Question

A 1-kg mass is attached to a spring with stiffness 45N/m. The damping constant for the system is 6 N-sec/m. The mass is pulled 1 m to the right of the equilibrium position and released. Find the equation of motion in phase-shift form. When will the mass first return to its equilibriom position, and at what velocity?

Answer 1

k=45 N/m m= 1 kg B-6 Ns/m

When the mass is released, there are three internal forces acting on the mass and 0 external forces.

The internal forces are:

Force due to mass =
dt2
=
dt2
      [$\because$ mass m = 1 kg]

Force due to damper =
dx dx
   [$\because$ damping constant B = 6 Ns/m]

Force due to spring = [$\because$ spring stiffness k = 45 N/m]

Where x is the displacement.

At any instant, total internal forces = total external forces

$\Rightarrow \frac{d^2x}{dt^2} + 6\frac{dx}{dt} + 45x = 0$

Taking Laplace transform,

s2x(s) _ sx(0) _ x'(0)-6(X(s)-x(0)) +45x(s) = 0

ex(s) _ s _ 0 + 6(sX (s)-1) + 45x(s) = 0
[$\because$ initial displacement x(0)=1,
                                                                        and initial change in displacement = initial velocity = x'(0)=0]

s"X(s) + 6SX(s) + 45x (s) = s + 6

$\Rightarrow X(s) = \frac{s+6}{s^2 + 6s + 45} = \frac{s+3+3}{(s+3)^2 + 6^2} = \frac{s+3}{(s+3)^2 + 6^2} + \frac{1}{2}\left( \frac{6}{(s+3)^2 + 6^2} \right )$

$\Rightarrow x(t) = e^{-3t}cos(6t) + \frac{1}{2}e^{-3t}sin(6t)$

$\Rightarrow x(t) = e^{-3t}cos(6t) + \frac{1}{2}e^{-3t}cos\left(6t-\frac{\pi}{2}\right)$

Let the mass return to equilibrium position i.e. x=0 at time t=t₀ seconds

$\Rightarrow e^{-3t_0}cos(6t_0) + \frac{1}{2}e^{-3t_0}cos\left(6t_0-\frac{\pi}{2}\right)=0$

cos(60) sin(6to)
    [ $\because e^{-3t_0}\neq 0$ ]

$\Rightarrow tan(6t_0)=-2$

$\Rightarrow 6t_0=tan^{-1}(2) = 2.0344$     [minimum angle in radian >0 whose tan is 2]

$\Rightarrow t_0=0.3391\; s = 339.1 \; ms$

dt2

dt2

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dx dx

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s2x(s) _ sx(0) _ x'(0)-6(X(s)-x(0)) +45x(s) = 0

ex(s) _ s _ 0 + 6(sX (s)-1) + 45x(s) = 0

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