended from a string of
known length L and the period of oscillation T, after
the bob is displaced from its equilibrium position, is
measured. Similarly, eqn. 11 gives g in terms of T and
L;

g=4π^(2 L/T)……..14
 EXAMPLES
1. A particle in SHM has a period of 0.4s. if the maximum speed attained during the motion
is 0.3m/s, what is the amplitude of the motion?
Solution
T = 0.4, Vmax= 0.3m/s, ω=2𝜋/T=2𝜋/0.4=15.71 rad/s
Vmax= ωA , A=Vmax/ω=0.3/15.71=0.019m
2. A boy attaches a 1.3kg toy to the free end of a clamped horizontal
spring lying on a frictionless surface. If the force constant of the
spring is 200N/m, what is the period of the resulting motion when
the spring is pulled and released?

Solution
M= 1.3kg
k= 200N/m
T=2π√m/k= 2 π(1.3/200)1/2=0.51s
 TUTORIAL QUESTIONS
TUTORIAL QUESTIONS
1. A simple pendulum has a length of 4m and executes vibrations in 80.8s. what is the
value of the acceleration due to gravity g where the pendulum is located?
2. A simple pendulum motion is represented in SI units as y= 0.08 sin (0.400 𝜋𝑡). What
are the amplitude and period of the motion?
3. A simple pendulum performs SHM in a vertical plane with an amplitude of 5cm. At
time zero, the bob has a displacement of 5cm from its equilibrium position. If it
passes through the equilibrium position 0.500s later, determine the period of the
motion.
4. Suppose you wish to construct a simple pendulum of period 1.0s. if g=9.8m/s2, what
is the length of the pendulum and what will be the new period if this length is
doubled?
 GRAVITATION
The scientists report from their studies that
the planets move in circular orbits about the
sun. All the planets lie nearly in one plane.
Also, that the axis of the earth is at an angle
231/2 to the normal to the plane of its orbit.
The axis points toward the pole star and the
earth rotates about it once in every 24 hours.
Another scientist, Kepler came up with an
analysis and concluded with these three laws
called the Kepler’s law;
1. The planets move in ellipses around the sun
as a focus. It is sometimes called the law of
orbits.
2. The line joining the sun and the planet sweeps
out equal areas in equal times. It is also called
law of areas.
3. The ratio R3/T2 is the same for each planet
where R is the average radius of the orbit and T
is the period of time per revolution. It is called
the law of period. These three laws give a
complete description of the motion of the
planets around the sun.
Newton’s Law of Universal Gravitation
Newton established a theory in 1687 known as Law of
universal gravitation. It states that every particle in the
Universe attracts every other particle with a force that is
directly proportional to the product of their masses and
inversely proportional to the square of the distance
between them. If the particles have masses m1 and m2 and
are separated by a distance r, the law of gravitation can be
stated as:
Fgα or Fg= Gm1m2/r2………..1, Fg is the gravitational force
Where G is a fundamental physical constant called the
gravitational constant. G can be expressed in N.m2kg-2 and
careful measurement shows that G= 6.67X10-11N.m2Kg-2.
Likewise, a body of mass m which lies at a
distance r from the center of the earth is
attracted towards the earth with a gravitational
force is:

Fg= GmME/r2……..2
ME is the mass of the earth. If gr is the
acceleration due to gravity at the distance r
(from the center of the earth), this gravitational
force is also equal to the weight of the body i.e
Fg= GmME/r2= mgr
or gr = GME/r2 …………….3
The acceleration due to gravity thus varies
inversely as the square of the distance from
the center of the earth. For a body which lies
near the surface of the earth, i.e. at a
distance from the center of the earth which
is approximately equal to the radius of the
earth, RE,
Fg = GmME/Re2 = mg
where g is the acceleration due to gravity on
the surface. Thus,
g = GME/RE2 ………………………4
It shows that the acceleration due to gravity
near the earth’s surface is dependent on the
mass and radius of the earth. The same is
true of the acceleration due to gravity near
surface of any other planet. Since the earth
is not spherical but spheroidal in shape, the
distance from the center of the earth to the
surface varies from the equator to the pole.
The numerical value of the acceleration due
to gravity therefore varies as one move from
the equator to the poles.
QUESTIONS
1. What is the speed of a satellite in a circular orbit
just above the surface of the earth if the radius of
the earth is 6.4x106 m and g = 9.8m/s2.
2. Suppose the moon is moving in a circular path of
radius 3.8x105 km about the earth. The period of
revolution is 27.5 days. Calculate the moon’s
acceleration.

3. Assume that the earth revolves around the sun in

a circular orbit of radius 1.49x 108 km and has a period
of 365. 25 days. Determine a. the orbital speed of the
earth b. the acceleration of the earth relative to the sun
c. the gravitational force between the sun and the earth
if the mass of the earth is 5.98x 1024 kg.