m-d-2-x-dt-2-f-k-dx-dt-x-0-x-0-x-0-v-0-x-t-

Question Number 4800 by 123456 last updated on 13/Mar/16

m \frac{d^{2} x}{{d t}^{2}} = f - k \frac{d x}{d t}

x (0) = x_{0}

x^{'} (0) = v_{0}

x (t) = ?

Answered by Yozzii last updated on 13/Mar/16

m \overset{. .}{x} + k \dot{x} - f = 0

A u x i l i a r y q u a d r a t i c e q u a t i o n l e a d s t o

{m p}^{2} + k p - f = 0

\Rightarrow p = \frac{- k \pm \sqrt{k^{2} - 4 m (- f)}}{2 m}

p = \frac{- k \pm \sqrt{k^{2} + 4 m f}}{2 m}

I f k^{2} + 4 m f > 0 \Rightarrow p h a s t w o d i s t i n c t v a l u e s .

S o, t h e g e n e r a l s o l u t i o n i s g i v e n b y

x (t) = {A e}^{p_{1} t} + {B e}^{p_{2} t}

w h e r e p_{1} = \frac{\sqrt{k^{2} + 4 m f} - k}{2 m},

p_{2} = \frac{- k - \sqrt{k^{2} + 4 m f}}{2 m},

A a n d B a r e c o n s t a n t s .

S i n c e x (0) = x_{0} \Rightarrow x_{0} = A + B

\Rightarrow A = x_{0} - B \dots \dots . (1)

D i f f e r e n t i a t i n g x (t) w . r . t t g i v e s

x^{^{'}} (t) = {A p}_{1} e^{p_{1} t} + {B p}_{2} e^{p_{2} t}

N o w, x^{^{'}} (0) = v_{0} .

∴ v_{0} = {A p}_{1} + {B p}_{2} \dots \dots \dots (2)

S u b s t i t u t i n g A f r o m (1) i n t o (2) y i e l d s

v_{0} = p_{1} (x_{0} - B) + {B p}_{2}

v_{0} = p_{1} x_{0} + B (p_{2} - p_{1})

\Rightarrow B = \frac{v_{0} - p_{1} x_{0}}{p_{2} - p_{1}}

N o t e t h a t p_{2} - p_{1} = \frac{- \sqrt{k^{2} + 4 m f}}{m} \neq 0 u n l e s s

k^{2} = - 4 m f \Rightarrow m f ⩽ 0 \Rightarrow k = m = f = 0

o r k = \pm 2 \sqrt{- m f} w h e r e m f < 0 . m \neq 0

s i n c e t h e n p_{2} - p_{1} i s u n d e f i n e d .

S o k = 0 i f f f = 0 a n d m \neq 0 .

∴ A = x_{0} - \frac{v_{0} - p_{1} x_{0}}{p_{2} - p_{1}} = \frac{x_{0} p_{2} - v_{0}}{p_{2} - p_{1}} .

∴ x (t) = {\frac{x_{0} p_{2} - v_{0}}{p_{2} - p_{1}}} e^{p_{1} t} + {\frac{v_{0} - p_{1} x_{0}}{p_{2} - p_{1}}} e^{p_{2} t}

o r x (t) = \frac{m (v_{0} - p_{2} x_{0}) e^{p_{1} t}}{\sqrt{k^{2} + 4 m f}} + \frac{m (p_{1} x_{0} - v_{0}) e^{p_{2} t}}{\sqrt{k^{2} + 4 m f}}

o r x (t) = \frac{1}{\sqrt{k^{2} + 4 m f}} ({\frac{2 v_{0} m + x_{0} (k + \sqrt{k^{2} + 4 m f})}{2}} e^{t (\frac{\sqrt{k^{2} + 4 m f} - k}{2 m})} + {\frac{x_{0} (\sqrt{k^{2} + 4 m f} - k) - 2 {m v}_{0}}{2}} e^{- t (\frac{k + \sqrt{k^{2} + 4 m f}}{2 m})})

C h e c k i n g :

x (0) = \frac{x_{0} p_{2} - v_{0}}{p_{2} - p_{1}} + \frac{v_{0} - p_{1} x_{0}}{p_{2} - p_{1}} = \frac{x_{0} (p_{2} - p_{1})}{p_{2} - p_{1}} = x_{0}

x^{^{'}} (0) = \frac{1}{p_{2} - p_{1}} (p_{2} p_{1} x_{0} - p_{1} v_{0} + p_{2} v_{0} - p_{1} p_{2} x_{0}) = v_{0}

Commented by Yozzii last updated on 13/Mar/16

I f k^{2} + 4 m f = 0 \Rightarrow a r e p e a t e d r o o t f o r

t h e a u x i l i a r y q u a d r a t i c e q u a t i o n e x i s t s

a n d i s g i v e n b y p = \frac{- k}{2 m} (m \neq 0) . T h e g e n e r a l

s o l u t i o n f o r x (t) i s t h e r e f o r e

x (t) = (A t + B) e^{p t}

W e a r e g i v e n t h a t x (0) = x_{0} a n d x^{^{'}} (0) = v_{0} .

∴ x_{0} = 0 + B \Rightarrow B = x_{0} .

D i f f e r e n t i a t i n g x (t) w . r . t t

\Rightarrow x^{^{'}} (t) = {p e}^{p t} (A t + B) + {A e}^{p t}

∴ v_{0} = p B + A \Rightarrow A = v_{0} - p B = v_{0} + \frac{{k x}_{0}}{2 m}

∴ x (t) = {(v_{0} + \frac{{k x}_{0}}{2 m}) t + x_{0}} e^{- k t / 2 m} (m \neq 0) .

Commented by Yozzii last updated on 13/Mar/16

I f k^{2} + 4 m f < 0 w e g e t t h e g e n e r a l

s o l u t i o n o f x (t) t o b e

x (t) = e^{- k t / 2 m} (A c o s \frac{t \sqrt{- (k^{2} + 4 m f)}}{2 m} + B s i n \frac{t \sqrt{- (k^{2} + 4 m f)}}{2 m}) .

S i n c e x (0) = x_{0}

\Rightarrow x_{0} = (1) (A \times 1 + 0) \Rightarrow A = x_{0} .

A l s o, x^{^{'}} (0) = v_{0} .

x^{^{'}} (t) = \frac{- k}{2 m} e^{- k t / 2 m} (x_{0} c o s \frac{t \sqrt{- (k^{2} + 4 m f)}}{2 m} + B s i n \frac{t \sqrt{- k^{2} - 4 m f}}{2 m})

+ e^{- k t / 2 m} (- x_{0} \frac{\sqrt{- k^{2} - 4 m f}}{2 m} s i n \frac{t \sqrt{- k^{2} - 4 m f}}{2 m} + B \frac{\sqrt{- k^{2} - 4 m f}}{2 m} c o s \frac{t \sqrt{- k^{2} - 4 m f}}{2 m})

∴ v_{0} = \frac{- {k x}_{0}}{2 m} + \frac{B \sqrt{- k^{2} - 4 m f}}{2 m}

\Rightarrow B = \frac{2 {m v}_{0} + {k x}_{0}}{\sqrt{- k^{2} - 4 m f}}

∴ x (t) = e^{- k t / 2 m} (x_{0} c o s \frac{t \sqrt{- k^{2} - 4 m f}}{2 m} + \frac{2 {m v}_{0} + {k x}_{0}}{\sqrt{- k^{2} - 4 m f}} s i n \frac{t \sqrt{- k^{2} - 4 m f}}{2 m}) (m \neq 0) .

Answered by Dnilka228 last updated on 14/Mar/16

f (0) \neq \cos f

W h y ?

f + {(\frac{e}{2})}^{\sqrt{f}} = \tan e_{α} + α_{e}

\cos f = e + 2 \times 2 n

n = n

n = \sqrt{45}

\sqrt{45} + \frac{n}{e} = 0 %

b u t

(5 - 2) + (2 - 5) \div 0 = \infty

\infty = \infty

0 \neq \infty

\sqrt{x} i s x \div \cos f

α + β t o o \neq \infty

\frac{Φ e}{φ e} + \cos f = \frac{c o s f (1) \div c o s f (2)}{D}

Commented by FilupSmith last updated on 16/Mar/16

\infty \neq \infty

Proof :

a = {0, 1, 2, 3, \dots, n}, n \in N + {0}, 0 ⩽ n ⩽ \infty

b = {0, 0.01, 0.001, \dots, 0.2, 0.02, \dots, 1, \dots, r}, r \in Z, 0 ⩽ r ⩽ \infty

Each set contains all whole numbers ⩾ 0

∴ a \subseteq b

It is a clear observation that ∣ b ∣>∣ a ∣

So, assuming each are infinite, b is still

bigger .

Commented by FilupSmith last updated on 16/Mar/16

You wrote :

(5 - 2) + (2 - 5) \div 0 = \infty

t h i s i s i n c o r r e c t

3 - \frac{3}{0} \neq \infty

3 \lim_{x \to \infty} (1 - \frac{1}{x}) = L

A^{+} = \lim_{x \to 0 +} (1 - \frac{1}{x})

A^{-} = \lim_{x \to 0 -} (1 - \frac{1}{x})

A^{+} = 1 - \infty

A^{+} = - \infty

A^{-} = 1 - (- \infty)

A^{-} = + \infty

∴ A^{\pm} = \mp \infty

L^{\pm} = 3 A^{\pm}

∴ L^{\pm} = \mp \infty