Question 2.2: Find the response (in terms of force Fk and deflection x1) o...

Separating variables in Eq. (2.10)

\frac{dF_k}{dt}=kv_{21}          (2.10)

and integrating with respect to time, we have

\int_{F_k(0)}^{F_k(t)}{dF_k}=k\int_{0}^{t}{v_1dt}=(8000)(20)\int_{0}^{t}{dt},
F_k(t)-0=(8000)(20)t-0,
or
F_k(t)=160,000t lb           (2.12)

We may use the definition

v_1\equiv \frac{dx_1}{dt} .       (2.13)

Separating variables in Eq. (2.13) and integrating again with respect to time yields

\int_{x_1(0)}^{x_1(t)}{dx_1}=\int_{0}^{t}{v_1dt}=\int_{0}^{t}{20dt,}
or
x_1(t)-0=20t-0,
or
x_1(t)=20t in.        (2.14)

The results are shown in Fig. 2.7.
Again, the essential role played by integration in finding the dynamic response of a system is evident. From an energy point of view, the rate at which potential energy is stored in the spring is equal to the rate at which the steadily increasing force does work on the spring as it deflects the spring:

Separating variables and integrating yields

\int_{E _p(0)}^{E_p(t)}{dE _p}=\int_{0}^{t}{F_kv_1dt}=\left(\frac{1}{k} \right)\int_{0}^{t}{F_k\left(\frac{dF_k}{dt} \right) dt }=\left(\frac{1}{k} \right) \int_{F_k(0)}^{F_k(t)}{F_kdF_k} ,

so that

E _p(t)=\left(\frac{1}{2k} \right)F^2_k.               (2.15)

Thus it takes time for the work input to add to the accumulated energy stored in the spring. A spring is a T-type element, storing energy as a function of the square of its T variable F_k.
Note again, as for changing velocity in Example 2.1, it would not be realistic here to try to apply a step change in force to a spring; such a force source would have to move at a very, very great velocity to deflect the spring suddenly, which would require a very, very great power source. In general it can be stated that inputs that would suddenly add to the stored energy in a system are not realistic and cannot be achieved in the natural world.