Modelling free mechanical oscillations

2 Modelling free mechanical oscillations

Consider the following schematic diagram of a shock absorber:

Figure 6

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The equation of motion can be described in terms of the vertical displacement $x$ of the mass.

Let $m$ be the mass, $k \frac{d x}{d t}$ the damping force resulting from the dashpot and $n x$ the restoring force resulting from the spring. Here, $k$ and $n$ are constants.

Then the equation of motion is

$m \frac{d^{2} x}{d t^{2}} = - k \frac{d x}{d t} - n x .$

Suppose that the mass is displaced a distance $x_{0}$ initially and released from rest. Then at $t = 0$ , $x = x_{0}$ and $\frac{d x}{d t} = 0$ . Writing the differential equation in standard form gives

$m \frac{d^{2} x}{d t^{2}} + k \frac{d x}{d t} + n x = 0.$

We shall see that the nature of the oscillations described by this differential equation depends crucially upon the relative values of the mechanical constants $m, k$ and $n$ . This will be explored in subsequent Tasks.

Task!

Find and solve the auxiliary equation of the differential equation

$m \frac{d^{2} x}{d t^{2}} + k \frac{d x}{d t} + n x = 0.$

Putting $x = e^{λ t}$ , the auxiliary equation is $m λ^{2} + k λ + n = 0.$

Hence $λ = \frac{- k \pm \sqrt{k^{2} - 4 m n}}{2 m} .$

The value of $k$ controls the amount of damping in the system. We explore the solution for various values of $k$ .

2.1 Case 1: No damping

If $k = 0$ then there is no damping. We expect, in this case, that once motion has started it will continue for ever. The motion that ensues is called simple harmonic motion . In this case we have

$λ = \frac{\pm \sqrt{- 4 m n}}{2 m},$ that is, $λ = \pm \sqrt{\frac{n}{m}} i where i^{2} = - 1.$

and the solution for the displacement $x$ is:

$x = A cos (\sqrt{\frac{n}{m}} t) + B sin (\sqrt{\frac{n}{m}} t)$ where $A, B$ are arbitrary constants.

Task!

Impose the initial conditions $x = x_{0}$ and $\frac{d x}{d t} = 0$ at $t = 0$ to find the unique solution to the ODE:

$\frac{d x}{d t} = - \sqrt{\frac{n}{m}} A sin (\sqrt{\frac{n}{m}} t) + \sqrt{\frac{n}{m}} B cos (\sqrt{\frac{n}{m}} t)$

When $t = 0, \frac{d x}{d t} = 0$ so that $\sqrt{\frac{n}{m}} B = 0 so that B = 0.$

Therefore $x = A cos (\sqrt{\frac{n}{m}} t) .$

Imposing the remaining initial condition: when $t = 0, x = x_{0}$ so that $x_{0} = A$ and finally:

$x = x_{0} cos (\sqrt{\frac{n}{m}} t) .$

2.2 Case 2: Light damping

If $k^{2} - 4 m n < 0$ , i.e. $k^{2} < 4 m n$ then the roots of the auxiliary equation are complex:

$λ_{1} = \frac{- k + i \sqrt{4 m n - k^{2}}}{2 m} λ_{2} = \frac{- k - i \sqrt{4 m n - k^{2}}}{2 m}$

Then, after some rearrangement:

$x = e^{- k t ∕ 2 m} [A cos p t + B sin p t]$ in which $p = \sqrt{4 m n - k^{2}} ∕ 2 m$ .

Task!

If $m = 1, n = 1$ and $k = 1$ find $λ_{1}$ and $λ_{2}$ and then find the solution for the displacement $x$ .

$λ = \frac{- 1 + i \sqrt{4 - 1}}{2} = - 1 ∕ 2 \pm i \sqrt{3} ∕ 2$ . Hence $x = e^{- t ∕ 2} [A cos \frac{\sqrt{3}}{2} t + B sin \frac{\sqrt{3}}{2} t]$ .

Impose the initial conditions $x = x_{0}, \frac{d x}{d t} = 0$ at $t = 0$ to find the arbitrary constants and hence find the solution to the ODE:

Differentiating, we obtain

$\frac{d x}{d t} = - \frac{1}{2} e^{- t ∕ 2} [A cos \frac{\sqrt{3}}{2} t + B sin \frac{\sqrt{3}}{2} t] + e^{- t ∕ 2} [- \frac{\sqrt{3}}{2} A sin \frac{\sqrt{3}}{2} t + \frac{\sqrt{3}}{2} B cos \frac{\sqrt{3}}{2} t]$

At $t = 0$ ,

$x = x_{0} = A$ (i)

$\frac{d x}{d t} = 0 = - \frac{1}{2} A + \frac{\sqrt{3}}{2} B$ (ii)

Solving (i) and (ii) we obtain

$A = x_{0} B = \frac{\sqrt{3}}{3} x_{0}$ then $x = x_{0} e^{- t ∕ 2} [cos \frac{\sqrt{3}}{2} t + \frac{\sqrt{3}}{3} sin \frac{\sqrt{3}}{2} t] .$ The graph of $x$ against $t$ is shown in Figure 7. This is the case of light damping. As the damping in the system decreases (i.e. $k \to 0$ ) the number of oscillations (in a given time interval) will increase. In many mechanical systems these oscillations are usually unwanted and the designer would choose a value of $k$ to either reduce them or to eliminate them altogether. For the choice $k^{2} = 4 m n$ , known as the critical damping case, all the oscillations are absent.

Figure 7

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2.3 Case 3: Heavy damping

If $k^{2} - 4 m n > 0$ , i.e. $k^{2} > 4 m n$ , then there are two real roots of the auxiliary equation, $λ_{1}$ and $λ_{2}$ :

$λ_{1} = \frac{- k + \sqrt{k^{2} - 4 m n}}{2 m} λ_{2} = \frac{- k - \sqrt{k^{2} - 4 m n}}{2 m}$

Then

$x = A e^{λ_{1} t} + B e^{λ_{2} t} .$

Task!

If $m = 1, n = 1$ and $k = 2.5$ find $λ_{1}$ and $λ_{2}$ and then find the solution for the displacement $x$ .

$λ = \frac{- 2.5 \pm \sqrt{6.25 - 4}}{2} = - 1.25 \pm 0.75$

Hence $λ_{1}, λ_{2} = - 0.5, - 2$ and so $x = A e^{- 0.5 t} + B e^{- 2 t}$ Impose the initial conditions $x = x_{0}, \frac{d x}{d t} = 0$ at $t = 0$ to find the arbitrary constants and hence find the solution to the ODE.

Differentiating, we obtain

$\frac{d x}{d t} = - 0.5 A e^{- 0.5 t} - 2 B e^{- 2 t}$

At $t = 0$ ,

$x = x_{0} = A + B$ (i)

$\frac{d x}{d t} = 0 = - 0.5 A - 2 B$ (ii)

Solving (i) and (ii) we obtain $A = \frac{4}{3} x_{0} B = - \frac{1}{3} x_{0}$ then $x = \frac{1}{3} x_{0} (4 e^{- 0.5 t} - e^{- 2 t}) .$

The graph of $x$ against $t$ is shown below. This is the case of heavy damping.

Other cases are dealt with in the Exercises at the end of the Section.

2 Modelling free mechanical oscillations

Task!

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2.1 Case 1: No damping

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2.2 Case 2: Light damping

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2.3 Case 3: Heavy damping

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