This is our presentation (by Aulia Rodlia and friends)

1. Name’s of Groups :
1. Arventa
2. Atika
3. Aulia
4. Imam
5. M. Ababil
6. Sebma

2. Basic Competition :
• Analizing the connection of elasticity
with simple harmonic motion.

3. Indicator :
• Determining connection of elasticity
material’s character with simple
harmonic motion.

4. Simple Harmonic Motion
Simple harmnic motion is defined as a
motion that is always influenced by the
force whose its magnitude is directly
proportional to the distance of a point and
the direction always goes to that point.
In a simple harmonic motion, the
magnitutude of the restored force on the
spring is proportional to the object distance
from its state of equilibrium.
Mathematically, it can be written as follows.
F = kx

5. Besides the spring, the restored force also
works on the harmonic motion of a simple
pendulum.
The restored force acting upon the harmonic
motion of a simple pendulum is a component of
weight that is perpendicular to the rope. Thus, the
magnitude of the restored force on the simple
pendulum can be determined as follows:
F = mg sin ɵ

6. Period and Frequency on Spring
Period (T) is the time required to do one
complete vibration.
Based in the figure above, period is the time
required by the load to move upward from A to O
to B, then down from B to O and back to A. The
motion to the travel the path of A-O-B-O-A is
called one vibration.

7. Frequency (f) is the number of oscillations
made by the load in one second. Based on the
figure, frequency is the number of motions to
travel the part A-O-B-O-A made by the load in
one second.
1 1 t n
T= ⇔ f = atau T = n ⇔ f = t
f T

8. The magnitude of restored force on a spring
that makes the simple harmonic motioncan be
expressed as follows :
F = -ky
The restored force acted upon an object can
also be calculated from the acceleration of a body
based on Newton’s second law, that is:
F = m ay
F = -m ω2 y
From the equations F = -ky dan F = -m ω2 can
be obtain. ω =
2 k
m
k
ω=
m

9. Because ω = 2π , then the equation above
T
becomes :
2π k m
= ⇒ T = 2π
T m k
Where :
T = period of the spring vibration (s)
m = the load mass (kg)
k = spring force contant (N/m)
1 k
f =
2π m

10. The Deviation of Simple
Harmonic Motion
If we study the deviation against time graph
(y-t graph) from simple harmonic motion, we will
know that the equation of simple harmonic motion
is a sinusiodal function (with constant frequency
and amplitude)

11. Mathematically, the equation of deviation for y-t
sinusoidal graph as in the figure above can be
expressed by the following equation:
y = A sin ωt
2π
ω= = 2πf
Because T , then
2π
y = A sin t
T
y = A sin 2πf
Where :
y = deviation (m) t = time (s)
A = amplitude (m)
ω = angular velocity(rad/s)
T = period (s)

12. We can also determine the equation of simple
harmonic motion deviation by using the uniform
circular motion method. A harmonic motion (harmonic
oscillation) can be discribed as a point that moves in
circular motion with radius R.
Then
y = R sin θ

13. 2π
Because θ = ω t and ω = , then
T
2π
y = R sin t
T
Because R (radius of circle) equal to A
(amplitude), then 2π
y = A sin t
T
If at time t = 0 the object has a phase angle of
θ0, then the equation for simple harmonic motion
deviation is as follows :
y = A sin θ
y = A sin ( ωt +θ0 )
 2π 
y = A sin  t +θ0 
T 

14. The Velocity of Simple Harmonic
Motion
dy
Vy =
dt
d
V y = [ A sin (ωt +θ0 ) ]
dt
V y = ωA cos(ωt +θ0 )
Because cos( ωt + θ 0 ) = 1 , so that the maximum
value from V y = ωA , then
Vm = ωA
V y = Vm cos θ = Vm cos( ωt + θ 0 )

15. We can also determine the velocity of simple
harmonic motion by using the method of uniform
circular motion.
V = ωR
The velocity of simple harmonic motion is the
projection of linear velocity of an object to y axis.

16. From the figure, we can obtain the following
equation. V y = V cos θ
V y = V cos ωt
V y = ωR cos ωt ⇔ R = A
V y = ωA cos ωt
The equation above can be arranged as follows.
V y = ω A2 cos 2 ωt
(
V y = ω A 1 − sin ωt
2 2
)
V y = ω A2 − A2 sin 2 ωt
V y = ω A2 − y 2

17. Sample problem :
An object of 4 kg in mass in hung on a
spring having a constant of 100 N/m.
What is the period and frequency of the
spring if the object is given a small
devaition (pulled then released)?

18. Solution
Because :
m = 4 kg
k = 100 N/m
Question :
Frequency and periodof spring
Answer :
m 4kg 1
T = 2π = 2π = 2.3,14. s = 2,512 s
k 100 N / m 5
1 1
f = = = 0,398 Hz
T 2,512 s

19. Exercise :
1. A spring is hung with a load of 1,8 kg, so that
the spring length increment is 2 cm. What is
the period and frequency of the spring
oscillation?
2. From the figure below, calculate the
amplitude, period, and frequency of harmonic
motion!
y(cm)
4
0
2 4 6 8 10 12 14 16 t (s)
-4

20. 3. What is the spring contant, if it is given force
of 400 N, and increses of 4 cm in length?
4. A harmonic motion at time t = 0 has a zero
deviation. If the maximum deviation is 10 cm
and period of oscillation is 0,5 s, determine the
equation of velocity!
5. Calculate the amplitude and period of a particle
that moves in harmonic motion if it has velocity
of 8 m/s at a distance of 2 m from the center
and 4 m/s at a distance of 3 m from the
center!