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

Theorem

Problem Definition

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:

$\mathbf x_s = \dfrac {\mathbf F_0} {\sqrt {\paren {k^2 - \omega^2 m^2}^2 + \omega^2 c^2} } \map \cos {\omega t - \phi}$

where:

$\phi = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }$

Proof

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

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

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

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

where:

$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$

$\alpha = \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$

$\phi = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }$

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

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

where:

$K \in \R: k > 0$

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

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

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

where:

$\alpha = \sqrt {a^2 - b^2}$

$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$

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

$\dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$

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

$\mathbf x_s = \dfrac {\mathbf F_0} {\sqrt {\paren {k^2 - \omega^2 m^2}^2 + \omega^2 c^2} } \map \cos {\omega t - \phi}$

where:

$\phi = \map \arctan {\dfrac {\omega c} {k - \omega^2 m} }$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.20$: Vibrations in Mechanical Systems: $(29)$