The Simple Harmonic Motion

The simple harmonic motion is alternating motion of a particle or motion of an object which is always towards its equilibrium point (i.e. restoring force), where its movement opposite with working resultant force. In our daily life, there are many examples of the simple harmonic motion such as oscillating a spring is attached a load vertically, oscillating a pendulum, a molecule inside a solid etc.

A. Period and Frequency

A mass on a spring will trace out a sinusoidal pattern as a function of time, movement mass from A to E (A-B-C-D-E) is called one complete vibration, while the time required to do one complete vibration is called Period symbolized by T. While frequency symbolized by f is the sum of vibration in one second.

To make more understand of the simple harmonic motion concept, study the figure beside. The first condition, the pendulum or the spring-mass system are in equilibrium. To make the simple harmonic motion, we must give a distortion force to both system. The pendulum for example, we give a distortion force by pull-right or left then it release. And so do for spring-mass system, we must pull-down slowly then it release so, the spring will exert a restoring force, which is a force that tends to restore it to the equilibrium position. In the case of the spring-mass system, this force is the elastic force, which is given by Hooke's Law, F = − kx, where F is the restoring force, x is the displacement, and k is the spring constant.. Since the restoring force is proportional to the displacement, the pendulum or spring-mass system are a simple harmonic oscillator.

Gerak Harmonik Sederhana

Gerak harmonik sederhana adalah gerak bolak balik sebuah partikel atau gerak sebuah benda yang selalu menuju titik keseimbangannya, dimana gerakannya berlawanan dengan resultan gaya yang bekerja. Dalam kehidupan sehari-hari, terdapat banyak contoh gerak harmonik sederhana seperti gerak bolak-balik pegas yang disertai beban secara vertikal, gerak bolak-balik sebuah bandul, gerak molekul dalam zat padat dll.

A. Perioda dan Frekuensi

Sebuah massa pada sebuah pegas akan meninggalkan jejak sebuah pola sinusoidal sebagai fungsi waktu, gerak massa dari A menuju E (A-B-C-D-E) disebut satu getaran penuh, sementara waktu yang dibutuhkan untuk melakukan satu getaran penuh disebut perioda disimbolkan dengan T. Sementara frekuensi diberi symbol f adalah jumlah getaran dalam satu detik.

Untuk lebih memahami konsep gerak harmonic sederhana, pelajari gambar disamping. Pada kondisi pertama system bandul atau pegas-beban dalam keadaan seimbang. Untuk membuat gerak harmonic sederhana kita harus memberi gaya usikan atau gangguan pada kedua system. Bandul misalnya kita beri gaya usik dengan menarik ke kanan atau ke kiri kemudian lepaskan. Juga untuk pegas-beban, kita harus menarik kebawah secara perlahan kemudian lepaskan sehingga pegas akan menggunakan sebuah gaya kembali (restoring force) yaitu sebuah gaya yang cenderung untuk mengembalikan pegas ke posisi seimbang. Dalam kasus system pegas-beban, gaya ini adalah gaya elastic yang diberikan oleh hukum Hooke, F = − kx, dimana F adalah gaya kembali (restoring force), x adalah perpindahan, dan k adalah konstanta pegas. Karena gaya kembali berbanding lurus dengan pertambahan panjang, system bandul atau pegas-beban adalah sebuah contoh gerak harmonic sederhana.

Mathematically the period and frequency can be state as follows:

T=1

f or f = 1

T

Where T = Period (s) F = frequency (Hz)

The spring which has constant spring (k) is vibrating due to loaded mass (m), the its period can be calculated with the formula:

Where

k = constant spring (N/m)m = mass of load (kg)

While oscillating the simple pendulum which has length of string (l), the period of its oscillate can be calculated with the formula:

Wherel = length of string (m)g = earth’s gravitational (m/s2)

Example:

1. A spring with constant force 1250 N/m is hanged vertically, at the end of string is given by load 1 kg, when it is pulled-down then released, what is the period and its frequency

Solution:The period system is:

The frequency system is:

2. A simple pendulum is swinging vertically with frequency 1 Hz, if earth’s gravitational 10 m/s2, what is length of string which is used

Solution:

Secara matematik perioda dan frekuensi dapat dinyatakan sebagai berikut:

T=2 π √ mk

T=2 π √ lg

T=2 π √ 11250T=2 π √ m

k⇔⇔ T=0 . 177 s

f =5 .65. HZ⇔⇔ f = 10 .177

f = 1T

T=1 s⇔T=11T=1

f

T=1

f or f = 1

T

dimana T = Perioda (s) F = frequency (Hz)

Pegas yang memiliki konstanta (k) bergerak harmonic yang disebabkan beban massa (m), periodanya dapat dihitung dengan rumus:

Dimana

k = constanta pegas (N/m)m = massa beban (kg)

Sementara gerak harmonic bandul sederhana yang memiliki panjang tali (l), perioda getarannya dapat dihitung dengan rumus:

Wherel = length tali (m)g = gravitasi bumi (m/s2)

Contoh:

1. Sebuah pegas dengan constanta 1250 N/m yang digantung vertical, pada ujung tali diberi beban 1kg, ketika pegas ini ditarik ke bawah kemudian dilepas, berapa perioda dan frekuensinya!

Solusi:Perioda sistem:

Frekuensi sistem:

2. Sebuah bandul sederhana berayun vertical dengan frekuensi 1Hz, jika gravitasi bumi 10 m/s2, berapa panjang tali yang digunakan

Solusi:

l = 0,2485 m l = 24,85 cm

T=2 π √ mk

T=2 π √ lg

T=2 π √ 11250T=2 π √ m

k⇔⇔ T=0 . 177 s

f =5 .65. HZ⇔⇔ f = 10 .177

f = 1T

T=1 s⇔T=11T=1

f

⇔ l=10×12

4 π2l= g .T 2

4 π 2

T 2=4 π2 lg

⇔T=2 π √ lg

⇔

The Simple Harmonic Motion Equations

A. Displacement of Simple Harmonic MotionAs we already know, a mass on a spring will trace out a sinusoidal pattern as a function of time. The motion equation for simple harmonic motion contains a complete description of the motion, and other parameters of the motion can be calculated from it

Base on figure above, the general of sinusoidal equation can be written as follows:

orWhere:

y = displacement (m)A = Amplitude (m)

ω = Angular frequency (rad/s) T = Period (s) t = time vibration (s)

Example:A particle is moving with the simple harmonic motion, if amplitude 4 cm and period 4 s, determine of displacement after 1/3 second latter

Solution:

↔ ↔ y = 4 sin ( 360o

12)

y = 4 sin 30o = 4 x 0.5 = 2 cm

l = 0,2485 m l = 24,85 cm

ω= 2πT

f = ω2π

Τ=1f

y=A sin ( 2 πT

. t )y=A sin (ω . t )

y=A sin ( 2 πT

. t ) y=4sin( 3600

4×1

3)

⇔ l=10×12

4 π2l= g .T 2

4 π 2

T 2=4 π2 lg

⇔T=2 π √ lg

⇔

Persamaan Gerak Harmonik Sederhana

A. Perpindahan Gerak Harmonic SederhanaSeperti telah diketahui, sebuah massa pada pegas akan meninggalkan jejak sebuah pola sinusoidal sebagai fungsi waktu. Persamaan gerak harmonic sederhana penuh berisi deskripsi gerak dan parameter lain tentang gerak dapat kita tentukan.

Berdasarkan gambar di atas, secara umum persamaannya dapat dituliskan sebagai berikut:

orWhere:

y = simpangan (m)A = Amplitudo (m)

ω = frekuensi sudut (rad/s) T = Perioda (s) t = waktu getar (s)

Contoh:Sebuah partikel bergerak harmonik sederhana, jika amplitude 4cm dan perioda 4s, tentukan simpangan gelombang setelah 1/3 detik kemudian!

Solusi:

↔ ↔ y = 4 sin ( 360o

12)

y = 4 sin 30o = 4 x 0.5 = 2 cm

Exercises:

1. A spring with constant force 2500 N/m is hanged vertically, at the end of string is given by load 5 kg, when it is pulled-down then released, what is the period and its frequency

2. A simple pendulum is swinging vertically with frequency 5 Hz, if earth’s gravitational 10 m/s2, what is length of string which is used

Τ=1f

f = ω2 π

ω= 2πT

y=A sin ( 2 πT

. t )y=A sin (ω . t )

y=A sin ( 2 πT

. t ) y=4sin( 3600

4×1

3)

3. Particle is moving with the simple harmonic motion, if amplitude 10 cm and period 8 s, determine of displacement after 0,5 second latter

4. An object has mass m is hung at the end of spring and vibrated with period 3s. Subsequently, at that object is increased by mass 1 kg and the period to be 5s. How much of mass m now?

5. How much length of simple pendulum that is vibrating 50 times in interval time 1 minute (use g = 10 m/s2)

B. Velocity of The Simple Harmonic MotionLinear velocity projection in one of the circle diameter produces velocity harmonic motion. Mathematically velocity of the simple harmonic motion is the first derivation from the displacement harmonic motion:

or

Where: vy = velocity of vibration (m/s)

A = Amplitude (s)ω = Angular frequency (rad/s)T = Period (s)t = time vibration (s)

Example:Equation of the simple harmonic motion expressed by y = 10 sin (0.5 πt). Determine: amplitude, period, frequency, displacement and velocity at time t = ½ second.Solution:

y = 10 sin (0.5 πt) can be changed by

y=10⋅sin⋅( 2 π4

⋅t )

Latihan:

1. Sebuah pegas dengan konstanta gaya 2500 N/m digantung vertika, pada ujung pegas diberi beban 5kg, pada saat pegas ditarik ke bawah kemudian dilepas, berapa perioda and frekuensinya

2. Sebuah bandul sederhana mengayun vertical dengan frekuensi 5Hz, Jika percepatan gravitasi bumi 10 m/s2, berapa panjang tali yang digunakan?

v y=dydt

v y=d ( A⋅sin(ω⋅t )

dt

v y=A⋅ω⋅cos(ω⋅t )

v y=2 πT

A⋅cos( 2πT

⋅t )

3. Partikel bergerak harmonic sederhana, jika amplitude 10 cm dan periode 8s, tentukan simpangan setelah 0,5 detik kemudian!

4. Sebuah benda bermassa m tergantung pada ujung pegas dan bergetar dengan periode 3s, selanjutnya, benda tersbut ditambah massa 1kg and periodanya menjadi 5s. Berapa massa m sekarang?

5. Berapa pnjang bandul sederhana yang berayun 50 kali dalam selang waktu 1 menit (gunakan g=10 m/s2)

B. Kecepatan Gerak Harmonik SederhanaProyeksi kecepatan linier dalam satu diameter lingkaran menghasilkan gerak kecepatan harmonic. Secara matematik kecepatan gerak harmonic sederhana adalah turunan pertama dari perpindahan gerak harmonik:

or

Dimana: vy = velocity getar (m/s)

A = Amplitude (s)ω = frequency sudut (rad/s)T = Period (s)t = waktu getar (s)

Contoh:Sebuah persamaan gerak harmonic sederhana dinyatakan dengan y = 10 sin (0.5 πt). Tentukan amplitude, perioda, frekuensi, perpindahan, dan kecepatan pada saat t = ½ second.Solusi:

y = 10 sin (0.5 πt) dapat dirubah menjadi

y=10⋅sin⋅( 2 π4

⋅t )

So

Amplitude is A = 10 cm

Period is T = 4 seconds

Frequency is f = 1/T = ¼ = 0,25 Hz

v y=dydt

v y=d ( A⋅sin(ω⋅t )

dt

v y=A⋅ω⋅cos (ω⋅t )

v y=2 πT

A⋅cos( 2πT

⋅t )

Displacement at t = ½ s is y = 10 sin (

2 π4

×12 )

y = 10 sin (3600

4×1

2 )

y = 10 sin 45o = 10

12 √2

y = 5√2 cm

Velocity at t = ½ s is v =

2 πT

⋅A⋅cos45o

v = 2π f A cos 45o

v = 2π 0.25 10

12 √2

v = 2,5 π √2 m/s

C. Acceleration of The Simple Harmonic Motion Linear acceleration projection in one of the circle diameter produces acceleration harmonic motion. Mathematically acceleration of the simple harmonic motion is the first derivation from the velocity harmonic motion:

a y=dv y

dt

a y=d( A⋅ω⋅cos( ω⋅t )

dt

a y=−A⋅ω2⋅sin(ω⋅t ) or

a y=−4 π2

T 2 ⋅A⋅sin( 2 πT

⋅t )

Where: ay = acceleration of vibration (m/s)

A = Amplitude (s)ω = Angular frequency (rad/s)T = Period (s)t = time vibration (s)

Jadi

Amplitude A = 10 cm

Period T = 4 seconds

Frequency f = 1/T = ¼ = 0,25 Hz

Perpindahan pada saat t = ½ s is y = 10 sin (

2 π4

×12 )

y = 10 sin (3600

4×1

2 )

y = 10 sin 45o = 10

12 √2

y = 5√2 cm

Kecepatan pada t = ½ s v =

2 πT

⋅A⋅cos45o

v = 2π f A cos 45o

v = 2π 0.25 10

12 √2

v = 2,5 π √2 m/s

C. Percepatan Gerak Harmonik Sederhana Proyeksi percepatan linier dalam satu diameter lingkaran menghasilkan percepatan gerak harmonic. Secara matematis percepatan gerak harmonic sederhana

a y=dv y

dt

a y=d( A⋅ω⋅cos( ω⋅t )

dt

a y=−A⋅ω2⋅sin(ω⋅t ) or

a y=−4 π2

T 2 ⋅A⋅sin( 2 πT

⋅t )

Dimana: ay = percepatan getaran (m/s2)

A = Amplitude (s)ω = frequency sudut (rad/s)T = Period (s)t = waktu getar (s)

note:mark “-“ shows that direction vector of acceleration (a) opposite with the displacement (y)

Example:1. A particle vibrated harmonic with amplitude 12 cm and period 0,5 seconds, determine:

a. velocity of particle when displacement is 6 cmb. maximum velocityc. acceleration of particle when displacement 10 cmd. maximum acceleratione. time required by particle to moved from equilibrium position to the point at 6 cm from

equilibrium position

Solution:a. velocity when y = 6 cm.

consider = α

6 = 12 sin α

sin α = 6/12 = 0.5

α = 30o

vy = 24 π √3m/s

b. maximum velocity value of maximum cosines is 1 so

vmax = 48π m/s

Note:tanda “-“menunjukkan bahwa arah vector percepatan berlawanan arah dengan perpindahan

y=A sin ( 2 πT

. t )

( 2πT

. t )y=12⋅sin( 2 πT

. t )

v y=2 πT

A⋅cos( 2πT

⋅t )

v y=2 πT

A⋅cosα

v y=2 π0 . 5

⋅12⋅cos⋅300

v y=2 π0 . 5

⋅12⋅12 √3

v y=2 πT

A⋅cos( 2πT

⋅t )

vmax=2π0. 5

⋅12⋅(1 )

Contoh:1. Sebuh partikel bergerak harmonic dengan amplitude 12 cm dan periode 0,5s, tentukan:

a. kecepatan partikel ketika perpindahan 6m b. kecepatan maksimumc. percepatan partikel ketika perpindahan 10cmd. percepatan maksimume. waktu yang dibutuhkan oleh partikel untuk berpindah dari posisi ke titik 6cm dari

posisi seimbang.Solusi:a. kecepatan ketika y = 6 cm.

anggap = α

6 = 12 sin α

sin α = 6/12 = 0.5

α = 30o

vy = 24 π √3m/s

b. keceptan maksimum nilai maksimum cosines adalah 1

vmax = 48π m/s

c. acceleration when y=10 cm

y=A sin ( 2 πT

. t )

( 2πT

. t )y=12⋅sin( 2 πT

. t )

v y=2 πT

A⋅cos( 2πT

⋅t )

v y=2 πT

A⋅cosα

v y=2 π0 . 5

⋅12⋅cos⋅300

v y=2 π0 . 5

⋅12⋅12 √3

v y=2 πT

A⋅cos( 2πT

⋅t )

vmax=2π0. 5

⋅12⋅(1 )

a y=4 π2

T 2 ⋅A⋅sin( 2 πT

⋅t )

ay = 160π2 m/s2

d. maximum acceleration can be reach if y = A

ay = 192π2 cm/s2

e. time required y = A sin θ θ = 30o

2. A spring is hung vertically, at its the end loaded a mass 5 kg. When harmonic motion is happen, its frequency 8 Hz. Determine the constant force of spring and acceleration when displacement 10 cm.

Solution:The constant force of spring

T=2π √ mk

T 2=4 π2 mk

k= 4 π2

T 2 m

k=4 π2 f 2mk=4 π2 82⋅5=1280⋅π2

N/m

c. percepatan ketika y=10 cm

a y=4 π2

T 2 ⋅y

a y=4 π2

0 . 52⋅10

a y=4 π2

T 2 ⋅y

a y=4 π2

0 .52⋅12

sin θ= yA

= 612

=0 . 5

θ=2 πT

⋅t

300=3600

0 .5⋅t

t=153600

=0 ,042⋅s

a y=4 π2

T 2 ⋅A⋅sin( 2 πT

⋅t )

ay = 160π2 m/s2

d.percepatan maksimum yang dapat dicapai jika y = A

ay = 192π2 cm/s2

e. waktu yang dibutuhkan y = A sin θ θ = 30o

2. Sebuah pegas tergantung vertical, di ujungnya dibebani massa 5kg. Ketika terjadi gerak harmonic, frekuensinya 8 Hz. Tentukan gaya konstanta pegas dan percepatan ketika perpindahan 10cm

Solusi:Konstanta pegas

T=2π √ mk

T 2=4 π2 mk

k= 4 π2

T 2 m

k=4 π2 f 2mk=4 π2 82⋅5=1280⋅π2

N/m

The acceleration

a y=4 π2

T 2 ⋅y

a y=4 π2

0 . 52⋅10

a y=4 π2

T 2 ⋅y

a y=4 π2

0 .52⋅12

sin θ= yA

= 612

=0 . 5

θ=2 πT

⋅t

300=3600

0 .5⋅t

t=153600

=0 ,042⋅s

a y=4 π2

T 2 ⋅y

ay = 4π2 f 2 y = 4π2 82 0.1 = 25.6 π2 m/s2

Exercises:

1. Equation of the simple harmonic motion expressed by y = 30 sin (0.25 πt). Determine: amplitude, period, frequency, displacement and velocity at time t = 0,6 second.

2. A particle vibrated harmonic with amplitude 50 cm and period 2 seconds, determine:a. velocity of particle when displacement is 10 cmb. maximum velocityc. acceleration of particle when displacement 30 cmd. maximum acceleratione. time required by particle to moved from equilibrium position to the point at 8 cm from equilibrium position

3. A spring is hung vertically, at its the end loaded a mass 10 kg. When harmonic motion is happen, its frequency 4 Hz. Determine the constant force of spring and acceleration when displacement 5 cm.

4. The load is hung at the end of spring vertically, when it occur simple harmonic motion its amplitude 10 cm and frequency 4 Hz. Calculate:a. Constant of springb. The maximum velocity c. The maximum accelerationd. Velocity when its displacement y = 5 cme. Acceleration when its displacement y = 6 cm

The Relationship Between Velocity And DisplacementAs we already know that equation of velocity simple harmonic motion is:

v y=A .ω . cos (ω .t )

The equation above can be changed as follows:

v y=A .ω √cos2 .(ω . t )

v y=A . ω √1−sin2(ω . t )

v y=ω√A2−A2 . sin2(ω . t )

v y=ω√ A2− y2

Where: vy = simple harmonic velocity (m/s)

ω = angular velocity (rad/s)A = Amplitude (m)y = displacement (m)

percepatana y=

4 π2

T 2 ⋅y

ay = 4π2 f 2 y = 4π2 82 0.1 = 25.6 π2 m/s2

Latihan:

1. Persamaan gerak harmonic sederhana dinyatakan dengan y = 30 sin (0.25 πt). Tentukanlah perioda, frekuensi, simpangan dan kecepatan pada saat t = 0,6s

2. Sebuah partikel bergerak harmonic dengan amplitude 50 cm dan perioda 2s, Tentukanlah:a. Kecepatan partikel pada saat simpangannya 10 cmb. Kecepatan maksimumc. Percepatan partikel pada saat simpangan 30 cmd. Percepatan maksimume. Waktu yang dibutuhkan partikel untuk bergerak dari keadaan seimbang ke posisi

8cm ari posisi seimbang3. Sebuah pegas tergantung vertical pada ujungnya dibebani massa 10 kg, saat gerak

harmonic terjadi, frekuensinya 4Hz. Tentukanlah konstanta gaya pegas dan percepatan saat simpangan 5 cm

4. Sebuah beban tergantung pada ujung pegas vertical, saat terjadi gerak harmonic amplitudonya 10 cm dan frekuensi 4 Hz , hitunglah:a. Konstanta pegasb. Kecepatan maksimumc. Perceoatan maksimumd. Kecepatan saat simpangannya 5 cme. Percepatan saat simpangannya 6 cm

Hubungan antara kecepatan dan perpindahanSeperti telah kita ketahui bahwa persamaan kecepatan gerak harmonic sederhana:

v y=A .ω . cos (ω .t )

Persamaan di atas dapat di rubah dengan:

v y=A .ω √cos2 .(ω . t )

v y=A . ω √1−sin2(ω . t )

v y=ω√A2−A2 . sin2(ω . t )

v y=ω√ A2− y2

dimana: vy = kecepatan gerak harmonik (m/s)

ω = kecepatan sudut (rad/s)A = Amplitudo (m)y = perpindahan (m)

Example:

A particle vibrated harmonically with period 0,2 seconds, when it is displacing 6 cm from equilibrium point, its vibrate velocity 4/5 time its maximum velocity. Determine the amplitude vibration and maximum accelerationSolution:

Amplitude vibration

v y=ω√ A2− y2

45

v max=2 πT √ A2−36

45 ( 2 π

TA )=2 π

T √ A2−36

45

( A )=√ A2−36

1625

( A )2=A2−36

925

A2=36

A2=259

×36

A = 10 cm

a. Maximum accelerationamax=ω2 A

amax=4 π2

T2 . A

= 4 π2

0 , 04.10

=π2⋅1000 = 12759 cm/s2

= 127,59 m/s2

Phase angle, phase and phase difference

a. Angle of Phase (θ )Phase angle is angle which has been travelled by the vibrant of an object, hence applying equation as follows:

θ=ω⋅t+θo

θ=2 π⋅t

T+θo

for θ0=0 , angle of phase is θ=ω⋅t=2 π⋅t

T

Contoh:

Sebuah partikel bergerak harmonic dengan perioda 0,2s, ketika berpindah 6cm dari titik seimbang, kecepatan getarannya 4/5 kali kecepatan maksimumnya. Tentukan amplitude getaran dan percepatan maksimum

Solusi:b. Amplitudo getaran

v y=ω√ A2− y2

45

v max=2 πT √ A2−36

45 ( 2 π

TA )=2 π

T √ A2−36

45

( A )=√ A2−36

1625

( A )2=A2−36

925

A2=36

A2=259

×36

A = 10 cmb. Percepatan maksimum

amax=ω2 A

amax=4 π2

T2 . A

= 4 π2

0 , 04.10

=π2⋅1000 = 12759 cm/s2

= 127,59 m/s2

Phase angle, phase and phase phase difference

a. Fase sudut (θ )Fase sudut adalah sudut yang telah ditempuh oleh benda getar, maka berlaku persamaan:

θ=ω⋅t+θo

θ=2 π⋅t

T+θo

untuk θ0=0 , sudut fasenya θ=ω⋅t=2 π⋅t

T

b. Phase (ϕ )

Angle of phase divided by one time angle rotation is called phase of vibration. It can formulated:

ϕ= θ2 π

ϕ=2 π t

T+θ0

2 π

ϕ= tT

+θ0

2 π

for θ0=0 , Phase vibration isϕ= t

T

c. Difference of Phase (Δϕ )

Δϕ=t2

T−

t 1

T=

t2−t1

T

Where: θ = angle of phase ω = angular velocity (rad/s) T = period (s) t = time vibration (s)

Δϕ = difference of phase

Example:

1. An object vibrated with frequency 0.5 Hz. Determine:a. Phase after 2 seconds, and 4 secondsb. Angle of phase after 2 seconds, and 4 secondsc. Difference of phase between 2 seconds until 4 seconds

Solution:

a. PhaseT= 1

T= 1

0 .5=2 s

t = 2sϕ=

t1

T=2

2=1

and t = 2 ϕ=

t1

T=4

2=2

b. Angle of phase

θ=2 πT

⋅t=2 π tT

for t = 2s θ=2 π t

T=2 π 2

2=2π

rad

for t = 4s θ=2 π t

T=2 π 4

2=4 π

red

b. Fase (ϕ )Sudut fase dibagi dengan satu kali sudut rotasi disebut fase getar yang dapat dirumuskan:

ϕ= θ2 π

ϕ=2 π t

T+θ0

2 π

ϕ= tT

+θ0

2 π

untuk θ0=0 , fase getarnyaϕ= t

T

c. Beda Fase (Δϕ )

Δϕ=t2

T−

t 1

T=

t2−t1

T

dimana: θ = sudut faseω = kecepatan sudut (rad/s)T = period (s)t = waktu getar (s)

Δϕ = beda fase

Contoh:

1. Sebuah benda bergerak harmonic dengan frekuensi 0.5 Hz. Tentukan:a. Fase setelah 2s dan 4s b. Fase sudut setelah 2s dan 4sc. Beda fase antara 2s sampai 4s

Solusi:

a. FaseT= 1

T= 1

0 .5=2 s

t = 2sϕ=

t1

T=2

2=1

dan t = 2 ϕ=

t1

T=4

2=2

b. Fase sudut

θ=2 πT

⋅t=2 π tT

Untuk t = 2s θ=2 π t

T=2 π 2

2=2 π

rad

Untuk t = 4s θ=2 π t

T=2 π 4

2=4 π

rad

c. Difference of angle

Δϕ=t2−t1

T=4−2

2=1

2. A particle vibrated harmonically with amplitude 20 cm at frequency 1 Hz. At the time of phase is 1/2, determine particle’s velocity

Solution:

or

v y=2 π .1 . 20. cos (2 π . 1

2)=40 π . cos(360 . 1

2)

v y=40 π . cos (180)=−40 π cm/s

Exercises:1. A particle vibrated harmonically with period 2 seconds, when it is displacing 10 cm

from equilibrium point, its vibrate velocity 1/2 time its maximum velocity. Determine the amplitude vibration and maximum acceleration

2. An object vibrated with frequency 2 Hz. Determine:a. Phase after 3 seconds, and 6 secondsb. Angle of phase after 3 seconds, and 6 secondsc. Difference of phase between 3 seconds until 6 seconds

3. A particle vibrated harmonically with amplitude 10 cm at frequency 5 Hz. At the time of phase is 1/4, determine particle’s velocity

4. A particle is moving simple harmonic motion with amplitude 6 cm and period 8 s. Calculate: a. the displacement after 4s, 6s and 8s

c. How much phase and angle of phase5. The particle vibrated harmonically with amplitude 4 cm and period 0,5s, when its

phase ¼ , determine of displacement, velocity and acceleration

v y=2 π⋅f⋅A⋅cos(2 π⋅ϕ )v y=2 πT

A⋅cos( 2 πT

⋅t )

c. Beda Fase

Δϕ=t2−t1

T=4−2

2=1

2. Sebuah partikel bergerak harmonic dengan amplitude 20cm pada frekuensi 1Hz. Pada saat fase sama dengan ½, tentukan kecepatan partikelnya!

Solusi:

or

v y=2 π .1 . 20. cos (2 π . 1

2)=40 π . cos(360 . 1

2)

v y=40 π . cos (180)=−40 π cm/s

Latihan:1. Sebuah partikel bergerak harmonic dengan perioda 2s, pada saat simpangannya 10cm

dari titik keseimbangan, kecepatan getarannya ½ kali kecepatan maksimumnya. Tentukan amplitude getaran dan percepatan maksimum

2. Sebuah benda bergetar dengan frekuensi 2Hz . Tentukanlah:a. Fase setelah 3s dan 6sb. Sudut fase setelah 3s dan 6sc. Perbedaan fase antara 3s sampai 6s

3. Sebuah partikel bergerak harmonic dengan amplitude 10cm pada frekuensi 5Hz. Pada saat fase ¼, tentukan kecepatan partikel tersebut!

4. Sebuah partikel bergerak harmonic dengan amplitude 6cm dan perioda 8s.Hitunglah: a. Simpangan setelah 4s, 6s dan 8s b. Berapa fase dan sudut fasenya

5. Partikel bergetar harmonic dengan amplitude 4cm dan perioda 0,5s, pada saat fasenya ¼, tentukan simpangan, kecepatan dan percepatannya!

v y=2 π⋅f⋅A⋅cos(2 π⋅ϕ )v y=2 πT

A⋅cos( 2 πT

⋅t )

Energy in Simple Harmonic MotionA mass on a spring transforms energy back and forth between kinetic and potential energy. If there were no dissipation, conservation of energy would dictate that the motion would continue forever.

Kinetic and Potential Energy

Kinetic energy (Ek) is energy of motion. The kinetic energy of an object is the energy it possesses because of its motion. The kinetic energy of a point mass m is given by

Ek=12

mv2=12

m⋅A2ω2 cos2(ω .t )

If k=m .ω2 is constant of vibration then,

Ek=12⋅k⋅A2 cos2 (ω . t )

Potential energy (Em) is energy stored in a spring. Energy is stored whether spring is compressed or stretched; amount stored given by:

EP=12

k . y2=12

k . A2 . sin2(ω⋅t )

Mechanical energy (Em) is energy due to the motion and position of an object.

EM=Ek+EP=12⋅k⋅A2 cos2 (ω . t )+ 1

2k . A2 . sin2 (ω⋅t )

EM=12

k . A2

Energi gerak harmonic sederhanaSebuah massa pada sebuah pegas mengubah energy kinetic dan energy potensial secara bergantian. Jika tidak ada energy yang hilang (disipasi), gerak akan terus berlanjut (continue) selamanya mematuhi hukum kekekalan energy.

Energi kinetic dan Potensial

Energy kinetik (Ek) adalah energy gerak. Energi kinetic sebuah benda adalah energy yang dimilki disebabkan gerakannya. Energi kinetic sebuah benda titik dirumuskan dengan:

Ek=12

mv2=12

m⋅A2ω2 cos2(ω .t )

Jika k=m . ω2 adalah kontanta getaran, maka

Ek=12⋅k⋅A2 cos2 (ω . t )

Energi Potensial (EP) adalah energy yang tersimpan dalam pegas baik yang ditekan maupun yang ditarik, jumlah energy yang tersimpan diberi rumus:

EP=12

k . y2=12

k . A2 . sin2(ω⋅t )

Energy mekanik (Em) adalah energy yang disebabkan gerak dan posisi dari sebuah

benda.

EM=Ek+EP=12⋅k⋅A2 cos2 (ω . t )+ 1

2k . A2 . sin2 (ω⋅t )

EM=12

k . A2

If we express kinetic and potential energy into the graph, hence its result as follows:

Now we will look at the energy with displacement:

Where: EK = kinetic energy (J)EP = potential energy (J)EM = mechanical energy or total energy (J)k = constant of vibrationω = angular velocity (rad/s)A = amplitude (m)m = mass (kg)t = time vibration (s)y = displacement (m)

Jika kita nyatakan energy kinetic dan potensial ke dalam grafik, maka hasilnya:

Sekarang kita akan melihat grafik energy dengan perpindahan:

Dimana: EK = energy kinetik (J)EP = energy potensial (J)EM = energy mekanik atau energy total (J)k = konstanta getarω = kecepatan sudut (rad/s)A = amplitude (m)m = massa (kg)t = waktu getar (s)y = perpindahan (m)

Example:

A particle has mass 500 grams vibrated harmonically with amplitude 2 cm and period 4 seconds. Determine kinetic, potential and mechanical energy after 4,5 seconds later.

Solution:

a. kinetic energyEk=

12

m⋅A2 ω2 cos2(ω . t )

Ek=12

m⋅A2(2 πf )2 cos2 ( 2 πT

. t )

Ek=12

m⋅A2 4 .π 2( 1T )

2

cos2( 2 πT

. t )

Ek=12

0,5⋅0 , 022 . 4 . 3 ,142 .( 14 )

2cos2( 2 π

4. 4,5 )

Ek = 0,000123 J

b. potential energyEP=

12

k . A2 . sin2(ω⋅t )

EP=12

m .ω2 . A2 .sin2( 2 πT

⋅t )

EP=12

m .(2 π . f )2 . A2 . sin2( 2 πT

⋅t )

EP=12

m . 4 . π2 . f 2. A2 . sin2 ( 2 πT

⋅t )

EP=12

.0,5. 4 .3 , 142( 14

)2. 0 , 022 . sin2( 2π4

⋅4,5)

EP = 0,000123 J

c. mechanical energy

J

EM=12

k . A2

EM= 12

.m . ω2 A2

EM= 12

. m .(2 πf )2 A2

EM=12×0,5×(2×3 , 14×1

4)2×0 ,022

EM=0 ,000246

Contoh:

Sebuah partikel memiliki massa 500g bergerak harmonic dengan amplitude 2cm dan periode 4s. Tentukan energy kinetic, potensial dan mekanik setelah 4,5s kemudian:

Solusi:

a. energy kinetikEk=

12

m⋅A2 ω2 cos2(ω . t )

Ek=12

m⋅A2(2 πf )2 cos2 ( 2 πT

. t )

Ek=12

m⋅A2 4 .π 2( 1T )

2cos2( 2 π

T. t )

Ek=12

0,5⋅0 ,022 . 4 . 3 ,142 .( 14 )

2

cos2( 2π4

. 4,5 )

Ek = 0,000123 J

b. Potensial energiEP=

12

k . A2 . sin2(ω⋅t )

EP=12

m .ω2 . A2 .sin2( 2 πT

⋅t )

EP=12

m .(2 π . f )2 . A2 . sin2( 2 πT

⋅t )

EP=12

m . 4 . π2 . f 2. A2 . sin2 ( 2 πT

⋅t )

EP=12

.0,5. 4 .3 , 142( 14

)2. 0 , 022 . sin2( 2π4

⋅4,5)

EP = 0,000123 J

c. Energy mekanik EM=1

2k . A2

EM= 12

.m . ω2 A2

EM=12

. m .(2 πf )2 A2

EM=12×0,5×(2×3 , 14×1

4)2×0 ,022

J

Exercises:

1. A particle has mass 100 grams vibrated harmonically with amplitude 4 cm and period 10 seconds. Determine kinetic, potential and mechanical energy after 6 seconds latter.

2. An object 80g is moving simple harmonic motion with maximum velocity 50 m/s and period 2s, determine amplitude, displacement, acceleration, maximum kinetic energy and the total energy

3. An object has mass 5kg is vibrating harmonically with 720 vibration per minute, amplitude 10cm. When its kinetic energy 10 J, determine potential energy and its displacement

4. An object has mass 50 g vibrated harmonically with time vibrate 5s. When object vibrated ½ s, its velocity 10 m/s. Determine of potential energy, kinetic energy and acceleration of vibration

5. See to the figure bellows, then determine ratio of time vibration from each composing!

A B

C

D

EM=0 ,000246

Latihan:

1. Sebuah partikel bermassa 100 g bergerak harmonic dengan amplitude 4cm dan periode 10 s. Tentukan energy kinetic, potensial, dan mekanik setelah 6s kemudian!

2. Sebuah benda 80g bergerak harmonic sederhana dengan kecepatan maksimum 50 m/s dan perioda 2s, tentukan amplitude, simpangan, percepatan, energy kinetic maksimum dan energy total!

3. Sebuah benda massa 5kg bergetar harmonic dengan 720 getaran per menit, amplitude 10cm. Pada saat energy kinetiknya 10J, tentukan energy potensial dan simpangannya

4. Sebuah benda massa 50g bergerak harmonic dengan waktu getar 5s. Pada saat benda bergetar ½ s, kecepatannya 10 m/s. Tentukan energy potensial, kinetic dan percepatannya

5. Perhatikan gambar di bawah ini, kemudian tentukan perbandingan waktu getar masing-masing susunan!

A B

C

D

Test of Competence

1. Spring of a car vibrates up and down with period√2 s when tire pass a barrier. Mass of car and driver is 300kg. If driver adds some passengers, mass of car and passengers become 600 kg, hence the new period of spring when the car pass the barrier is …

a. 2 √2 s d. 1 s

b. 2 s e. 1/√2 s

c. √2

2. The simple harmonic of spring, if mass of load which is hung at the end of spring is 1kg its period 2s, but if the load is added to be 4kg, hence its period is …a. ¼ s d. 4 sb. ½ s e. 8 sc. 1 s

3. An object moves simple harmonic at spring with constant force 80 N/m. Amplitude of vibration is 20 cm and maximum velocity 4 m/s, hence mass of above is …a. 1 kg d. 0,2 kgb. 0,8 kg e. 0,1 kgc. 0,4 kg

4. A ball has mass 20 g hung at spring. That ball pulled is down from its equilibrium point, then it is released. Obviously, it is going a simple vibration with frequency 32 Hz. If ball is changed by other mass 80 g, hence the frequency will happen is …a. 64 Hz d. 8 Hzb. 32 Hz e. 4 Hzc. 16 Hz

5. Two mass and springs system A and B are vibrate with frequencies fA and fB. If fA = 2fB and by assuming that both constant of force are same, so both of mass mA and mB meet relationship …

a.mA=

mB

4 d. mA=2. mB

b.mA=

mB

4 e. mA=4 . mB

c.mA=

mB

√2

6. A spring has length 20 cm hung vertically. Then edge of spring is given load 200 g. So, its length is increases 10 cm. The load is pulled down as far as 5 cm then released, so the load is vibrates harmonically. If g = 10 m/s2, so the frequency is …a. 0,5 Hz d. 18 Hzb. 1,6 Hz e. 62,8 Hzc. 5,0 Hz

7. The load 75 g hung vertically at a spring vibrates up-down with frequency 3 Hz. If the load was reduced 1/3 of it, then the frequency become …a. 3 Hz d. 3,7 Hzb. 3,2 Hz e. 4 Hzc. 3,5 Hz

8. An object has mass 0,1 kg moves harmonically with amplitude 0,1 m and period 0,2s. The maximum of force which is works at system about …a. 1 N d. 9,9 Nb. 5,5 N e. 12,4 Nc. 7,8 N

9. A particle is vibrates harmonically with period 6s and amplitude 10 cm. The speed of particle when it be 5 cm from equilibrium point is …a. 7,09 cm/s d. 11,07 cm/sb. 8,51 cm/s e. 19,12 cm/sc. 9,07 cm/s

10. An object moves harmonically with amplitude A. When its velocity equal a half from its maximum velocity, the displacement of vibration is …a. Zero d. 0,87 Ab. 0,5 A e. 1 Ac. 0,64 A

11. An object has mass 3kg vibrates harmonically with period 2s and amplitude 10cm. The force which is works at object when the displacement 6 cm is …a. 1,8 N d. 2,5 Nb. 2,0 N e. 3,0 Nc. 2,2 N

12. If the displacement y = 5cm, acceleration of simple vibration a = -5 cm/s2, then at the displacement 10 cm, its acceleration is … cm/s2

a. -25 c. -10 e. -1,25b. -20 d. -2,5

13. Two oscillators vibrates with the same phase at t = 0, both of each frequencies continues 10 Hz and 40 Hz. After 5/4 s both of wave has difference angle phase is …a. 0o c. 45o e. 180o

b. 30o d. 90o

14. A particle has mass 10 g vibrated harmonically with frequency 100 Hz and amplitude 8 cm. Potential energy when angle of phase 30o is … Joule.a. 0,12π2 c. 0,23π2 e. 0,45π2

b. 0,7π2 d. 0,32π2

15. An object vibrates harmonically with amplitude √3 cm. Kinetic energy at equilibrium point is 20 J. Amount of potential energy object when velocity is ½ of its maximum velocity is …a. Zero c. 10 J e. 20 Jb. 5 J d. 15 J

16. An object vibrates harmonically with amplitude 40 cm. If potential energy when maximum displacement is 10 J, so potential energy at the displacement of 20cm is … a. 0,5 J c. 2,5 J e. 10,0 Jb. 1,0 J d. 5,0 J

17. When kinetic energy of object which vibrates harmonically is equal its potential energy so, …a. Angle of phase 180o c. Angle of phase 45o e. acceleration is zerob. Phase is ¾ d. Phase is ¼

18. An object has mass 0,150 kg moves harmonically at the edge of spring which is have constant spring 200 N/m. When the object at 0,01 m from its equilibrium point, velocity of object become 0,2 m/s. The total energy of object when its position 0,005 m/s from its equilibrium position is …a. 0,003 J c. 0,030 J e. 0,073 Jb. 0,013 J d. 0,053 J

19. An object mass 50 g moves harmonically with amplitude 10 cm and period 0,2s. Amount of force which is work at system when its displacement of ½ amplitude is …a. 1,0 N c. 4,8 N e. 8,4 Nb. 2,5 N d. 6,9 N

20. A particle is doing move harmonically. The kinetic energy of particle is KE, potential energy is PE, and total energy is TE. When the particle on the centre of equilibrium position and amplitude position, hence the comparison of KE/TE and PE/TE are …a. 1/4 and 3/4 c. 3/4 and 1/4 e. 3/8 and 5/8 b. 1/4 and 1/2 d. 1/8 and 7/8