Q7 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.

Shock absorbers in automobiles and aircraft can be described as forced overdamped mass-spring systems. Derive an expression analogous to equation (8) for the general solution to the differential equation (1) when $b^{2} > 4 mk$ .

Step-by-Step Solution

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Answer

Therefore, the general solution to the differential equation is:

$y (t) = c_{1} e^{(- \frac{b}{2 m} + \frac{1}{2 m} \sqrt{b^{2} - 4 mk}) t} + c_{2} e^{(- \frac{b}{2 m} - \frac{1}{2 m} \sqrt{b^{2} - 4 mk}) t} + \frac{F_{0}}{{(k - m γ^{2})}^{2} + b^{2} γ^{2}} [(k - m γ^{2}) \cos (γ t) + b γ \sin (γ t)]$

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 of the forcing function and it is given by $A (γ) = F_{0} M (γ)$ , where

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

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. And the corresponding homogeneous equation is $y_{h} (t) = Asin (ω t + ϕ), ω : = \sqrt{\frac{k}{m}}$

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

Referring to equation (8): one gets,

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

In the overdamped motion case, the auxiliary equation has two real roots.

$r_{1} = - \frac{b}{2 m} + \frac{1}{2 m} \sqrt{b^{2} - 4 mk} r_{2} = - \frac{b}{2 m} - \frac{1}{2 m} \sqrt{b^{2} - 4 mk}$

Since $b^{2} > 4 mk$ the homogeneous solution to $m \frac{d^{2} y}{{dt}^{2}} + b \frac{dy}{dt} + ky = F_{0} \cos (γ t)$ and it becomes:

$y_{h} (t) = c_{1} e^{r_{1} t} + c_{2} e^{r_{2} t} y_{p} (t) = \frac{F_{0}}{{(k - m γ^{2})}^{2} + b^{2} γ^{2}} [(k - m γ^{2}) \cos (γ t) + b γ \sin (γ t)]$

Then, the general solution is;

Q6 E

Q8 E

Other exercises in this chapter

Q5 E

In the following problems, take g=32ft/sec2 for the U.S. Customary System and g=9.8m/sec2for the MKS system.An undamped system is governed bymd2ydt2+ky=F0c

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Q6 E

In the following problems, take g=32ft/sec2 for the U.S. Customary System and g=9.8m/sec2 the MKS system.Derive the formula for given in (21).

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Q8 E

In the following problems, take g=32ft/sec2 for the U.S. Customary System and g=9.8m/sec2 for the MKS system.The response of an overdamped system to a

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Q10 E

In the following problems, take g=32ft/sec2 for the U.S. Customary System and g=9.8m/sec2for the MKS system.Show that the period of the simple harmonic mo

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