Q44E

Question

A mass-spring system is driven by an external force $g (t) = 2 \sin 3 t + 10 \cos 3 t$ . The mass equals 1, the spring constant equals 5, and the damping coefficient equals 2. If the mass is initially located at $y (0) = - 1$ , with an initial velocity $y' (0) = 5$ , find its equation of motion.

Step-by-Step Solution

Verified

Answer

The equation of the motion is $y = e^{- t} \sin (2 t) + \sin (3 t) - \cos (3 t) .$

1Step 1: Use the given information for writing the differential equation.

Given that,

The mass equals m = 1,

The spring constant equals c =5,

And the damping coefficient equals b =2.

The differential equation is,

$\begin{matrix} my'' + by' + cy = g (t) \\ y'' + 2 y' + 5 y = 2 \sin 3 t + 10 \cos 3 t . ..... (1) \end{matrix}$

Write the homogeneous differential equation of the equation (1),

$y'' + 2 y' + 5 y = 0$

2Step 2: Find the complementary solution of the given equation.

The auxiliary equation for the above equation,

$m^{2} + 2 m + 5 = 0$

Solve the auxiliary equation,

$\begin{matrix} m^{2} + 2 m + 5 = 0 \\ m = \frac{- 2 \pm \sqrt{4 - 20}}{2} \\ m = \frac{- 2 \pm \sqrt{- 16}}{2} \\ m = - 1 \pm 2 i \end{matrix}$

The roots of the auxiliary equation are,

$m_{1} = - 1 + 2 i, m_{2} = - 1 - 2 i$

The complementary solution of the given equation is,

$y_{c} = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t)$

3Step 3: Find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

$y_{p} (t) = Asin (3 t) + Bcos (3 t) . ..... (2)$

Now find the first and second derivatives of the above equation,

$\begin{matrix} y_{p}' (t) = 3 Acos (3 t) - 3 Bsin (3 t) \\ y_{p}'' (t) = - 9 Asin (3 t) - 9 Bcos (3 t) \end{matrix}$

Substitute the value of $y_{p} (t), y_{p}' (t)$ and $y_{p}'' (t)$ in the equation (1),

$\begin{matrix} y'' + 2 y' + 5 y = 2 \sin 3 t + 10 \cos 3 t \\ - 9 Asin (3 t) - 9 Bcos (3 t) + 2 [3 Acos (3 t) - 3 Bsin (3 t)] + 5 [Asin (3 t) + Bcos (3 t)] = 2 \sin 3 t + 10 \cos 3 t \\ (- 4 A - 6 B) \sin (3 t) + (- 4 B + 6 A) \cos (3 t) = 2 \sin 3 t + 10 \cos 3 t \end{matrix}$

Comparing all coefficients of the above equation,

$\begin{matrix} - 4 A - 6 B = 2 . ...... (3) \\ - 4 B + 6 A = 10 . ..... (4) \end{matrix}$

Solve the above equations,

$\begin{matrix} 3 (- 4 A - 6 B) = 2 \times 3 \\ 2 (- 4 B + 6 A) = 10 \times 2 \end{matrix}$

$\begin{array}{l} - 12 A - 18 B = 6 \\ 12 A - 8 B = 20 \\ \bar{\begin{array}{l} - 26 B = 26 \\ B = - 1 \end{array}} \end{array}$

Substitute the value of B in the equation (3),

$\begin{matrix} - 4 A - 6 (- 1) = 2 \\ A = 1 \end{matrix}$

Substitute the value of A and B in the equation (2),

$\begin{matrix} y_{p} (t) = Asin (3 t) + Bcos (3 t) \\ y_{p} (t) = \sin (3 t) - \cos (3 t) \end{matrix}$

Therefore, the particular solution of equation (1),

$y_{p} (t) = \sin (3 t) - \cos (3 t)$

4Step 4: Find the general solution and use the given initial condition.

Therefore, the general solution is,

$\begin{matrix} y = y_{c} (t) + y_{p} (t) \\ y = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t) + \sin (3 t) - \cos (3 t) . ..... (5) \end{matrix}$

Given the initial condition,

$y (0) = - 1, y' (0) = 5$

Substitute the value of y = -1 and t = 0 in the equation (3),

$\begin{matrix} y = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t) + \sin (3 t) - \cos (3 t) \\ - 1 = c_{1} e^{- (0)} \cos (2 (0)) + c_{2} e^{- (0)} \sin (2 (0)) + \sin (3 (0)) - \cos (3 (0)) \\ - 1 = c_{1} - 1 \\ c_{1} = 0 \end{matrix}$

Now find the derivative of the above equation,

$y' = c_{1} [- 2 e^{- t} \sin (2 t) - e^{- t} \cos (2 t)] + c_{2} [2 e^{- t} \cos (2 t) - e^{- t} \sin (2 t)] + 3 \cos (3 t) + 3 \sin (3 t)$

Substitute the value of y’ = 5 and t = 0 in the above equation,

$\begin{matrix} y' = c_{1} [- 2 e^{- t} \sin (2 t) - e^{- t} \cos (2 t)] + c_{2} [2 e^{- t} \cos (2 t) - e^{- t} \sin (2 t)] + 3 \cos (3 t) + 3 \sin (3 t) \\ 5 = c_{1} [- 2 e^{- (0)} \sin (2 (0)) - e^{- (0)} \cos (2 (0))] + c_{2} [2 e^{- (0)} \cos (2 (0)) - e^{- (0)} \sin (2 (0))] \\ + 3 \cos (3 (0)) + 3 \sin (3 (0)) \\ 5 = c_{1} [- 1] + c_{2} [2] + 3 \\ - c_{1} + 2 c_{2} = 2 . ...... (6) \end{matrix}$

Substitute the value of $c_{1}$ in the equation (6),

$\begin{matrix} - c_{1} + 2 c_{2} = 2 \\ 0 + 2 c_{2} = 2 \\ c_{2} = 1 \end{matrix}$

Substitute the value of $c_{1}$ and $c_{2}$ in the equation (5),

$\begin{matrix} y = c_{1} e^{- t} \cos (2 t) + c_{2} e^{- t} \sin (2 t) + \sin (3 t) - \cos (3 t) \\ y = (0) e^{- t} \cos (2 t) + (1) e^{- t} \sin (2 t) + \sin (3 t) - \cos (3 t) \\ y = e^{- t} \sin (2 t) + \sin (3 t) - \cos (3 t) \end{matrix}$

Thus, the equation of motion is:

$y = e^{- t} \sin (2 t) + \sin (3 t) - \cos (3 t)$

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