Answer:
Let [tex]A[/tex], [tex]B[/tex], and [tex]k[/tex] denote three constants (with the requirement that [tex]k > 0[/tex].) The following assumes that the mass of this object is [tex]m[/tex]. Assume that [tex]x(t)[/tex] denotes the position of the object at time [tex]t[/tex].
(a) [tex]\displaystyle x^\prime(t) = -\sqrt{\frac{k}{m}}\, A\, \sin\left(t\, \sqrt{\frac{k}{m}}\right) + \sqrt{\frac{k}{m}}\, B\, \cos\left(t\, \sqrt{\frac{k}{m}}\right)[/tex].
(b) [tex]\displaystyle x^{\prime\prime}(t) = -\left(\frac{k}{m}\right)\, A\, \cos\left(t\, \sqrt{\frac{k}{m}}\right) - \left(\frac{k}{m}\right)^{2}\, B\, \sin\left(t\, \sqrt{\frac{k}{m}}\right)[/tex]
Explanation:
The differential equation for a simple harmonic motion might take the following form:
[tex]\displaystyle \frac{d^{2} x}{d t^{2}} = -\frac{k}{m}\, x[/tex].
The minus sign on the right-hand side highlights the fact that the displacement and acceleration of the object should be in opposite directions.
Notice how this equation is in the form of a homogeneous second-order ODE:
[tex]x^{\prime\prime}(t) + \underbrace{P(t)}_{0}\, x(t) + \underbrace{Q(x)}_{\sqrt{k / m}} = 0[/tex]
Let [tex]r[/tex] be a constant. One possible solution to this homogeneous second-order ODE would be in the form [tex]x(t) = e^{r\, t}[/tex], such that [tex]x^{\prime}(t) = r\, e^{r\, t}[/tex] whereas [tex]x^{\prime\prime}(t) = r^{2}\, e^{r\, t}[/tex].
Substitute into the original ODE to obtain:
[tex]\displaystyle \underbrace{r^{2}\, e^{r t}}_{x^{\prime\prime}(t)} + \frac{k}{m}\, \underbrace{e^{r t}}_{x(t)} = 0[/tex].
Rearrange the equation and solve for [tex]r[/tex].
[tex]\displaystyle e^{r t}\, \left(r^{2} + \frac{k}{m}\right) = 0[/tex]
Notice that [tex]e^{r\, t} > 0[/tex]. Hence, it must be true that [tex]\displaystyle r^{2} + \frac{k}{m} = 0[/tex]. Solve for [tex]r[/tex] given that [tex]k > 0[/tex]:
[tex]\displaystyle r_{1, 2} = \pm i\sqrt{\frac{k}{m}}[/tex], where [tex]i[/tex] is the imaginary unit.
The two particular solutions for the ODE would be:
[tex]x_1(t) = e^{\left(i\,\sqrt{k/m}\right)\, t}[/tex] and [tex]x_2(t) = e^{\left(-i\,\sqrt{k/m}\right)\, t}[/tex].
Apply Euler's Formula to rewrite both solutions in terms of trigonometric functions:
[tex]\displaystyle x_1(t) = e^{\left(i\,\sqrt{k/m}\right)\, t}= \sqrt{\frac{k}{m}} \left(\cos\left( \sqrt{\frac{k}{m}} t\right) + + i\, \sin\left( \sqrt{\frac{k}{m}} t\right)\right)[/tex].
[tex]\displaystyle x_2(t) = e^{\left(-i\,\sqrt{k/m}\right)\, t}= -\sqrt{\frac{k}{m}} \left(\cos\left( -\sqrt{\frac{k}{m}} t\right) + i\, \sin\left( -\sqrt{\frac{k}{m}} t\right)\right)[/tex].
The general solution would be in the form:
[tex]\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{\frac{k}{m}} t\right) + B\, \left(i\, \sin\left(\sqrt{\frac{k}{m}} t\right)\right)\right)[/tex],
Where [tex]C_1[/tex] and [tex]C_2[/tex] are constants (not necessarily real numbers.)
Since position is supposed to assume a real value for any real [tex]t[/tex], set [tex]B[/tex] to a multiple of [tex]i[/tex] such that the general solution is real-valued:
[tex]\displaystyle x(t) = C_1\, x_1(t) + C_2\, x_2(t) = A \cos\left(\sqrt{\frac{k}{m}} t\right) + B\, \sin\left(\sqrt{\frac{k}{m}} t\right)\right)[/tex].
Differentiate to obtain general expressions for velocity (first derivative) and acceleration (second derivative.)
