hcju/proofwikiqa huggingface dataset:
Note that: :$\paren {x^2 + x} + \paren {2 - x^2} - \paren {2 + x} = 0$ so $\map {y_1} x$ such that $y_1 = {y_1}' = {y_1}''$ satisfies $(1)$. Hence: :$y_1 = e^x$ is a particular solution of $(1)$. $(1)$ can be expressed as: :$(2): \quad y'' + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = 0$ which is in the form: :$y'' + \map P x y' + \map Q x y = 0$ where: :$\map P x = \dfrac {2 - x^2} {x^2 + x}$ :$\map Q x = \dfrac {2 + x} {x^2 + x}$ From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another: :$\map {y_2} x = \map v x \, \map {y_1} x$ where: :$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$ is also a particular solution of $(1)$. We have that: {{begin-eqn}} {{eqn | l = \int P \rd x       | r = \int \paren {\dfrac {2 - x^2} {x^2 + x} } \rd x       | c =  }} {{eqn | r = \int \dfrac 2 {x \paren {x + 1} } \rd x - \int \paren {\dfrac x {x + 1} } \rd x       | c =  }} {{eqn | r = 2 \, \map \ln {\frac x {x + 1} } - \int \paren {\dfrac x {x + 1} } \rd x       | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$ }} {{eqn | r = 2 \, \map \ln {\frac x {x + 1} } - x + \map \ln {x + 1}       | c = Primitive of $\dfrac x {a x + b}$ }} {{eqn | r = \map \ln {\frac {x^2} {x + 1} } - x       | c =  }} {{eqn | r = \map \ln {\frac {e^{-x} x^2} {x + 1} }       | c =  }} {{eqn | r = -\map \ln {\frac {e^x \paren {x + 1} } {x^2} }       | c =  }} {{eqn | ll= \leadsto       | l = e^{-\int P \rd x}       | r = e^{\map \ln {\frac {e^x \paren {x + 1} } {x^2} } }       | c =  }} {{eqn | r = \frac {e^x \paren {x + 1} } {x^2}       | c =   }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = v       | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x       | c = Definition of $v$ }} {{eqn | r = \int e^{-2 x} \paren {\frac {e^x \paren {x + 1} } {x^2} } \rd x       | c =  }} {{eqn | r = \int \frac {e^{-x} \paren {x + 1} } {x^2} \rd x       | c =  }} {{eqn | r = \int \frac {e^{-x} } x \rd x + \int \frac {e^{-x} } {x^2} \rd x       | c =  }} {{eqn | r = \int \frac {e^{-x} } x \rd x + \frac {-e^{-x} } x - \int \frac {e^{-x} } x \rd x       | c = Primitive of $\dfrac {e^{a x} } {x^n}$ }} {{eqn | r = -\frac {e^{-x} } x       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = y_2       | r = v y_1       | c = Definition of $y_2$ }} {{eqn | r = \paren {-\frac {e^{-x} } x} e^x       | c =  }} {{eqn | r = -\frac 1 x       | c =  }} {{end-eqn}} From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution: :$y = C_1 e^x + \dfrac {C_2} x$ {{qed}}
By Newton's Second Law of Motion, the force on $C$ equals its mass times its acceleration: :$\mathbf F = m \mathbf a$ By Acceleration is Second Derivative of Displacement with respect to Time: :$\mathbf a = \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}$ By Hooke's Law: :$\mathbf F = -k \mathbf x$ So: {{begin-eqn}} {{eqn | l = m \mathbf a       | r = -k \mathbf x       | c =  }} {{eqn | ll= \implies       | l = m \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}       | r = -k \mathbf x       | c =  }} {{eqn | ll= \implies       | l = \dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + \dfrac k m \mathbf x       | r = 0       | c =  }} {{end-eqn}} {{qed}}
From Motion of Cart attached to Wall by Spring, the horizontal position of $C$ is given as: :$\dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac k m \mathbf x = 0$ As both $k$ and $m$ are positive, $\dfrac k m$ can be expressed as: :$\dfrac k m = \alpha^2$ for some $\alpha \in \R_{>0}$. Hence: :$\dfrac {\d^2 x} {\d t^2} + \alpha^2 x = 0$ and so from Linear Second Order ODE: $y'' + k^2 y = 0$: :$x = C_1 \cos \alpha t + C_2 \sin \alpha t$ Hence the result. {{qed}}
From Position of Cart attached to Wall by Spring, the horizontal position of $C$ is given as: :$(1): \quad x = C_1 \cos \alpha t + C_2 \sin \alpha t$ where: :$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$ :$\alpha = \sqrt {\dfrac k m}$ Differentiating $(1)$ {{WRT}} $x$ gives: :$(2): \quad x' = -C_1 \alpha \sin \alpha t + C_2 \alpha \cos \alpha t$ Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = x_0       | r = C_1 \cos 0 + C_2 \sin 0       | c =  }} {{eqn | ll= \implies       | l = x_0       | r = C_1       | c =  }} {{end-eqn}} Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -C_1 \alpha \sin 0 + C_2 \alpha \cos 0       | c =  }} {{eqn | ll= \implies       | l = 0       | r = C_2       | c =  }} {{end-eqn}} Hence: :$x = x_0 \cos \alpha t$ where $\alpha = \sqrt {\dfrac k m}$. {{qed}}
From Position of Cart attached to Wall by Spring, the horizontal position of $C$ is given as: :$(1): \quad x = C_1 \cos \alpha t + C_2 \sin \alpha t$ where: :$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$ :$\alpha = \sqrt {\dfrac k m}$ From the definition of frequency: {{begin-eqn}} {{eqn | l = \nu       | r = \dfrac \alpha {2 \pi}       | c =  }} {{eqn | r = \dfrac 1 {2 \pi} \sqrt {\dfrac k m}       | c =  }} {{eqn | r = Q \sqrt k       | c = where $Q = \dfrac 1 {2 \pi \sqrt m}$ }} {{end-eqn}} From Square Root is Strictly Increasing, the square root is strictly increasing on positive real numbers. Hence, if the mass is constant (as is implicit in the statement of the problem), increasing $k$ causes an increase in $\nu$. {{qed}}
From Position of Cart attached to Wall by Spring, the horizontal position of $C$ is given as: :$(1): \quad x = C_1 \cos \alpha t + C_2 \sin \alpha t$ where: :$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$ :$\alpha = \sqrt {\dfrac k m}$ From the definition of frequency: {{begin-eqn}} {{eqn | l = \nu       | r = \dfrac \alpha {2 \pi}       | c =  }} {{eqn | r = \dfrac 1 {2 \pi} \sqrt {\dfrac k m}       | c =  }} {{eqn | r = Q \dfrac 1 {\sqrt m}       | c = where $Q = \dfrac {\sqrt k} {2 \pi}$ }} {{end-eqn}} From Square Root is Strictly Increasing, the square root is strictly increasing on positive real numbers. From Reciprocal Function is Strictly Decreasing, the reciprocal function is strictly decreasing on positive real numbers. Hence, if the force constant of $S$ is constant (as is implicit in the statement of the problem), increasing $m$ causes a decrease in $\nu$. {{qed}}
Let $\mathbf F_s$ be the force on $C$ exerted by the spring $S$. Let $\mathbf F_d$ be the damping force. By Newton's Second Law of Motion, the total force on $C$ equals its mass times its acceleration: :$\mathbf F_s + \mathbf F_d = m \mathbf a$ By Acceleration is Second Derivative of Displacement with respect to Time: :$\mathbf a = \dfrac {\d^2 \mathbf x} {\d t^2}$ By definition, the velocity $\mathbf v$ is defined as: :$\mathbf v = \dfrac {\d \mathbf x} {\d t}$ By Hooke's Law: :$\mathbf F = -k \mathbf x$ By definition of the damping force: :$\mathbf F_d = - c \mathbf v$ So: {{begin-eqn}} {{eqn | l = m \mathbf a       | r = -k \mathbf x - c \mathbf v       | c =  }} {{eqn | ll= \leadsto       | l = m \dfrac {\d^2 \mathbf x} {\d t^2}       | r = -k \mathbf x - c \dfrac {\d \mathbf x} {\d t}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac c m \dfrac {\d \mathbf x} {\d t} + \dfrac k m \mathbf x       | r = 0       | c =  }} {{end-eqn}} {{qed}}
{{:Position of Cart attached to Wall by Spring under Damping/Preliminary}} === Overdamped === {{:Position of Cart attached to Wall by Spring under Damping/Overdamped}}{{qed|lemma}} === Critically Damped === {{:Position of Cart attached to Wall by Spring under Damping/Critically Damped}}{{qed|lemma}} === Underdamped === {{:Position of Cart attached to Wall by Spring under Damping/Underdamped}}{{qed}}
=== Sufficient Condition === Let us assume that :$H$ is directed. Let $x, y \in H$. By definition of directed: :$\exists z \in H: x \preceq z \land y \preceq z$ By definition :$z$ is upper bound of $\left\{ {x, y}\right\}$ By definitions of supremum and join: :$x \vee y = \sup\left\{ {x, y}\right\} \preceq z$ Thus by definition of lower set: :$x \vee y \in H$ {{qed|lemma}} === Necessary Condition === Let us assume that :$\forall x, y \in H: x \vee y \in H$ Let $x, y \in H$. By assumption: :$x \vee y \in H$ By definition of supremum: :$x \vee y$ is upper bound of $\left\{ {x, y}\right\}$ Thus :$x \preceq x \vee y \land y \preceq x \vee y$ Thus by definition :$H$ is directed. {{qed}}
From Motion of Cart attached to Wall by Spring under Damping, the horizontal position of $C$ is given as: :$\dfrac {\d^2 \mathbf x} {\d t^2} + \dfrac c m \dfrac {\d \mathbf x} {\d t} + \dfrac k m \mathbf x = 0$ With the given substitutions $a$ and $b$, this resolves to: :$\dfrac {\d^2 \mathbf x} {\d t^2} + 2 b \dfrac {\d \mathbf x} {\d t} + a^2 \mathbf x = 0$ This is a homogeneous linear second order ODE with constant coefficients. Recall that $m_1$ and $m_2$ are the roots of the auxiliary equation: :$m^2 + 2 b + a^2 = 0$ By Solution to Quadratic Equation with Real Coefficients: {{begin-eqn}} {{eqn | l = m_1, m_2       | r = \dfrac {- 2 b \pm \sqrt {\paren {2 b}^2 - 4 a^2} } 2       | c =  }} {{eqn | r = -b \pm \sqrt {b^2 - a^2}       | c =  }} {{end-eqn}} From the initial problem definition, we have that $k, m, c \in \R_{>0}$. Hence $a, b \in \R_{>0}$. Hence the nature of $m_1$ and $m_2$ is dependent upon whether $b > a$, $b = a$ or $b < a$.
When $b > a$, we have $b^2 - a^2 > 0$ and so $m_1$ and $m_2$ are real and distinct. So from Solution of Constant Coefficient Homogeneous LSOODE: Real Roots of Auxiliary Equation: :$\mathbf x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$ where {{begin-eqn}} {{eqn | l = m_1       | r = -b + \sqrt {b^2 - a^2}       | c =  }} {{eqn | l = m_2       | r = -b - \sqrt {b^2 - a^2}       | c =  }} {{end-eqn}}
When $b = a$, we have $b^2 - a^2 = 0$ and so: :$m_1 = m_2 = -b = -a$ So from Solution of Constant Coefficient Homogeneous LSOODE: Equal Real Roots of Auxiliary Equation: :$C_1 e^{-a x} + C_2 x e^{-a x}$
When $b < a$, we have $b^2 - a^2 < 0$ and so: {{begin-eqn}} {{eqn | l = m_1       | r = -b + i \sqrt {a^2 - b^2}       | c =  }} {{eqn | l = m_2       | r = -b - i \sqrt {a^2 - b^2}       | c =  }} {{end-eqn}} So from Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation: :$\mathbf x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right)$ where: :$\alpha = \sqrt {a^2 - b^2}$
From Position of Cart attached to Wall by Spring under Damping: Overdamped: :$(1): \quad x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$ where $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$. It remains to evaluate $C_1$ and $C_2$ under the given conditions. Differentiating $(1)$ {{WRT}} $t$ gives: :$(2): \quad x' = C_1 m_1 e^{m_1 t} + C_2 m_2 e^{m_2 t}$ Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = x_0       | r = C_1 e^0 + C_2 e^0       | c =  }} {{eqn | ll= \leadsto       | l = x_0       | r = C_1 + C_2       | c =  }} {{end-eqn}} Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = C_1 m_1 e^0 + C_2 m_2 e^0       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = C_1 m_1 + C_2 m_2       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = C_1 m_1 + \left({x_0 - C_1}\right) m_2       | c =  }} {{eqn | ll= \leadsto       | l = C_1 \left({m_1 - m_2}\right)       | r = -x_0 m_2       | c =  }} {{eqn | ll= \leadsto       | l = C_1 \left({m_1 - m_2}\right)       | r = -\dfrac {x_0 m_2} {m_1 - m_2}       | c =  }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = x_0       | r = C_1 + C_2       | c =  }} {{eqn | ll= \leadsto       | l = C_2       | r = x_0 - C_1       | c =  }} {{eqn | r = x_0 + \dfrac {x_0 m_2} {m_1 - m_2}       | c =  }} {{eqn | r = x_0 \left({1 + \dfrac {m_2} {m_1 - m_2} }\right)       | c =  }} {{eqn | r = x_0 \left({\dfrac {\left({m_1 - m_2}\right) + m_2} {m_1 - m_2} }\right)       | c =  }} {{eqn | r = \dfrac {x_0 m_1} {m_1 - m_2}       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = x       | r = -\dfrac {x_0 m_2} {m_1 - m_2} e^{m_1 t} + \dfrac {x_0 m_1} {m_1 - m_2} e^{m_2 t}       | c =  }} {{eqn | r = \dfrac {x_0} {m_1 - m_2} \left({m_1 e^{m_2 t} - m_2 e^{m_1 t} }\right)       | c =  }} {{end-eqn}} {{qed}}
From Position of Cart attached to Wall by Spring under Damping: Critically Damped: :$x = C_1 e^{-a t} + C_2 t e^{-a t}$ where $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$. It remains to evaluate $C_1$ and $C_2$ under the given conditions. Differentiating $(1)$ {{WRT}} $t$ gives: :$(2): \quad x' = -a C_1 e^{-a t} + C_2 e^{-a t} - a C_2 t e^{-a t}$ Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = x_0       | r = C_1 e^0 + C_2 e^0 \times 0       | c =  }} {{eqn | ll= \leadsto       | l = x_0       | r = C_1       | c =  }} {{end-eqn}} Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -a C_1 e^0 + C_2 e^0 - a C_2 e^0 \times 0       | c =  }} {{eqn | ll= \leadsto       | l = 0       | r = -a C_1 + C_2       | c =  }} {{eqn | ll= \leadsto       | l = C_2       | r = a C_1        | c =  }} {{eqn | r = a x_0       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = x       | r = x_0 e^{-a t} + a x_0 t e^{-a t}       | c =  }} {{eqn | r = x_0 e^{-a t} \left({1 + a t}\right)       | c =  }} {{end-eqn}} {{qed}}
From Position of Cart attached to Wall by Spring under Damping: Underdamped: :$(2): \quad x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right)$ where $C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$. It remains to evaluate $C_1$ and $C_2$ under the given conditions. Differentiating $(1)$ {{WRT}} $t$ gives: :$(2): \quad x' = -b e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) + e^{-b t} \left({-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}\right)$ Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = x_0       | r = e^0 \left({C_1 \cos 0 + C_2 \sin 0}\right)       | c =  }} {{eqn | r = C_1       | c =  }} {{end-eqn}} Setting the initial condition $x' = 0$ when $t = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = 0       | r = -b e^0 \left({C_1 \cos 0 + C_2 \sin 0}\right) + e^0 \left({-\alpha C_1 \sin 0 + \alpha C_2 \cos 0}\right)       | c =  }} {{eqn | r = -b C_1 + \alpha C_2       | c =  }} {{eqn | ll= \leadsto       | l = C_2       | r = \frac {b C_1} \alpha       | c =  }} {{eqn | r = \frac {b x_0} \alpha       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = x       | r = e^{-b t} \left({x_0 \cos \alpha t + \frac {b x_0} \alpha \sin \alpha t}\right)       | c =  }} {{eqn | r = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)       | c =  }} {{end-eqn}} {{qed}}
From Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation, the general solution of $(1)$ is: :$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$ where: :$\alpha = \sqrt {a^2 - b^2}$ This is a homogeneous linear second order ODE with constant coefficients. Let $m_1$ and $m_2$ be the roots of the auxiliary equation: :$m^2 + 2 b + a^2 = 0$ From Solution to Quadratic Equation with Real Coefficients: {{begin-eqn}} {{eqn | l = m_1       | r = -b + i \sqrt {a^2 - b^2}       | c =  }} {{eqn | l = m_1, m_2       | r = -b - i \sqrt {a^2 - b^2}       | c =  }} {{end-eqn}} So from Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation: :$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$ where: :$\alpha = \sqrt {a^2 - b^2}$ The following assumptions are made: :We may label a particular point in time $t = 0$ at which the derivative of $x$ {{WRT}} $t$ is itself zero. :We allow that at this arbitrary $t = 0$, the value of $x$ is assigned the value $x = x_0$. This corresponds, for example, with a physical system in which the moving body is pulled from its equilibrium position and released from stationary at time zero. Differentiating $(1)$ {{WRT}} $t$ gives: :$\quad x' = -b e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + e^{-b t} \paren {-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}$ Setting the initial condition $x = x_0$ when $t = 0$ in $(1)$: {{begin-eqn}} {{eqn | l = x_0       | r = e^0 \paren {C_1 \cos 0 + C_2 \sin 0}       | c =  }} {{eqn | r = C_1       | c =  }} {{end-eqn}} Setting the initial condition $x' = 0$ when $t = 0$: {{begin-eqn}} {{eqn | l = 0       | r = -b e^0 \paren {C_1 \cos 0 + C_2 \sin 0} + e^0 \paren {-\alpha C_1 \sin 0 + \alpha C_2 \cos 0}       | c =  }} {{eqn | r = -b C_1 + \alpha C_2       | c =  }} {{eqn | ll= \leadsto       | l = C_2       | r = \frac {b C_1} \alpha       | c =  }} {{eqn | r = \frac {b x_0} \alpha       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = x       | r = e^{-b t} \paren {x_0 \cos \alpha t + \frac {b x_0} \alpha \sin \alpha t}       | c =  }} {{eqn | n = 2       | r = e^{-b t} \dfrac {x_0} \alpha \paren {\alpha \cos \alpha t + b \sin \alpha t}       | c =  }} {{end-eqn}} From Multiple of Sine plus Multiple of Cosine:Cosine Form, $(2)$ can be expressed as: {{begin-eqn}} {{eqn | l = x       | r = \dfrac {x_0} \alpha e^{-b t} \paren {\sqrt {\alpha^2 + b^2} \, \map \cos {\alpha t + \arctan \dfrac {-b} \alpha} }       | c =  }} {{eqn | r = \dfrac {x_0 \sqrt {\sqrt{a^2 - b^2}^2 + b^2} } \alpha e^{-b t} \, \map \cos {\alpha t + \arctan \dfrac {-b} \alpha}       | c =  }} {{eqn | r = \dfrac {x_0 \sqrt {a^2 - b^2 + b^2} } \alpha e^{-b t} \, \map \cos {\alpha t + \arctan \dfrac {-b} \alpha}       | c =  }} {{eqn | r = \dfrac {x_0 \, a} \alpha e^{-b t} \, \cos {\alpha t + \arctan \dfrac {-b} \alpha}       | c =  }} {{eqn | r = \dfrac {x_0 \, a} \alpha e^{-b t} \, \cos {\alpha t - \arctan \dfrac b \alpha}       | c = Tangent Function is Odd }} {{eqn | r = \dfrac {x_0 \, a} \alpha e^{-b t} \, \cos {\alpha t - \theta}       | c = where $\theta = \arctan \dfrac b \alpha$ }} {{end-eqn}} {{qed}}
Let the position of $S$ be described in the canonical form: :$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$ where $\alpha = \sqrt {a^2 - b^2}$. The zeroes of $(1)$ occur exactly where: :$\map \cos {\alpha t - \theta} = 0$ Thus the period $T$ of $\map \cos {\alpha t - \theta}$ is given by: :$\alpha T = 2 \pi$ and so: :$T = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$ {{qed}}
Let: :$a^2 = \dfrac k m$ :$2 b = \dfrac c m$ From Position of Cart attached to Wall by Spring under Damping: Underdamped: $x = x_0$ at $t = 0$: :$x = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)$ where $\alpha = \sqrt {a^2 - b^2}$. Let $T$ be the period of oscillation of $C$. Then: {{begin-eqn}} {{eqn | l = T       | r = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }       | c = Period of Oscillation of Underdamped System is Regular }} {{eqn | r = \dfrac {2 \pi} {\sqrt {\dfrac k m - \left({\dfrac c {2 m} }\right)^2} }       | c =  }} {{eqn | r = \dfrac {2 \pi} {\sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} } }       | c =  }} {{end-eqn}} {{qed}}
Let the position of $S$ be described in the canonical form: :$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$ where $\alpha = \sqrt {a^2 - b^2}$. From Period of Oscillation of Underdamped System is Regular: :$T = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$ where $T$ is the period of the movement of $S$. By definition of natural frequency: :$\nu = \dfrac 1 T = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$ {{qed}}
From Period of Oscillation of Underdamped Cart attached to Wall by Spring: :$T = \dfrac {2 \pi} {\sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} } }$ where $T$ is the period of oscillation of $C$. Then by definition of the natural frequency of $C$: {{begin-eqn}} {{eqn | l = \nu       | r = \dfrac 1 T       | c =  }} {{eqn | r = \dfrac 1 {\left({\dfrac {2 \pi} {\sqrt {\frac k m - \frac {c^2} {4 m^2} } } }\right)}       | c =  }} {{eqn | r = \dfrac 1 {2 \pi} \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }       | c =  }} {{end-eqn}} {{qed}}
Let $\mathbf F_s$ be the force on $C$ exerted by the spring $S$. Let $\mathbf F_d$ be the damping force. Let $\mathbf F_e$ be the external force on $C$. By Newton's Second Law of Motion, the total force on $C$ equals its mass times its acceleration: :$\mathbf F_s + \mathbf F_d + \mathbf F_e = m \mathbf a$ By Acceleration is Second Derivative of Displacement with respect to Time: :$\mathbf a = \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}$ By definition, the velocity $\mathbf v$ is defined as: :$\mathbf v = \dfrac {\mathrm d \mathbf x}{\mathrm d t}$ By Hooke's Law: :$\mathbf F = -k \mathbf x$ By definition of the damping force: :$\mathbf F_d = - c \mathbf v$ By definition of the external force: :$\mathbf F_e = \mathbf F_0 \cos \omega t$ So: {{begin-eqn}} {{eqn | l = m \mathbf a       | r = -k \mathbf x - c \mathbf v + \mathbf F_0 \cos \omega t       | c =  }} {{eqn | ll= \implies       | l = m \dfrac {\mathrm d^2 \mathbf x}{\mathrm d t^2}       | r = -k \mathbf x - c \dfrac {\mathrm d \mathbf x}{\mathrm d t} + \mathbf F_0 \cos \omega t       | c =  }} {{eqn | ll= \implies       | l = m \dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + c \dfrac {\mathrm d \mathbf x} {\mathrm d t} + k \mathbf x       | r = \mathbf F_0 \cos \omega t       | c =  }} {{end-eqn}} {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 + 2 b m + a^2 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = - b + \sqrt {b^2 - a^2}$ :$m_2 = - b - \sqrt {b^2 - a^2}$ As $b^2 < a^2$, this converts into: :$m_1 = - b + i \sqrt {a^2 - b^2}$ :$m_2 = - b - i \sqrt {a^2 - b^2}$ These are complex and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$ {{qed}} Category:Examples of Constant Coefficient Homogeneous LSOODEs 9k5xhzn1dsruhtz976rp02pe76he2ci
From Forced Vibration of Cart attached to Wall by Spring, the motion of $C$ is described by the second order ODE: :$(1): \quad m \dfrac {\d^2 \mathbf x} {\d t^2} + c \dfrac {\d \mathbf x} {\d t} + k \mathbf x = \mathbf F_0 \cos \omega t$ Setting: :$a^2 = \dfrac k m$ :$2 b = \dfrac c m$ this can be written as: :$(2): \quad \dfrac {\d^2 \mathbf x} {\d t^2} + 2 b \dfrac {\d \mathbf x} {\d t} + a^2 \mathbf x = \dfrac {\mathbf F_0} m \cos \omega t$ We are given that $C$ is underdamped, and so $b < a$. From Linear Second Order ODE: $y'' + 2 b y' + a^2 y = K \cos \omega x$ for $b < a$, $(2)$ has the general solution: :$\mathbf x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega t - \phi}$ where: :$\alpha = \sqrt {a^2 - b^2}$ :$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$ Substituting back: :$a^2 = \dfrac k m$ :$2 b = \dfrac c m$ :$K = \dfrac {\mathbf F_0} m$ we obtain: {{begin-eqn}} {{eqn | o =        | r = \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega t - \phi}       | c =  }} {{eqn | r = \dfrac {\mathbf F_0} m \frac 1 {\sqrt {\paren {\dfrac c m}^2 \omega^2 + \paren {\dfrac k m - \omega^2}^2} } \map \cos {\omega t - \phi}       | c =  }} {{eqn | r = \dfrac {\mathbf F_0} m \frac 1 {\sqrt {\dfrac 1 {m^2} c^2 \omega^2 + \dfrac 1 {m^2} \paren {k - \omega^2 m}^2} } \map \cos {\omega t - \phi}       | c =  }} {{eqn | r = \dfrac {\mathbf F_0} m \frac 1 {\dfrac 1 m \sqrt {c^2 \omega^2 + \paren {k - \omega^2 m}^2} } \map \cos {\omega t - \phi}       | c =  }} {{eqn | r = \dfrac {\mathbf F_0} {\sqrt {c^2 \omega^2 + \paren {k - \omega^2 m}^2} } \map \cos {\omega t - \phi}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \phi       | r = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }       | c =  }} {{eqn | r = \map \arctan {\dfrac {\dfrac c m \omega} {\dfrac k m - \omega^2} }       | c =  }} {{eqn | r = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }       | c = multiplying top and bottom by $m$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \alpha       | r = \sqrt {a^2 - b^2}       | c =  }} {{eqn | r = \sqrt {\dfrac k m - \paren {\dfrac c {2 m} }^2}       | c =  }} {{eqn | r = \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }       | c =  }} {{end-eqn}} {{qed}}