5.4.3 Symbolic Solutions of the Spring Equation

We are ready now to find symbolic solutions of the differential equation,

where \(k\) and \(m\) are arbitrary positive constants. Then we can use the initial values for \(x\) and \(x'\) to complete our search for formulas for the solutions we generated graphically on the preceding page. As usual, we attack this problem by first solving an easier inverse problem. That is, we calculate derivatives of functions that would appear to have the right form to satisfy the differential equation.

In the next Activity we use \(\omega\) for the constant multiplier of \(t\). The symbol \(\omega\) is a lower case Greek omega — it is a conventional symbol for a period-altering scale factor in trigonometric functions.

Activity 4

1. Show that

   ${\displaystyle \frac{d^2}{d{t}^2}\sin (\omega t)=-{\omega }^2\sin (\omega t)}$

   and

   ${\displaystyle \frac{d^2}{d{t}^2}\cos (\omega t)=-{\omega }^2\cos (\omega t)}$.

2. Explain why \(x(t)=\sin(\omega t)\) and \(x(t)=\cos(\omega t)\) are both solutions of the differential equation

   ${\displaystyle \frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x\mathrm{.}}$

3. Explain why any function of the form \(x(t)=A\,\sin(\omega t)+B\,\cos(\omega t)\) is a solution of the same differential equation.

Comment on Activity 4

Notice the parallel between Activity 4 and what we did in Chapter 2. There we looked for functions to satisfy an equation of the form

first derivative is proportional to the function

and found that the solutions were all of the form \(A\,e^{kt}\). In particular, the scale factor for the independent variable was the proportionality constant in the differential equation. Now we are solving

second derivative is negatively proportional to the function

and finding a similar result with sines and cosines. In particular, the scale factor for the independent variable is the square root of the absolute value of the proportionality constant.

Second derivative is positively proportional to the function

is a different, but also similar, situation — see Problems 11 and 12.

Our original differential equation,

has precisely the form

if we identify \(\omega^2\) with \(k/m\), that is, if we set \(\omega=\sqrt{\frac{k}{m}}\). And we can do that, because we know that both \(k\) and \(m\) are positive. Thus, according to Activity 4, any function of the form

is a solution of

Now we need to determine what combination of coefficients \(A\) and \(B\) — if any — will give us one function that also satisfies the initial conditions for displacement and velocity. We start by substituting \(t=0\) and \(x=x_0\):

Since \(\sin(0)=0\) and \(\cos(0)=1\), this equation tells us that \(B\) must be the starting displacement \(x_0\).

Since \(A\) dropped out of the equation when we substituted \(t=0\), we know nothing yet about the value of \(A\). But we have another initial condition! In order to use it, we need a formula for \(v(t)\), which you can provide by differentiating the formula for \(x(t)\).

Checkpoint 2

Conclusion: The unique solution of the spring-mass initial value problem,

\(\frac{d^2x}{dt^2}=-\frac{k}{m}x\),   with  \(x(0)=x_0\)  and  \(x'(0)=0\),

is

\(x(t)=x_0 \, \cos\left(\sqrt{\frac{k}{m}} \, t\right)\).

In order to generate a numerical solution on the preceding page, we had to give numerical values for the starting displacement and for the ratio of spring constant to mass. We chose both of those numbers to be \(1\), so the solution we generated was just \(x=\cos(t)\). The velocity function was then the derivative of \(x=\cos(t)\), that is, \(v=-\sin(t)\). And that was exactly what we concluded in our discussion of Activity 3.