Note : All calculations will be done in SI.
It means that
{ m = 64 l b s ≈ 29.03 k g Δ x = 8 f e e t ≈ 2.44 m l 0 = 3 f e e t ≈ 0.92 m g ≈ 9.81 m / s 2 \left\{\begin{array}{l}
m=64\,\,lbs\approx 29.03\,kg\\[0.3cm]
\Delta x=8\,\,feet\approx 2.44\,\,m\\[0.3cm]
l_0=3\,\,feet\approx 0.92\,\,m\\[0.3cm]
g\approx 9.81\,\,m/s^2
\end{array}\right. ⎩ ⎨ ⎧ m = 64 l b s ≈ 29.03 k g Δ x = 8 f ee t ≈ 2.44 m l 0 = 3 f ee t ≈ 0.92 m g ≈ 9.81 m / s 2
1 STEP : Find the stiffness of the spring
(more information : https://en.wikipedia.org/wiki/Hooke%27s_law)
In our case,
m g = k ⋅ Δ x → k = m g Δ x ≈ 29.03 ⋅ 9.81 2.44 ≈ 116.72 N m mg=k\cdot\Delta x\to k=\frac{mg}{\Delta x}\approx\frac{29.03\cdot9.81}{2.44}\approx116.72\,\frac{N}{m} m g = k ⋅ Δ x → k = Δ x m g ≈ 2.44 29.03 ⋅ 9.81 ≈ 116.72 m N
2 STEP : Let's describe all the forces that act on our body
The force of gravity − F g = m g Hooke’s force − F s = − k x damping force − F d = − 8 ⋅ v ≡ − 8 ⋅ d x d t an external force − f ( t ) = 6 t 2 \text{The force of gravity} - F_g=mg\\[0.3cm]
\text{Hooke's force}- F_s=-kx\\[0.3cm]
\text{damping force}-F_d=-8\cdot v\equiv-8\cdot\frac{dx}{dt}\\[0.3cm]
\text{an external force}- f(t)=6t^2 The force of gravity − F g = m g Hooke’s force − F s = − k x damping force − F d = − 8 ⋅ v ≡ − 8 ⋅ d t d x an external force − f ( t ) = 6 t 2 3 STEP : Let's write the equation of motion (Newton's second law) for a given body, together with the initial conditions
m a ⃗ = F g ⃗ + F s ⃗ + F d ⃗ + f ⃗ ( t ) x ( 0 ) = l 0 ≈ 0.92 m x ′ ( 0 ) = 0 m\vec{a}=\vec{F_g}+\vec{F_s}+\vec{F_d}+\vec{f}(t)\\[0.3cm]
x(0)=l_0\approx0.92\,m\\[0.3cm]
x'(0)=0 m a = F g + F s + F d + f ( t ) x ( 0 ) = l 0 ≈ 0.92 m x ′ ( 0 ) = 0 Suppose that all movement occurs in the vertical direction, the axis is directed along the action of the force of gravity. Here is the corresponding figure.
Note: since the forces of Hooke, dumping and one more act in the same direction, for the convenience of reading the diagram, I will draw from NOT from one point.Since the direction of the external force is not indicated, then I myself will choose its direction.
O y : m ⋅ a ⏞ x ′ ′ = m g − 8 v ⏞ x ′ − 116.72 ⋅ x + 6 t 2 29.03 ⋅ x ′ ′ + 8 ⋅ x ′ + 116.72 ⋅ x = 29.03 ⋅ 9.81 + 6 t 2 Oy : m\cdot\overbrace{a}^{x''}=mg-8\overbrace{v}^{x'}-116.72\cdot x+6t^2\\[0.3cm]
29.03\cdot x''+8\cdot x'+116.72\cdot x=29.03\cdot 9.81+6t^2 O y : m ⋅ a x ′′ = m g − 8 v x ′ − 116.72 ⋅ x + 6 t 2 29.03 ⋅ x ′′ + 8 ⋅ x ′ + 116.72 ⋅ x = 29.03 ⋅ 9.81 + 6 t 2 This is an inhomogeneous second-order equation with constant coefficients. The solution has the form
x ( t ) = x h o m ( t ) + x p a r t ( 1 ) + x p a r t ( 2 ) , where x h o m ( t ) : 29.03 ⋅ x ′ ′ + 8 ⋅ x ′ + 116.72 ⋅ x = 0 x ( t ) = e λ t → x ′ ( t ) = λ ⋅ e λ t → x ′ ′ ( t ) = λ 2 ⋅ e λ t 29.03 ⋅ λ 2 ⋅ e λ t + 8 ⋅ λ ⋅ e λ t + 116.72 ⋅ e λ t = 0 ∣ ÷ e λ t 29.03 ⋅ λ 2 + 8 ⋅ λ + 116.72 = 0 D = 8 2 − 4 ⋅ 29.03 ⋅ 116.72 ≈ 116.14 ⋅ i [ λ 1 = − 8 − 116.14 ⋅ i 2 ⋅ 29.09 ≈ − 0.14 − 2 i λ 2 = − 8 + 116.14 ⋅ i 2 ⋅ 29.09 ≈ − 0.14 + 2 i x h o m ( t ) = A 1 ⋅ e ( − 0.14 − 2 i ) t + A 2 ⋅ e ( − 0.14 + 2 i ) t → x h o m ( t ) = C 1 e − 0.14 t ⋅ cos 2 t + C 2 e − 0.14 t ⋅ sin 2 t x(t)=x_{hom}(t)+x_{part}^{(1)}+x_{part}^{(2)},\,\,\text{where}\\[0.3cm]
x_{hom}(t) : 29.03\cdot x''+8\cdot x'+116.72\cdot x=0\\[0.3cm]
x(t)=e^{\lambda t}\to x'(t)=\lambda\cdot e^{\lambda t}\to x''(t)=\lambda^2\cdot e^{\lambda t}\\[0.3cm]
\left.29.03\cdot \lambda^2\cdot e^{\lambda t}+8\cdot \lambda\cdot e^{\lambda t}+116.72\cdot e^{\lambda t}=0\right|\div e^{\lambda t}\\[0.3cm]
29.03\cdot \lambda^2+8\cdot \lambda+116.72=0\\[0.3cm]
\sqrt{D}=\sqrt{8^2-4\cdot29.03\cdot116.72}\approx 116.14\cdot i\\[0.3cm]
\left[\begin{array}{l}
\lambda_1=\displaystyle\frac{-8-116.14\cdot i}{2\cdot29.09}\approx-0.14-2i\\[0.3cm]
\lambda_2=\displaystyle\frac{-8+116.14\cdot i}{2\cdot29.09}\approx-0.14+2i
\end{array}\right.\\[0.3cm]
x_{hom}(t)=A_1\cdot e^{(-0.14-2i)t}+A_2\cdot e^{(-0.14+2i)t}\to\\[0.3cm]
\boxed{x_{hom}(t)=C_1e^{-0.14t}\cdot\cos{2t}+C_2e^{-0.14t}\cdot\sin{2t}} x ( t ) = x h o m ( t ) + x p a r t ( 1 ) + x p a r t ( 2 ) , where x h o m ( t ) : 29.03 ⋅ x ′′ + 8 ⋅ x ′ + 116.72 ⋅ x = 0 x ( t ) = e λ t → x ′ ( t ) = λ ⋅ e λ t → x ′′ ( t ) = λ 2 ⋅ e λ t 29.03 ⋅ λ 2 ⋅ e λ t + 8 ⋅ λ ⋅ e λ t + 116.72 ⋅ e λ t = 0 ∣ ∣ ÷ e λ t 29.03 ⋅ λ 2 + 8 ⋅ λ + 116.72 = 0 D = 8 2 − 4 ⋅ 29.03 ⋅ 116.72 ≈ 116.14 ⋅ i ⎣ ⎡ λ 1 = 2 ⋅ 29.09 − 8 − 116.14 ⋅ i ≈ − 0.14 − 2 i λ 2 = 2 ⋅ 29.09 − 8 + 116.14 ⋅ i ≈ − 0.14 + 2 i x h o m ( t ) = A 1 ⋅ e ( − 0.14 − 2 i ) t + A 2 ⋅ e ( − 0.14 + 2 i ) t → x h o m ( t ) = C 1 e − 0.14 t ⋅ cos 2 t + C 2 e − 0.14 t ⋅ sin 2 t
It remains to find partial solutions :
x p a r t ( 1 ) : 29.03 ⋅ x ′ ′ + 8 ⋅ x ′ + 116.72 ⋅ x = 29.03 ⋅ 9.81 x ( t ) = C o n s t → x ′ ( t ) = 0 → x ′ ′ ( t ) = 0 → 29.03 ⋅ 0 + 8 ⋅ 0 + 116.72 ⋅ C o n s t = 29.03 ⋅ 9.81 → C o n s t = 29.03 ⋅ 9.81 116.72 ≈ 2.44 → x p a r t ( 1 ) = 2.44 x p a r t ( 2 ) : 29.03 ⋅ x ′ ′ + 8 ⋅ x ′ + 116.72 ⋅ x = 6 t 2 x ( t ) = A t 2 + B t + C → x ′ ( t ) = 2 A t + b → x ′ ′ ( t ) = 2 A → 29.03 ⋅ 2 A + 8 ⋅ ( 2 A t + B ) + 116.72 ⋅ ( A t 2 + B t + C ) = 6 t 2 → 116.72 A t 2 + ( 116.72 B + 16 A ) t + ( 116.72 C + 8 B + 48.06 A ) = 6 t 2 { 116.72 A = 6 116.72 B + 16 A = 0 116.72 C + 8 B + 48.06 A = 0 → { A ≈ 0.05 B ≈ − 0.007 C ≈ − 0.02 x p a r t ( 2 ) = 0.05 t 2 − 0.007 t − 0.02 x_{part}^{(1)} : 29.03\cdot x''+8\cdot x'+116.72\cdot x=29.03\cdot 9.81\\[0.3cm]
x(t)=Const\to x'(t)=0\to x''(t)=0\to\\[0.3cm]
29.03\cdot 0+8\cdot 0+116.72\cdot Const=29.03\cdot 9.81\to\\[0.3cm]
Const=\frac{29.03\cdot 9.81}{116.72}\approx2.44\to\boxed{x_{part}^{(1)}=2.44}\\[0.3cm]
x_{part}^{(2)} : 29.03\cdot x''+8\cdot x'+116.72\cdot x=6t^2\\[0.3cm]
x(t)=At^2+Bt+C\to x'(t)=2At+b\to x''(t)=2A\to\\[0.3cm]
29.03\cdot 2A+8\cdot(2At+B)+116.72\cdot(At^2+Bt+C)=6t^2\to\\[0.3cm]
116.72At^2+(116.72B+16A)t+(116.72C+8B+48.06A)=6t^2\\[0.3cm]
\left\{\begin{array}{l}
116.72A=6\\[0.3cm]
116.72B+16A=0\\[0.3cm]
116.72C+8B+48.06A=0
\end{array}\right.\to \left\{\begin{array}{l}
A\approx 0.05\\[0.3cm]
B\approx-0.007\\[0.3cm]
C\approx-0.02
\end{array}\right.\\[0.3cm]
\boxed{x_{part}^{(2)}=0.05t^2-0.007t-0.02} x p a r t ( 1 ) : 29.03 ⋅ x ′′ + 8 ⋅ x ′ + 116.72 ⋅ x = 29.03 ⋅ 9.81 x ( t ) = C o n s t → x ′ ( t ) = 0 → x ′′ ( t ) = 0 → 29.03 ⋅ 0 + 8 ⋅ 0 + 116.72 ⋅ C o n s t = 29.03 ⋅ 9.81 → C o n s t = 116.72 29.03 ⋅ 9.81 ≈ 2.44 → x p a r t ( 1 ) = 2.44 x p a r t ( 2 ) : 29.03 ⋅ x ′′ + 8 ⋅ x ′ + 116.72 ⋅ x = 6 t 2 x ( t ) = A t 2 + Bt + C → x ′ ( t ) = 2 A t + b → x ′′ ( t ) = 2 A → 29.03 ⋅ 2 A + 8 ⋅ ( 2 A t + B ) + 116.72 ⋅ ( A t 2 + Bt + C ) = 6 t 2 → 116.72 A t 2 + ( 116.72 B + 16 A ) t + ( 116.72 C + 8 B + 48.06 A ) = 6 t 2 ⎩ ⎨ ⎧ 116.72 A = 6 116.72 B + 16 A = 0 116.72 C + 8 B + 48.06 A = 0 → ⎩ ⎨ ⎧ A ≈ 0.05 B ≈ − 0.007 C ≈ − 0.02 x p a r t ( 2 ) = 0.05 t 2 − 0.007 t − 0.02
Conclusion, general solution is
x ( t ) = x h o m ( t ) + x p a r t ( 1 ) + x p a r t ( 2 ) x ( t ) = e − 0.14 t ⋅ ( C 1 cos 2 t + C 2 sin 2 t ) + 0.05 t 2 − 0.007 t + 2.42 x(t)=x_{hom}(t)+x_{part}^{(1)}+x_{part}^{(2)}\\[0.3cm]
x(t)=e^{-0.14t}\cdot\left(C_1\cos{2t}+C_2\sin{2t}\right)+0.05t^2-0.007t+2.42 x ( t ) = x h o m ( t ) + x p a r t ( 1 ) + x p a r t ( 2 ) x ( t ) = e − 0.14 t ⋅ ( C 1 cos 2 t + C 2 sin 2 t ) + 0.05 t 2 − 0.007 t + 2.42
Now let's find the constants, for this we use the initial conditions:
x ( 0 ) ≈ 0.92 = 1 ⋅ ( C 1 ⋅ 1 + C 2 ⋅ 0 ) + 2.42 → C 1 = 0.92 − 2.42 = − 1.5 → C 1 = − 1.5 x ′ ( t ) = − 0.14 e − 0.14 t ⋅ ( C 1 cos 2 t + C 2 sin 2 t ) + + e − 0.14 t ⋅ ( − 2 C 1 sin 2 t + 2 C 2 cos 2 t ) + 0.1 t − 0.007 x ′ ( 0 ) = 0 = − 0.14 ⋅ C 1 + 2 C 2 − 0.007 → 2 C 2 = 0.14 ⋅ 1.5 + 0.007 = 0.217 → C 2 = 0.1085 x(0)\approx0.92=1\cdot(C_1\cdot1+C_2\cdot0)+2.42\to\\[0.3cm]
C_1=0.92-2.42=-1.5\to\boxed{C_1=-1.5}\\[0.3cm]
x'(t)=-0.14e^{-0.14t}\cdot\left(C_1\cos{2t}+C_2\sin{2t}\right)+\\[0.3cm]
+e^{-0.14t}\cdot\left(-2C_1\sin{2t}+2C_2\cos{2t}\right)+0.1t-0.007\\[0.3cm]
x'(0)=0=-0.14\cdot C_1+2C_2-0.007\to\\[0.3cm]
2C_2=0.14\cdot1.5+0.007=0.217\to \boxed{C_2=0.1085} x ( 0 ) ≈ 0.92 = 1 ⋅ ( C 1 ⋅ 1 + C 2 ⋅ 0 ) + 2.42 → C 1 = 0.92 − 2.42 = − 1.5 → C 1 = − 1.5 x ′ ( t ) = − 0.14 e − 0.14 t ⋅ ( C 1 cos 2 t + C 2 sin 2 t ) + + e − 0.14 t ⋅ ( − 2 C 1 sin 2 t + 2 C 2 cos 2 t ) + 0.1 t − 0.007 x ′ ( 0 ) = 0 = − 0.14 ⋅ C 1 + 2 C 2 − 0.007 → 2 C 2 = 0.14 ⋅ 1.5 + 0.007 = 0.217 → C 2 = 0.1085
Final result :
x ( t ) = e − 0.14 t ( − 1.5 cos ( 2 t ) + 0.1085 sin ( 2 t ) ) + 0.05 t 2 − 0.007 t + 2.42 \boxed{x(t)=e^{-0.14t}\left(-1.5\cos(2t)+0.1085\sin(2t)\right)+0.05t^2-0.007t+2.42} x ( t ) = e − 0.14 t ( − 1.5 cos ( 2 t ) + 0.1085 sin ( 2 t ) ) + 0.05 t 2 − 0.007 t + 2.42
ANSWER :
(a) model
29.03 ⋅ x ′ ′ + 8 ⋅ x ′ + 116.72 ⋅ x = 29.03 ⋅ 9.81 + 6 t 2 x ( 0 ) = l 0 ≈ 0.92 m x ′ ( 0 ) = 0 29.03\cdot x''+8\cdot x'+116.72\cdot x=29.03\cdot 9.81+6t^2\\[0.3cm]
x(0)=l_0\approx0.92\,m\\[0.3cm]
x'(0)=0 29.03 ⋅ x ′′ + 8 ⋅ x ′ + 116.72 ⋅ x = 29.03 ⋅ 9.81 + 6 t 2 x ( 0 ) = l 0 ≈ 0.92 m x ′ ( 0 ) = 0 (b) solution
x ( t ) = e − 0.14 t ( − 1.5 cos ( 2 t ) + 0.1085 sin ( 2 t ) ) + 0.05 t 2 − 0.007 t + 2.42 x(t)=e^{-0.14t}\left(-1.5\cos(2t)+0.1085\sin(2t)\right)+0.05t^2-0.007t+2.42 x ( t ) = e − 0.14 t ( − 1.5 cos ( 2 t ) + 0.1085 sin ( 2 t ) ) + 0.05 t 2 − 0.007 t + 2.42
(c) graph
For small times 0 ≤ t ≤ 10 0\le t\le 10 0 ≤ t ≤ 10
For large times 0 ≤ t ≤ 50 0\le t\le 50 0 ≤ t ≤ 50
Comments