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 Bryant Pearson. 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 Bryant Pearson. in Brunei

Mar 22,2025

UpStudy AI · Accepted Answer

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

We first consider the homogeneous equation associated with the given differential equation:
$$
\frac{d^{2} x}{d t^{2}} + 2 \frac{d x}{d t} + 5 x = 0.
$$
Assume a solution of the form $$ x(t) = e^{rt} $$. Substituting into the homogeneous equation gives:
$$
r^2 e^{rt} + 2r e^{rt} + 5e^{rt} = 0.
$$
Dividing through by $$ e^{rt} $$ (which is never zero) yields the characteristic equation:
$$
r^2 + 2r + 5 = 0.
$$
Solve the quadratic using the quadratic formula:
$$
r = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(5)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i.
$$
Thus, the general solution of the homogeneous equation is
$$
x_h(t) = e^{-t} \Bigl(C_1 \cos 2t + C_2 \sin 2t \Bigr),
$$
where $$ C_1 $$ and $$ C_2 $$ are arbitrary constants.

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

We now seek a particular solution $$ x_p(t) $$ to the non-homogeneous differential equation
$$
\frac{d^{2} x}{d t^{2}} + 2 \frac{d x}{d t} + 5 x = 10 \cos 2t.
$$
Since the forcing term is $$ 10 \cos 2t $$ and the homogeneous solution is multiplied by $$ e^{-t} $$, there is no resonance. Therefore, we propose a particular solution of the form:
$$
x_p(t) = A \cos 2t + B \sin 2t.
$$
Differentiate $$ x_p(t) $$:
$$
x_p'(t) = -2A \sin 2t + 2B \cos 2t,
$$
$$
x_p''(t) = -4A \cos 2t - 4B \sin 2t.
$$
Substitute $$ x_p(t) $$, $$ x_p'(t) $$, and $$ x_p''(t) $$ into the differential equation:
$$
(-4A \cos 2t - 4B \sin 2t) + 2(-2A \sin 2t + 2B \cos 2t) + 5(A \cos 2t + B \sin 2t).
$$
Simplify the expression by grouping like terms.

For the $$ \cos 2t $$ terms:
$$
-4A \cos 2t + 4B \cos 2t + 5A \cos 2t = ( -4A + 4B + 5A ) \cos 2t = ( A + 4B ) \cos 2t.
$$

For the $$ \sin 2t $$ terms:
$$
-4B \sin 2t - 4A \sin 2t + 5B \sin 2t = ( -4B - 4A + 5B ) \sin 2t = ( B - 4A ) \sin 2t.
$$

Thus, the left-hand side becomes:
$$
( A + 4B ) \cos 2t + ( B - 4A ) \sin 2t.
$$
This must equal the right-hand side $$ 10 \cos 2t $$. Therefore, we equate the coefficients:
$$
A + 4B = 10,
$$
$$
B - 4A = 0.
$$

Solve the second equation first:
$$
B = 4A.
$$
Substitute $$ B = 4A $$ into the first equation:
$$
A + 4(4A) = A + 16A = 17A = 10 \quad \Longrightarrow \quad A = \frac{10}{17}.
$$
Then,
$$
B = 4A = \frac{40}{17}.
$$
Thus, 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 of the differential equation is the sum of the homogeneous and particular solutions:
$$
x(t) = x_h(t) + x_p(t) = e^{-t} \Bigl(C_1 \cos 2t + C_2 \sin 2t \Bigr) + \frac{10}{17} \cos 2t + \frac{40}{17} \sin 2t.
$$
The general solution is: $$ x(t) = e^{-t} (C_1 \cos 2t + C_2 \sin 2t) + \frac{10}{17} \cos 2t + \frac{40}{17} \sin 2t. $$

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