Q14 E

Question

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 $2 \sin (2 t + \frac{π}{4})$ 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 $\frac{1}{π}$ and 0.2.

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 y 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

(21) $y_{p} (t) = \frac{F_{0}}{2 m ω} tsin ω t$ . 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 $F = 2 \sin (2 t + \frac{π}{4})$ .

So, $F_{0} = 2$ and $γ = 2$ . Since, $k = 40$ and $b = 3$ .

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

Simplify the external force,

$F = 2 \sin (2 t + \frac{π}{4}) = 2 (\sin (2 t) \cos (\frac{π}{4}) + \cos (2 t) \sin (\frac{π}{4})) = 2 (\frac{\sqrt{2}}{2} \sin (2 t) + \frac{\sqrt{2}}{2} \cos (2 t)) = \sqrt{2} \sin (2 t) + \sqrt{2} \cos (2 t)$

Substitute the values in equation (2).

$8 \frac{d^{2} y}{{dt}^{2}} + 3 \frac{dy}{dt} + 40 y = \sqrt{2} \sin (2 t) + \sqrt{2} \cos (2 t)$ … (3)

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

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

Now find the derivative of equation (3).

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

Substitute the values in equation (2).

$8 (- 4 A_{1} \cos 2 t - 4 A_{2} \sin 2 t) + 3 (- 2 A_{1} \sin 2 t + 2 A_{2} \cos 2 t) + 40 (A_{1} \cos 2 t + A_{2} \sin 2 t) = \sqrt{2} \sin (2 t) + \sqrt{2} \cos (2 t) (- 32 A_{1} \cos 2 t - 32 A_{2} \sin 2 t) + (- 6 A_{1} \sin 2 t + 6 A_{2} \cos 2 t) + (40 A_{1} \cos 2 t + 40 A_{2} \sin 2 t) = \sqrt{2} \sin (2 t) + \sqrt{2} \cos (2 t) (8 A_{1} + 6 A_{2}) \cos (2 t) + (- 6 A_{1} + 8 A_{2}) \sin (2 t) = \sqrt{2} \sin (2 t) + \sqrt{2} \cos (2 t)$