Simple Harmonic Motion

1. 1
CHAPTER 2
SIMPLE HARMONIC MOTION
2.1 Definitionof SimpleHarmonicMotion
 Simple harmonicmotion, occurswhenthe accelerationisproportionaltodisplacementbutthey
are inopposite directions.
 Simple HarmonicMotion(SHM).The motionthatoccurs whenan objectisacceleratedtowards
a mid-point.The size of the accelerationisdependentuponthe distance of the objectfromthe
mid-point.Verycommontype of motion,eg.seawaves,pendulums,spring.
 Simple harmonicmotionoccurswhenthe force Facting onan objectisdirectlyproportional to
the displacementx of the object,butinthe opposite direction.
 Mathematical statementF= - kx
 The force iscalleda restoringforce because italwaysactson the objectto returnit to its
equilibriumposition.
2.2 Descriptive terms
a. The amplitude A isthe maximumdisplacementfromthe equilibriumposition.
b. The periodT isthe time forone complete oscillation.Aftertime Tthe motionrepeatsitself.In
general x(t) =x (t + T).
c. The frequencyf isthe numberof oscillationspersecond.The frequencyequalsthe reciprocal of
the period. f = 1/T.
d. Althoughsimple harmonicmotionisnotmotioninacircle,it isconvenienttouse angular
frequency bydefining= 2pf = 2p/T
2.3 Simple HarmonicMotion
A bodyat simple harmonicmotionif :
a) The accelerationalwaysdirectedtowardafixedpointsontheirpath.
b) The accelerationisproportional toitsdisplacementfromafixedpointsandalwaysdirectedtoward
that points.

2. 2
Example forsimple harmonicmotion
a) When a mass hanging from the spring is deflected it will move with simple harmonic motion.
b) Motion of the piston in a cylinder is close simple harmonic motion.
c) Weight of the pendulum moves with simple harmonic motion if the angle is small.

3. 3
2.4 Simple HarmonicMotionDiagram
Q
θ
V
Vt
Vo
θ
θ
x
a
P
o
o
P
V
a
X
ω
Q
iii) space diagramii) vectordiagrami
Figure 2.1

4. 4
Figure 2.1 (i) shows a point P rotating with constant velocity , in a circular
path of radius , a
Linear velocity , V = a
From figure 2.1 (ii) , the horizontal component of velocity V , the velocity of point Q
Vq = V sin  = a sin 
From space diagram,
PQ = (a2 – x2)
 sin  = (a2 - x2)
a
 Vq = (a) (a2 - x2)
a
= () (a2 - x2)
P
V

O
,
Q
P , Q
X
P
V

O, Q
i) Vq maksimum=a ii) Vq minimum=0
Figure : 2.2

5. 5
Whenx = 0 , Vq ismaximum,
 Vq maks = (a2
– 0) = a (figure 2.2 i)
Whenx = a , Vq is minimum
 Vq min= (a2
– a2
) = 0 (figure 2.2 ii)
If P rotate with constant angular speed  , it also has a central
acceleration f , goes to center of rotation O.
f = a 2
o
P
V
a
X
ω
Q
P
f = a2
f q
Q
0
(ii) Vectordiagram
P
Q0
a
x
(i)
Figure : 2.3
(iii) space diagram

6. 6
(fromfigure 2.3 ii : vectordiagram )
central acceleration of the horizontal component f , the acceleration of point Q
fq = a 2
kos
fromfigure 2.3 iii , kos  = x
a
 fq = (a 2
) x
a
= x 2
Whenx = 0 , fq isminimum
 fq min = 02
= 0
Whenx = a , fq is maximum
 fq maks = a 2
2.5 PeriodicTime ,FrequencyandAmplitud
Periodic time or period , T is the time taken by the point Q to make a
swing back and forth to complete . And that time isequal to the time taken
by the OP toturn a rotation 2  radians with angularspeed  rad/s .
 Periodictime ,T= 2

But , fq = x2
dan  =  fq
X
 T = 2  x = 2  distance
fq acceleration

7. 7
Frequency, n is the number of a complete cycle of oscillation in the
penetration by the point Q in the second. The unit is the Hertz ( Hz), ie
one cycle per second .
Frequency , n =  Hz
2
n = 1/T
=1______________ Hz
2 distance/acceleration
Amplitude , a is the maximum displacement of point Q from a fixed
point O. The distance , 2a traveled by the point Q is known as a
stroke or swing .
EXAMPLE 1
A point moves with simple harmonic motion with the acceleration of 9
m/s2 and velocity of 0.92 m/s when it is 65 mm from the center
of travel. Find :
i. amplitude ,
ii. time of the periodic motion

8. 8
EXAMPLE 2
A particle moving with simple harmonic motion has a periodic time of
0.4 s and it was back and forth between two points is 1.22 m.
Determine :
i. The frequency and amplitude of the oscillation .
ii. Velocity and acceleration of the particle when it is 400 mm from the
center of oscillation .
iii.Velocity and maximum acceleration of the movement .
EXAMPLE 3
A mass of body 1.5 kg moving with simple harmonic motion is towards
to the end of the swing . At the time he was at A, 760 mm from the
center of oscillation , velocity and acceleration is 9 m/s and 10 m/s2
, respectively. Find :
a) frequency and amplitude of the oscillation ,
b) the maximum acceleration and the inertia of the body when it is at
the end of the swing ,
c) the time has elapsed for it to go and back to A.

9. 9
2.6 Elastic system - mass and spring.
mg
d
Stiffnessof
spring,S
Ked.tak
tegang
Ked.tegang
dan pegun
Ked.keseim
bangan,o
Ked.terpesong
Daya spring
x
Mf
f
mg

10. 10
The figure shows a body of mass M kg supported by a spring of
stiffness S N/m . Static spring deflection is d meters , then:
Mg = Sd
If the body is in the pull- down x meters from the equilibrium position O
( a fixed point) and then released , so
Body weight + inertia force = total spring force
Mg + Mf = Sd + Sx
Where, Mg = Sd
 Mf = Sx
f = (S/M) x
The reference S/M is constant for a system under consideration.
So , the acceleration f is proportional to the distance x from the
equilibrium position O . This indicates that the body is moving with simple
harmonic motion .
From SHM,
f = x2
=
√
𝑆
𝑀

11. 11
So, time periodic , T = 2

∴ 𝑇 = 2𝜋√
𝑀
𝑆
Mg = Sd dan 𝑀
𝑆
= 𝑑
𝑔
∴ 𝑇 = 2𝜋 √(
𝑑
𝑔
)
EXAMPLE 4
A body of mass 14 kg is suspended by springs up from the end
attached to a rigid support . The body produces a static deflection of
25 mm . It is in the pull down as far as 23 mm and then
released . Find :
i. Initial acceleration of the body ,
ii. the periodic time oscillations ,
iii. the maximum spring force ,
iv. the velocity and acceleration of the body when it is 12 mm
from the equilibrium position .

12. 12
EXAMPLE 5
Two body , each type per 6.5 kg , suspended on a vertical
spring with stiffness 2.45 kN/m. A body is removed and this causes
the system to oscillate. Find :
i. The maximum extension spring ,
ii. the periodic time and amplitude of the oscillation ,
iii. the velocity and acceleration of the mass when it is at the
center of the amplitude ,
iv. the total energy of the oscillations .
2.7 Simple pendulum
 , 
Mff

mg
mf
P


13. 13
The diagram shows a simple pendulum of length cord, L and weight B
with mass M. Amplitude of the oscillation  is small, not exceeding
120 , the angular displacement  and the three forces , the heavy weight ,
cord tension and inertia forces , is in equilibrium( figure ii) From the
force triangle .
sin 𝜃 =
𝑀𝑓
𝑀𝑔
= 𝑓
𝑔
sin θ ≅ θ , sebab θ adalah kecil
 𝜃 =
𝑓
𝑔
length of arc , x = L (s = r)
 = 𝑥
𝐿
∴
𝑓
𝑔
=
𝑥
𝐿
𝑓 = 𝑥 (
𝑔
𝐿
) for pendulum
f = x2 for SHM

14. 14
This shows that the oscillations of a simple pendulum is simple
harmonic.
2 = 𝑔
𝐿
and  =
√(
𝑔
𝐿
)
𝑇 =
2𝜋
𝜔
for SHM
∴ 𝑇 = √(
𝐿
𝑔
) for pendulum
Periodic time and frequency of a simple pendulum is independent of
the mass of the pendulum and the angle oscillation (if this angle
exceeds 120) .
If  is the angular speed line generating SHM .
angular speed of the pendulum
 =  (2
- 2
) and  maximum = 
angular acceleration of the pendulum
 = 2
 and maximum = 2

Periodic time , T = 2
𝑎𝑛𝑔𝑙𝑒 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑎𝑛𝑔𝑙𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
= 2 ((
𝜃
𝛼
)

15. 15
EXAMPLE 6
The amplitude of a simple pendulum is 7 0 and the periodic time is 5s ,
find:
i. maximum linear velocity of the pendulum weight and the
maximum angular speed of the pendulum cord ,
ii. maximum linear acceleration of the pendulum weight and
maximum angular acceleration of the pendulum cord .
EXAMPLE 7
A pendulum clock required rhythm second, with periodic time 2 s ,
was found late 80 s a day. The pendulum is shortened so that it
rhythm seconds exactly. Find the difference in the length of the
pendulum clock .
2.8 SIMPLE CONE

16. 16
figure shows a body B of mass M kg of rotate with  rad/s on the
vertical axis A-A and the angle  is assumed small. The forces acting
on the body is shown in the figure.
Tan  = Mr2 = r2
Mg g
But, tan  ≅ 𝜃 (Because 𝜃 is too small)
  = r2
g


A
P
A

17. 17
And from space diagram,
Sin  = 𝑟
𝐿
 = 𝑟
𝐿
∴ r2 = r and
𝜔 = √(
𝑔
𝐿
)
g L
This is the minimum value of  for a conical pendulum , and the value
is equal to the value  for a simple pendulum and the pendulum is
in equilibrium.
From the force triangle
Tan  = Mr2
= r2
Mg g
From the space diagram
Tan  = r / h
∴ r2
= r and  = (g/h)
g h

18. 18
Periodic time , T = 2𝜋
𝜔
for
∴ 𝑇 = 2𝜋√(
ℎ
𝑔
)
to get the tensile cord P, from triangle of forces,
sin  = 𝑀𝑟𝜔
𝑃
2
and from spacediagram,
sin  = 𝑟
𝐿
∴
𝑟
𝐿
=
𝑀𝑟𝜔
𝑃
2
P = ML2 = 𝑀𝑔𝐿
ℎ
EXAMPLE 8
Cord length a conical pendulum is 200 mm and weight is 2.4 kg. Find the rotation of the
pendulum at the moment leads to upward from its equilibrium position. If the cone
pendulum is rotating with a 88.8 rpm, determine:
i. Periodic time ,
ii. high-pendulum,
iii. the tension of the cord.
EXAMPLE 9

19. 19
Maximum permissible tension of the cord used for a conical pendulum
is 7 times heavier weight . If the cord length is 1.2 m , find:
i. angular speed in revolutions per minute ,
ii. high- pendulum,
iii. the change of high pendulum , when the angular speed dropped
to 20% .