Respuesta :
Answer:
x = 0.96 e⁻ᵗ - 0.16 e⁻⁶ᵗ
Explanation:
Let the position of the mass at any time be x.
The general formula for the elastic material is given as
mx" + cx' + kx = 0
m = mass of spring = 2 kg
c = damping constant = 14 kg/s
k = spring constant
F = kx
3.6 = 0.3k
k = 12 N/m
mx" + cx' + kx = 0
Becomes
2x" + 14x' + 12x = 0
x" + 7x' + 6x = 0
x(0) = 0.8 m, x'(0) = 0 m/s (It was initially at rest)
We then solve this differential equation
x" + 7x' + 6x = 0
x = eˢᵗ (s is an arbitrary number)
x' = seˢᵗ
x" = s²eˢᵗ
x" + 7x' + 6x = 0
Becomes
s²eˢᵗ + 7seˢᵗ + 6eˢᵗ = 0
s² + 7s + 6 = 0
s² + s + 6s + 6 = 0
s(s + 1) + 6(s + 1) = 0
(s + 1)(s + 6) = 0
s = -1 or -6
So,
x = Ae⁻ᵗ + Be⁻⁶ᵗ
At t = 0, x = 0.8
0.8 = A + B (eqn 1)
At t = 0, x' = 0
x' = -Ae⁻ᵗ - 6Be⁻⁶ᵗ
0 = -A - 6B
A = -6B
From eqn 1
A + B = 0.8
-6B + B = 0.8
-5B = 0.8
B = -0.16
A = -6B = -6 × -0.16 = 0.96
x = 0.96 e⁻ᵗ - 0.16 e⁻⁶ᵗ
Hope this Helps!!!
Answer:
[tex]x(t)=\frac{1}{25} e^{-6t}+\frac{19}{25} e^{-t}+te^{-t}[/tex]
Explanation:
Given that,
mass, m = 2kg
constant, c = 14kg/s
K = 3.6 / 0.3
= 12
[tex]F(t)= 10e^{-t}[/tex]
Differentiate equation
[tex]mx''+cx'+kx=f(t)\\\\2x''+14x'+12x=10e^{-t}\\\\x''+7x'+6x=5e^{-t}\\\\x(0)=0.8,x'(0)=0[/tex]
Characteristic equation
[tex]r^2+7r+6=0\\\\r^2+6r+r+6=0\\\\r(r+6)+1(r+6)=0\\r=-1,r=-6[/tex]
[tex]x_t=c_1e^{-6t}+c_2e^{-t}[/tex]
Let
[tex]x_p=Ate^{-t}[/tex]
[tex]x_p'=A(-te^{-t}+e^{-t}\\\\x_p''=A(te^{-t}-2e^{-t})\\\\x_p''+7x_p'+6x_p=5e^{-t}\\\\A(te^{-t}-2e^{-t}+7A(-te^{-t}+e^{-t})+6Ate^{-t}=5e^{-t}[/tex]
like terms cancel each other
[tex]-2e^{-t}+7Ae^{-t}=5e^{-t}[/tex]
[tex]5Ae^{-t}=5e^{-t}\\\\A=1[/tex]
[tex]x(t)=c_1e^{-6t}+c_2e^{-t}+te^{-t}\\\\x(0)=0.8\\\\c_1+c_2=0.8\\\\x'(0)=0\\\\-6c_1-c_2+1=0\\\\c_1+c_2=0.8----(1)\\\\6c_1+c_2=1----(2)[/tex]
subtract both equation
[tex]-5c=-0.2\\c_1=\frac{1}{25}[/tex]
[tex]c_2=\frac{8}{10} -\frac{1}{25} \\\\c_2=\frac{4}{5} -\frac{1}{25} \\\\c_2=\frac{19}{25}[/tex]
[tex]x(t)=\frac{1}{25} e^{-6t}+\frac{19}{25} e^{-t}+te^{-t}[/tex]
