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Transient Vibrations: Response of Spring–Mass System to a Step Function

(a) Schematic

(b) FBD/MAD (x, x¨ > 0)

Consider a simple spring–mass system subjected to a rectangular step load as shown in Figure 7.1(a). We would like to find the response of the mass to this transient loading condition. Figure 7.1(b) shows the associated FBD/MAD. The equation of motion obtained using Newton’s laws is

    \begin{equation*} \stackrel{+}{\rightarrow}\sum F = ma:\qquad -k x + F_0 = m \ddot{x} \end{equation*}


(7.2)   \begin{equation*} \label{eq:StepFunctionEOM} m \ddot{x} + kx = F_0, \qquad\qquad t>0. \end{equation*}

As usual, the solution is composed of the homogeneous and particular solutions

    \begin{equation*} x(t) = A \sin{pt} + B \cos{pt} + x_P(t). \end{equation*}

For the RHS in (7.2), the particular solution is

    \begin{equation*} x_P(t) = \frac{F_0}{k} \end{equation*}

so the total solution becomes

    \begin{equation*} x(t) = A \sin{pt} + B \cos{pt} + \frac{F_0}{k}. \end{equation*}

If we start with the initial conditions

    \begin{equation*} x(0) = 0, \qquad \dot{x}(0) = 0, \end{equation*}

the constants *Note that the total solution (including the particular solution) must be used to determine the arbitrary constants which arise from the homogeneous solution.* become

    \begin{equation*} A = 0,\qquad B = -\frac{F_0}{k}, \end{equation*}

so that the total solution is

(7.3)   \begin{equation*} \label{eq:StepFunctionSolution} \boxed{x(t) = \frac{F_0}{k} \Bigl[ 1 - \cos{pt} \Bigr].} \end{equation*}

Figure 7.2: Response of simple spring–mass system to applied step load

This response is illustrated in Figure 7.2. We can see that the maximum displacement of the mass is 2\frac{F_0}{k} which is twice the deflection that would have resulted if the load were applied statically. Now consider what happens if the step load is “turned off” at some time \hat{t}. After this time, since there is no longer any applied force, the system undergoes free vibrations given by

    \begin{equation*} x(t) = \hat{A} \sin{pt} + \hat{B} \cos{pt}, \qquad\qquad t>\hat{t}, \end{equation*}

where \hat{A} and \hat{B} need to be evaluated at the “initial” conditions (t=\hat{t}\,) which depend on the time \hat{t} at which the force was removed. The response of the mass while the step load is applied is given in equation (7.3). Similarly, the velocity of the mass during this period is, by differentiation,

    \begin{equation*} \dot{x}(t) = \frac{F_0 P}{k} \sin{pt}. \end{equation*}

Therefore, at the time \hat{t} when the load is removed, the “initial” conditions are

    \begin{equation*} x(\hat{t}) = \frac{F_0}{k} \Bigl[ 1 - \cos{p\hat{t}} \Bigr], \qquad \dot{x}(\hat{t}) = \frac{F_0}{k}{\sin{p\hat{t}}} \end{equation*}

which can be used to find \hat{A} and \hat{B}.

Consider two specific examples:

    1. {\hat{t} = \dfrac{2 \pi}{p}}

    \begin{alignat*}{2} x(\hat{t}) &= \frac{F_0}{k} \Bigl[ 1 - \cos{2 \pi} \Bigr] & &= 0, \\[2mm] \dot{x}(\hat{t}) &= \frac{F_0 p}{k} \sin{2 \pi}=0. \end{alignat*}

As a result,

    \begin{equation*} \hat{A} = 0, \qquad \hat{B} = 0, \end{equation*}

so the response in this case is

    \begin{equation*} x(t) = \begin{cases} \dfrac{F_0}{k\rule[-3mm]{0pt}{0pt}} \Bigl[ 1 - \cos{pt}], & t \le \dfrac{2 \pi}{p},\\ 0, \qquad\qquad\qquad, & t > \dfrac{2 \pi}{p}. \\ \end{cases} \end{equation*}

    1. {\hat{t} = \dfrac{\pi}{p}}

    \begin{alignat*}{2} x(\hat{t}) &= \frac{F_0}{k} \Bigl[ 1 - \cos{\pi} \Bigr] & &= 2 \frac{F_0}{k}, \\[2mm] \dot{x}(\hat{t}) &= \frac{F_0 p}{k} \sin{\pi} & &=0. \end{alignat*}

As a result,

    \begin{equation*} \hat{A} = 0, \qquad \hat{B} = \frac{2 F_0}{k}, \end{equation*}

so the response in this case is

    \begin{equation*} x(t) = \begin{cases} \dfrac{F_0}{k\rule[-5mm]{0pt}{0pt}} \Bigl[ 1 - \cos{pt} \Bigr], & t \le \dfrac{\pi}{p},\\ -\dfrac{2 F_0}{k} \cos{pt}, \qquad\qquad & t > \dfrac{\pi}{p}. \\ \end{cases} \end{equation*}

These responses are shown in Figure 7.3. As these responses clearly show, the transient nature of the applied loading can have a significant effect on the resulting response.

(a) τ =2π/p

(b) τ = π/p

Figure 7.3: Response of simple spring–mass system to applied step load which is subsequently removed

The previous responses can be understood on a physical basis. In the first case, F_0 is applied to a system originally in its equilibrium position. When F_0 is applied, the effect is to shift the new equilibrium position a distance \frac{F_0}{k} as shown in Figure 7.4.

Figure 7.4: Shift of equilibrium position

At time t=0 (immediately after the application of F_0), the mass is apparently in a position (top of Figure 7.4}) where it is displaced from its new equilibrium position. If we measure the displacement of the mass from the new equilibrium position as y and neglect both the applied force F_0 and the new equilibrium stretch in the spring (as we did regularly when gravity was present), we then have a {free} vibration problem with the initial conditions

    \begin{equation*} y(0) = -\frac{F_0}{k},\qquad \dot{y} = 0, \end{equation*}

so the free vibration response is

    \begin{equation*} y(t) = -\frac{F_0}{k} \cos{pt}. \end{equation*}

However, since x = y + \frac{F_0}{k}, we can write the response in terms of x as

    \begin{equation*} x(t) = \frac{F_0}{k} \Bigl[ 1 - \cos{pt} \Bigr] \end{equation*}

as before.

The two cases shown in Figure 7.3 can be understood on similar principles. When the force F_0 is removed, the equilibrium position of the system returns to its original position. The time at which the force F_0 is removed determines the initial conditions for the subsequent free vibration problem, which leads to the responses shown.

Response of Spring–Mass System to an Impulse

Figure 7.5: Impulse Loading

We now consider the response of a spring–mass system subjected to an impulsive loading shown in Figure 7.5. (Note that the quantity F_0 \Delta t is the area under the force-time curve and is known as the {impulse} applied to the system.) Here the time \hat{t} at which the load is removed is small (\Delta t) so that

    \begin{equation*} p\hat{t} = p \Delta t \ll 1. \end{equation*}

In such a case,

    \begin{equation*} \cos{p\hat{t}} \approx 1,\\[1mm] \end{equation*}

    \begin{equation*} \sin{p\hat{t}} \approx p\hat{t} = p\,\Delta t. \end{equation*}

The “initial” conditions are approximately

so the initial conditions (after impulse has been applied) become

    \begin{equation*} x(0) \approx x(\hat{t}) = 0, \end{equation*}

    \begin{equation*} \\[1mm] \dot{x}(0) \approx \dot{x}(\hat{t}) = \frac{F_0 \Delta t}{m}. \end{equation*}

The constants \hat{A} and \hat{B} are then

    \begin{equation*} \hat{A} = \frac{F_0 \Delta t}{m p}, \qquad \hat{B} = 0, \end{equation*}

so the response becomes

(7.4)   \begin{equation*} \label{eq:ImpulseResponse} \boxed{x(t) = \frac{F_0 \Delta t}{m p} \sin{pt}, \qquad t>0.} \end{equation*}

This response will play an important role later when we consider the method of convolution.


We can obtain this result another way by applying the basic principle of impulse and momentum for our system.

Figure 7.6: Impulse–momentum diagram for mass m

Figure 7.6 shows an impulse–momentum diagram for our simple spring mass system when the impulsive load is applied. The system is initially at rest and after the impulse has been applied has a resulting velocity v_f. (Note that the force of the spring is not included in the diagram. The assumption is that the impulse is applied fast enough that the mass does not have time to move while the impulse is being applied. Since the mass does not move, the force in the spring remains unchanged. We say that the force of the spring is {non-impulsive}.)
Applying the principle of impulse and momentum shows that

    \begin{equation*} \stackrel{+}{\rightarrow}\qquad m\cdot(0) + F_0 \Delta t = m v_f \end{equation*}

so we get

    \begin{equation*} v_f = \frac{F_0 \Delta t}{m}. \end{equation*}

After the impulse is over, we have a free vibration problem with the initial conditions

    \begin{equation*} x(0) \approx 0, \qquad \dot{x}(0) = v_f = \frac{F_0 \Delta t}{m}. \end{equation*}

The two constants are

    \begin{equation*} A = \frac{F_0 \Delta t}{m p}, \qquad B=0, \end{equation*}

so the response becomes

    \begin{equation*} x(t) = \frac{F_0 \Delta t}{m p} \sin{pt} \end{equation*}

as in equation (7.4).

We can also consider the response to some additional useful forcing functions.

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