Sebuah benda bergetar harmonik sederhana dengan persamaan y = 5 sin ( 3 t +  /6)y dalam meter, t dalam detik, dan besaran sudut dalam radian. Tentukan :a. Amp

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Diketahui :
persamaan simpangan getararan
Y = 5 sin (3π.t + π/6) dimana y dalam meter
dimana : 
ω = 3π 

A. Amplitudo, A = ____?
A = 5 meter

# frekuensi, f = __?
ω = 2π.f
3π = 2π.f
f = 3/2 Hz = 1,5 Hz

# Periode, T = __?
T = 1/f
T = 1/(3/2)
T = 2/3 detik

B. Kecepatan getar sesaat, V = ___?
V = dy/dt = ω.A. cos (ω.t + θ)
V = 3π. 5. cos (3π.t + π/6)
V = 15π cos (3π.t + π/6)

Percepatan getar sesaat, a = __?
a = dv/dt = - ω².A sin (ω.t + θ)
a = - (3π)². 5. sin (3π.t + π/6)
a = - 45.π². sin (3π.t + π/6)

Saat t = 2 detik
C. Simpangan, Y = ___?
Y = 5 sin (3π.t + π/6)
Y = 5 sin (3π.2 + π/6)
Y = 5 sin (π/6)
Y = 5. 1/2
Y = 5/2 meter = 2,5 meter

Kecepatan, V = __?
V = 15π cos (3π.t + π/6)
V = 15π cos (3π.2 + π/6)
V = 15π cos (π/6)
V = 15π .(0,5.√3)
V = 7,5π.√3 m/s

Percepatan, a = ___?
a = - 45.π². sin (3π.t + π/6)
a = - 45.π². sin (3π.2 + π/6)
a = - 45.π². sin (π/6)
a = - 45.π². (1/2)
a = - 22,5.π² m/s²


D. Kecepatan maksimumnya, V max =__?
Vmax = ω.A
Vmax = ω.A
Vmax = 3π. 5
Vmax = 15.π m/s

Percepatan maksimumnya, a max = ___
a max = ω².A
a max = (3π)². 5
a max = 45.π² m/s²


saat., t = 1 detik dan m = 100 gram = 0,1 Kg
E. Ek = ___?
Cari dulu kecepatan saat t = 1 detik, V(1) = ___?
V = 15π cos (3π.t + π/6)
V(1) = 15π cos (3π.1 + π/6)
V(1) = 15π cos (π + π/6)
V(1) = - 15π cos (π/6)
V(1) = - 15π (0,5.√3)
V(1) = -7,5.π.√3 m/s

Maka :
Ek = 1/2.m.V²
Ek = 1/2. 0,1. (-7,5.π.√3 )²
Ek = 0,05. (168,75.π²)
Ek = 8,4375.π² Joule

saat., t = 1 detik dan m = 100 gram = 0,1 Kg
E. Ep = ___?
Cari dulu simpangan saat t = 1 detik, y(1) = ___?
Y = 5 sin (3π.t + π/6)
Y(1) = 5 sin (3π.1 + π/6)
Y(1) = 5 sin (π + π/6) 
Y(1) = 5 sin (π/6)
Y(1) = 5 .(0,5)
Y(1) = 2,5 meter

Maka 
Ep = 1/2. k.Y²
Ep = 1/2. m.ω².Y²
Ep = 1/2. 0,1. (3π)². (2,5)²
Ep = 0,05. (9π²). (6,25)
Ep =  2,8125.π² Joule

F. Energi totalnya, Em = __?
Cara 1
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Em = Ep + Ek
Em = 2,8125.π² + 8,4375.π²
Em = 11,25.π² Joule

Cara 2
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Em = 1/2. k.A²
Em = 1/2. m.ω². A²
Em = 1/2. 0,1. (3π)². (5)²
Em = 0,05. (9.π²). (25)
Em = 11,25.π² Joule


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FZA