Name’s of Groups :

1. Arventa2. Atika3. Aulia4. Imam 5. M. Ababil6. Sebma

Basic Competition :

• Analizing the connection of elasticity with simple harmonic motion.

Indicator :• Determining connection of elasticity

material’s character with simple harmonic motion.

Simple Harmonic MotionSimple 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

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

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.

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.

atau Tf

fT

11

t

nf

n

tT

The magnitude of restored force on a spring that makes the simple harmonic motioncan be expressed as follows :

F = -kyThe 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 yFrom the equations F = -ky dan F = -m ω2

can be obtain.

m

k

m

k

2

Because , then the equation above becomes :

Where :T = period of the spring vibration (s)m = the load mass (kg)k = spring force contant (N/m)

T

2

k

mT

m

k

T2

2

m

kf

21

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)

Mathematically, the equation of deviation for y-t sinusoidal graph as in the figure above can be expressed by the following equation:

Because , then

Where :y = deviation (m) t = time (s)A = amplitude (m)ω = angular velocity(rad/s)T = period (s)f = frequency (Hz)

tAy sinf

T 2

2

fAy

tT

Ay

2sin

2sin

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 sinRy

Because and , then

Because R (radius of circle) equal to A (amplitude), then

If at time t = 0 the object has a phase angle

of θ0, then the equation for simple harmonic motion deviation is as follows :

T

2t

tT

Ry2

sin

tT

Ay2

sin

0

0

2sin

sin

sin

tT

Ay

tAy

Ay

The Velocity of Simple Harmonic Motion

Because , so that the maximum value from , then

0

0

cos

sin

tAV

tAdt

dV

dt

dyV

y

y

y

1cos 0 t

0coscos

tVVV

AV

mmy

m

AVy

We can also determine the velocity of simple harmonic motion by using the method of uniform circular motion.

The velocity of simple harmonic motion is the projection of linear velocity of an object to y axis.

RV

From the figure, we can obtain the following equation.

The equation above can be arranged as follows.

tAV

ARtRV

tVV

VV

y

y

y

y

cos

cos

cos

cos

22

222

22

22

sin

sin1

cos

yAV

tAAV

tAV

tAV

y

y

y

y

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)?

Solution Because :m = 4 kgk = 100 N/mQuestion :Frequency and periodof springAnswer :

HzsT

f

ssmN

kg

k

mT

398,0512,2

11

512,25

1.14,3.2

/100

422

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)

8 16

4

0

-4

1210642 14 t (s)

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!