HARMONIC PENDULUM OSCILLATIONS

A. ILLUSTRATION

A pendulum is a weight suspended from a pivot so that it can swing freely. When a pendulum is

displaced sideways from its resting equilibrium position, it is subject to a restoring force due to gravity

that will accelerate it back toward the equilibrium position. When released, the restoring force

combined with the pendulum's mass causes it to oscillate about the equilibrium position, swinging

back and forth. The time for one complete cycle, a left swing and a right swing, is called the period. A

pendulum swings with a specific period which depends (mainly) on its length.

From its discovery around 1602 by Galileo Galilei the regular motion of pendulums was used for

timekeeping, and was the world's most accurate timekeeping technology until the 1930s. Pendulums

are used to regulate pendulum clocks, and are used in scientific instruments such as accelerometers and

seismometers. Historically they were used as gravimeters to measure the acceleration of gravity in

geophysical surveys, and even as a standard of length. The word 'pendulum' is new Latin, from the

Latin pendulus, meaning 'hanging'.

The simple gravity pendulum is an idealized mathematical model of a pendulum. This is a weight (or

bob) on the end of a massless cord suspended from a pivot, without friction. When given an initial

push, it will swing back and forth at a constant amplitude. Real pendulums are subject to friction and

air drag, so the amplitude of their swings declines.

The period of swing of a simple gravity pendulum depends on its length, the local strength of gravity,

and to a small extent on the maximum angle that the pendulum swings away from vertical, θ0, called

the amplitude.[8] It is independent of the mass of the bob. If the amplitude is limited to small swings,

The period T of a simple pendulum, the time taken for a complete cycle, is:

where L is the length of the pendulum and g is the local acceleration of gravity.

For small swings the period of swing is approximately the same for different size swings: that is, the

period is independent of amplitude. This property, called isochronism, is the reason pendulums are so

useful for timekeeping. Successive swings of the pendulum, even if changing in amplitude, take the

same amount of time.

B. MATHEMATICAL MODELING

A pendulum with arm l, swing θ≪. If the deviation x, determine the equation θ as a function

of time (t). Look at the picture of gravity f=m .g. if θ≪, this means that sin θ=tan θ=θ

Large deviation x=l tan θ=lθ

Menurut Hukum II Newton,

F= ma= - m.g sin θ

Dimana: m = massa bandul

a = percepatan bandul

pada sumbu x didapat:

md2 xd t2

=−mgsin θ . . . . . . . . . . . . .1)

Dari hubungan x=l θ, kita dapat mencari:

dxdt

=l d θdt,d2 xd t2

=l d2θdt2

............................2)

Sisipkan harga 2) ke dalam 1) di peroleh:

mld2θdt2

+mgsin θ=0 kedua ruas di kalikan 1ml

d2θdt 2

+ gl

sinθ =0

Gunakan hampiran sin θ :

d2tdt 2

+ glθ=0

C. MATHEMATICAL MODEL SOLUTION

d2tdt 2

+ glθ=0

Persamaan bantu: r2 + gl

= 0

Akarnya : r1,2 = ±√ gl iSolusi umumnya : θ=c1cos √ gl t+c2sin√ gl t d2t

dt 2+ glθ ………………3)

Jika diketahui pada saat t=0 , θ=0° dan dxdt

=V 0. Tentukan persamaan θ( t)

Sisipkan syarat awal t=0 , θ=0 ke dalam persamaan 3) diperoleh C1=0 atau θ=C2 sin√ gl t dan

dθdt

=[√ gl ]C2 cos[√ gl t ] ....................4)

Pada persamaan 2) diketahui dxdt

=l dθdt

atau dθdt

=( dxdx )l

dan persamaan 4) menjadi:

( dxdt )l

=[√ gl ]C2cos [√ gl ] tUntuk t=0 ,

dxdt

=V 0.

Jadi,

V 0

l=[√ gl ]C2

C2=[V 0 √ lg ]l

Kemudian, sisipkan harga C2 ke dalam persamaan 4),

Diperoleh penyelesaian:

θ=[V 0 √ lg ]l

sin√ gl t

Contoh Soal dan Penyelesaian:

Diketahui seorang anak bermain ayunan dihalaman rumahnya. Jika berat total anak dengan bangku

ayunan 40 kg dan panjang tali ayunan adalah 2m dan percepatan gravitasi bumi adalah 10 m /s2.

Tentukanlah. a: solusi umum dari dari kasus tersebut!

b: jika diketahui saat t = 0,dan θ0=0°dan dxdt

=v0tentukanlah

persamaan dari θt .

Jawab:

Diketahui : massa : 40 kg

L(lengan momen) : 2 m

t : 3 detik

v0 : 0 m /s2

Sehingga

a. mLd2θd t2

+mg sinθ=0

dengan pendekatan

mLd2θd t2

+mgθ=0

40.2d2θd t2

+40.10θ=0

d2θd t2

+ 40.10 . θ

40.2=0

d2θd t2

+5θ=0

Sehingga solusi umumnya adalah

θ=C1 cos (√5 ) t+C 2sin (√5) t

b. jika diketahui pada saat t = 0,dan θ=¿ 30° dan dxdt

=v0maka diperoleh

30°=C1cos√5.0+C2 sin√5 .0.....(1)

30°=C1

C1 = π6

C1=3,146

=0,52333

Substitusi nilai C1 kedalam persamaan (1),sehingga

θ=0,52333 cos√5 t+C2sin√5 t....(2)

dθ2

dt 2=−0,52333.√5. sin√5 t+√5 .C2. cos √5 t

0=−0,52333.√5 sin√5 t+√5 .C2. cos√5t

√5 .C 2.cos √5 t=0,52333.√5 .sin√5 t

Untuk t = 0,maka

√5 .C 2.cos (√5 .0)=0,52333.(√5 .0)

C2=0

Kemudian substitusi nilai C1,C2 ke dalam (2)

Sehingga

θ=0,52333. cos(√5¿t )¿