Steady-State Solution to Cart attached to Wall by Spring under Forced Vibration

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Theorem

Problem Definition

CartOnSpringForcedVibration.png

Consider a cart $C$ of mass $m$ attached to a wall by means of a spring $S$.

Let $C$ be free to move along a straight line in a medium which applies a damping force $\mathbf F_d$ whose magnitude is proportional to the speed of $C$.

Let the force constant of $S$ be $k$.

Let the constant of proportion of the damping force $\mathbf F_d$ be $c$.

Let there be applied to $C$ an external force which varies as a function of time as:

$\mathbf F_e = \mathbf F_0 \cos \omega t$

where $\mathbf F_0$ is constant.

Let the displacement of $C$ at time $t$ from the equilibrium position be $\mathbf x$.


Then the steady state component of the horizontal position of $C$ at time $t$ can be expressed as:

$\mathbf x_s = \dfrac {\mathbf F_0} {\sqrt {\left({k^2 - \omega^2 m^2}\right)^2 + \omega^2 c^2} } \cos \left({\omega t - \phi}\right)$

where:

$\phi = \arctan \left({\dfrac {\omega c} {k - \omega^2 m} }\right)$


Proof

From Forced Vibration of Cart attached to Wall by Spring, the equation of motion of $C$ is:

$(1): \quad m \dfrac {\mathrm d^2 \mathbf x} {\mathrm d t^2} + c \dfrac {\mathrm d \mathbf x} {\mathrm d t} + k \mathbf x = \mathbf F_0 \cos \omega t$

From Position of Underdamped Cart attached to Wall by Spring under Forced Vibration, the general solution to $(1)$ is:

$\mathbf x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) + \dfrac {\mathbf F_0} {\sqrt {\left({k^2 - \omega^2 m^2}\right)^2 + \omega^2 c^2} } \cos \left({\omega t - \phi}\right)$

where:

$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
$\alpha = \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$
$\phi = \arctan \left({\dfrac {\omega c} {k - \omega^2 m} }\right)$


We have that $(1)$ is in the form:

$(2): \quad \dfrac {\mathrm d^2 y} {\mathrm d x^2} + 2 b \dfrac {\mathrm d y} {\mathrm d x} + a^2 x = K \cos \omega x$

where:

$K \in \R: k > 0$
$a, b \in \R_{>0}: b < a$

whose general solution from Second Order ODE: $y'' + 2 b y' + a^2 y = K \cos \omega x$ where $b < a$ is:

$(1): \quad y = e^{-b x} \left({C_1 \cos \alpha x + C_2 \sin \alpha x}\right) + \dfrac K {\sqrt {4 b^2 \omega^2 + \left({a^2 - \omega^2}\right)^2} } \cos \left({\omega x - \phi}\right)$

where:

$\alpha = \sqrt {a^2 - b^2}$
$\phi = \arctan \left({\dfrac {2 b \omega} {a^2 - \omega^2} }\right)$


By definition, the steady-state component of the general solution to $(2)$ is:

$\dfrac K {\sqrt {4 b^2 \omega^2 + \left({a^2 - \omega^2}\right)^2} } \cos \left({\omega x - \phi}\right)$


Thus by definition and comparison, the steady-state component of the general solution to $(1)$ is:

$\mathbf x_s = \dfrac {\mathbf F_0} {\sqrt {\left({k^2 - \omega^2 m^2}\right)^2 + \omega^2 c^2} } \cos \left({\omega t - \phi}\right)$

where:

$\phi = \arctan \left({\dfrac {\omega c} {k - \omega^2 m} }\right)$

$\blacksquare$


Sources