Q15 E

Question

In the following problems, take $g = 32 ft / \sec^{2}$ for the U.S. Customary System and $g = 9.8 m / \sec^{2}$ for the MKS system.

An 8-kg mass is attached to a spring hanging from the ceiling and allowed to come to rest. Assume that the spring constant is 40 N/m and the damping constant is 3 N/sec. At time t = 0, an external force $2 \sin (2 t) \cos (2 t)$ N is applied to the system. Determine the amplitude and frequency of the steady-state solution.

Step-by-Step Solution

Verified

Answer

Therefore, the frequency and amplitude of the steady–state solution is $\frac{2}{π}$ and $\frac{\sqrt{493}}{1972}$

1Step 1: General form

The general solution to (1) in the case $0 < b^{2} < 4 mk$ :

$y (t) = {Ae}^{- (\frac{b}{2 m}) t} \sin (\frac{\sqrt{4 mk - b^{2}}}{2 m} t + ϕ) + \frac{F_{0}}{\sqrt{{(k - m γ^{2})}^{2} + b^{2} γ^{2}}} \sin (γ t + θ)$

The angular frequency:

The amplitude of the steady-state solution to equation (1) depends on the angular frequency $γ$ of the forcing function and it is given by $A (γ) = F_{0} M (γ)$ , where

(13) $M (γ) : = \frac{1}{\sqrt{{(k - m γ^{2})}^{2} + b^{2} γ^{2}}}$ … (1)

The undamped system:

The system is governed by $m \frac{d^{2} y}{{dt}^{2}} + ky = F_{0} \cos γ t$ . And the homogenous solution of it is $y_{h} (t) = Asin (ω t + ϕ), ω : = \sqrt{\frac{k}{m}}$ :

And the corresponding homogeneous equation is $y_{p} (t) = \frac{F_{0}}{2 m ω} tsin ω t$ :

(21). And the correct form is $y_{p} (t) = A_{1} tcos ω t + A_{2} tsin ω t$ .

So, the general solution of the system is $y (t) = Asin (ω t + ϕ) + \frac{F_{0}}{2 m ω} tsin ω t$ .

2Step 2: Evaluate the equation

Given that, $m = 8$ . And $F = 2 \sin (2 t) \cos (2 t) = \sin (4 t)$ .

So, $F_{0} = 1$ and $γ = 4$ . Since, $k = 40$ and $b = 3$ .

Then, the differential equation is $8 \frac{d^{2} y}{{dt}^{2}} + 3 \frac{dy}{dt} + 40 y = \sin (4 t)$ … (2)

Since one needs to find the steady state solution of the equation. That is all we are looking only for $y_{p} (t)$ . Then,

$y_{p} (t) = A_{1} \cos 4 t + A_{2} \sin 4 t$ … (3)

Now find the derivative of equation (3).

$y_{p}^{'} (t) = - 4 A_{1} \sin 4 t + 4 A_{2} \cos 4 t y_{p}^{''} (t) = - 16 A_{1} \cos 4 t - 16 A_{2} \sin 4 t$

Substitute the values in equation (2).

$8 (- 16 A_{1} \cos 4 t - 16 A_{2} \sin 4 t) + 3 (- 4 A_{1} \sin 4 t + 4 A_{2} \cos 4 t) + 40 (A_{1} \cos 4 t + A_{2} \sin 4 t) = \sin (4 t) (- 128 A_{1} \cos 4 t - 128 A_{2} \sin 4 t) + (- 12 A_{1} \sin 4 t + 12 A_{2} \cos 4 t) + (40 A_{1} \cos 4 t + 40 A_{2} \sin 4 t) = \sin (4 t) (- 88 A_{1} + 12 A_{2}) \cos (4 t) + (- 12 A_{1} - 88 A_{2}) \sin (4 t) = \sin (4 t)$

Now equalize the like terms.

$- 88 A_{1} + 12 A_{2} = 0 A_{2} = \frac{22}{3} A_{1} - 12 A_{1} - 88 A_{2} = 1 - 12 A_{1} - 88 (\frac{22}{3} A_{1}) = 1 A_{1} = - \frac{3}{1972} A_{2} = - \frac{11}{986}$