MyPhysicsLab – Analytic Solution for Single Spring Simulation

We seek a solution x(t) that satisfies the differential equation x''(t) = − ^k⁄_m x(t) where

x = position of the block
t = time
k = spring stiffness
m = mass of the block

and the initial conditions x(0) = x₀ the position at time zero x'(0) = v₀ the velocity at time zero For convenience we first define a new constant c = √(k/m) so the differential equation becomes

x''(t) = −c² x(t)

(1)

We classify this differential equation as second order linear. It is second order because there is a second derivative: x''. It is linear because all the functions or derivatives of x appear by themselves, only multiplied by a constant. To solve equations of this type, begin by assuming the solution is of the form x(t) = e ^{r t} where r is a constant, but possibly a complex number. The other constant e is the well-known mathematical exponential constant e = 2.718... which has nice properties for doing calculus. We plug this proposed solution into the differential equation (1) and get r² e ^{r t} = −c² e ^{r t} where we took the second derivative of e ^{r t} to get the left hand side. Since e ^{r t} is never equal to zero, we can divide it out of this equation to get r² = −c² Since c > 0 the solution involves complex numbers and we have two values for r r = ±c i where i = √−1. We then have two solutions

x₁(t) = e ^{i c t}

(2a)

x₂(t) = e ^{−i c t}

(2b)

Try plugging either of these solutions into the differential equation (1) and you will see that they work! Remember that i is just a constant number. Then the rules for differentiation of exponential functions give us: x₁'(t) = (i c)e ^{i c t}
x₁''(t) = (i c) ² e ^{i c t} = −c² x₁(t) which matches the original differential equation (1). You might not know what it means to have an imaginary number as an exponent. If so, read on...

Non-imaginary Solution

To return from the world of imaginary (complex) numbers we use Euler's formula: e ^{i z} = cos z + i sin z Applying Euler's formula to our solutions – equations (2a) and (2b) – we have x₁(t) = e ^{i c t} = cos(c t) + i sin(c t) x₂(t) = e ^{−i c t} = cos(−c t) + i sin(−c t) Using some basic trig identities we can simplify the second solution to be x₂(t) = cos(c t) − i sin(c t) We still have the imaginary number i in these solutions. But now we can use a general property of linear differential equations:
a linear combination of solutions is also a solution A linear combination of things is simply the sum of those things multiplied by constants. For example a linear combination of x, y, z could be 2 x − 0.333 y + 13 z So we can form the following linear combinations which are also solutions but don't involve i: ¹⁄₂ x₁(t) + ¹⁄₂ x₂(t) = cos(c t)
− ⁱ⁄₂ x₁(t) + ⁱ⁄₂ x₂(t) = sin(c t) (remember that i is just a constant like any other number, so we can use it to form linear combinations). Now the general solution to the differential equation is given by a linear combination of the above solutions:

x(t) = a cos(c t) + b sin (c t)

(3)

where a, b are any constants. You can check that this still satisfies equation (1) by plugging it in.

Satisfying the Initial Conditions

We still need to satisfy the initial conditions. Using equation (3) above, the first initial condition x(0) = x₀ becomes a cos(0) + b sin(0) = x₀
a = x₀ To evaluate the second initial condition x'(0) = v₀ let's first calculate the derivative of our solution in equation (3). x'(t) = −a c sin(c t) + b c cos(c t) Now we can evaluate this at time t = 0 and use the second initial condition −a c sin(0) + b c cos(0) = v₀
b c = v₀
b = ^v₀⁄_c Now we've found the particular a, b which match the initial conditions. So the particular solution is

x(t) = x₀ cos(c t) + ^v₀⁄_c sin(c t)

(4)

Check the Solution

We will check that equation (4) is the solution by plugging it into the differential equation (1).

x''(t) = −c² x(t)

(1)

First we find the derivatives of equation (4). x'(t) = −x₀ c sin(c t) + v₀ cos(c t)
x''(t) = −x₀ c²cos(c t) − v₀ c sin(c t) This is the left hand side of equation (1). Now let's calculate the right hand side. −c² x(t) = −c² (x₀ cos(c t) + ^v₀⁄_c sin(c t))
= −c² x₀ cos(c t) − v₀ c sin(c t) and so we see that both sides match – so our solution (4) satisfies the differential equation (1).

Another Form of the Solution

The solution we found,

x(t) = x₀ cos(c t) + ^v₀⁄_c sin(c t)

(4)

may be puzzling to you, because if you played with the simulation you will see that no matter what you do to it, it always exhibits a simple sine motion. Yet the above solution seems to be more complex than that. The answer to the puzzle is that the solution can be simplified to just a sine function using a trig identity. You get a sine function whose phase is shifted. Assuming v₀ ≠ 0, we get: $x(t) = \sqrt{x_0^2 + (v_0/c)^2} \; \sin(c t + \tan^{-1} \frac{c x_0}{v_0})$ So we can see that the behavior is always a simple sine motion. If the velocity is non-zero at the start, then the sine wave is phase shifted.