QUESTION 8 [TOTAL MARKS: 21 MARKS]
A machine with a damping system is modelled as a spring-mass-damper system. The displacement satisfies the non-homogenous second order differential equation:

is the displacement (in metres) and is the time taken (in seconds). Determine the general function of .
[21 marks]

Ask by Burns Mccoy. in Brunei

Mar 22,2025

Question

QUESTION 8 [TOTAL MARKS: 21 MARKS]
A machine with a damping system is modelled as a spring-mass-damper system. The displacement satisfies the non-homogenous second order differential equation:

is the displacement (in metres) and is the time taken (in seconds). Determine the general function of .
[21 marks]

Ask by Burns Mccoy. in Brunei

Mar 22,2025

UpStudy AI · Accepted Answer

$$
\textbf{Step 1: Solve the Homogeneous Equation}
$$

The homogeneous part of the differential equation is
$$
\frac{d^2 x}{dt^2} + 2 \frac{dx}{dt} + 5x = 0.
$$
We assume a solution of the form
$$
x = e^{rt}.
$$
Substituting into the homogeneous equation gives the characteristic equation:
$$
r^2 + 2r + 5 = 0.
$$
Solve the quadratic using the quadratic formula:
$$
r = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i.
$$
Thus, the general solution for the homogeneous equation is:
$$
x_h(t) = e^{-t}\left(C_1 \cos 2t + C_2 \sin 2t\right),
$$
where $$ C_1 $$ and $$ C_2 $$ are arbitrary constants.

$$
\textbf{Step 2: Find a Particular Solution}
$$

Given the non-homogeneous differential equation:
$$
\frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+5 x=10 \cos 2t,
$$
a reasonable guess for a particular solution is:
$$
x_p(t)=A\cos 2t+B\sin 2t.
$$
Compute the derivatives:
$$
x_p'(t) = -2A\sin 2t + 2B\cos 2t,
$$
$$
x_p''(t) = -4A\cos 2t - 4B\sin 2t.
$$
Substitute $$ x_p $$, $$ x_p' $$, and $$ x_p'' $$ into the differential equation:
$$
(-4A\cos 2t - 4B\sin 2t) + 2(-2A\sin 2t + 2B\cos 2t) + 5(A\cos 2t+B\sin 2t)=10\cos 2t.
$$
Simplify by combining like terms:
$$
(-4A\cos 2t - 4B\sin 2t) + (-4A\sin 2t + 4B\cos 2t) + (5A\cos 2t+5B\sin 2t)=10\cos 2t.
$$
Group the terms for $$\cos 2t$$ and $$\sin 2t$$:
$$
\text{Coefficient of } \cos 2t: \quad -4A + 4B + 5A = (A + 4B),
$$
$$
\text{Coefficient of } \sin 2t: \quad -4B - 4A+ 5B = (B - 4A).
$$
Thus, we obtain:
$$
(A+4B)\cos 2t + (B-4A)\sin 2t = 10\cos 2t.
$$
For the equality to hold for all $$ t $$, equate the coefficients:
$$
A+4B=10, \quad B-4A=0.
$$

Solve the second equation:
$$
B = 4A.
$$
Substitute $$ B = 4A $$ into the first equation:
$$
A + 4(4A) = A + 16A = 17A = 10 \quad \Longrightarrow \quad A = \frac{10}{17}.
$$
Then,
$$
B = 4A = \frac{40}{17}.
$$
So, the particular solution is:
$$
x_p(t)=\frac{10}{17}\cos 2t+\frac{40}{17}\sin 2t.
$$

$$
\textbf{Step 3: Write the General Solution}
$$

The general solution $$ x(t) $$ is the sum of the homogeneous and particular solutions:
$$
x(t)=x_h(t)+x_p(t)=e^{-t}\left(C_1 \cos 2t + C_2 \sin 2t\right)+\frac{10}{17}\cos 2t+\frac{40}{17}\sin 2t.
$$
The general solution for $$ x(t) $$ is: $$ x(t) = e^{-t}(C_1 \cos 2t + C_2 \sin 2t) + \frac{10}{17} \cos 2t + \frac{40}{17} \sin 2t $$ where $$ C_1 $$ and $$ C_2 $$ are constants.

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