# Question #5331e

Oct 9, 2015

0.253 kg on both Earth and Mars.

#### Explanation:

At any time, the net force is gravitational force - spring force.
Thus, I write this equation: $m \frac{{d}^{2} x}{\mathrm{dt}} ^ 2 + 10 x = m g$
10 is the spring constant k.

The equation for this system is
$x \left(t\right) = {c}_{4} \cos \left(\sqrt{\frac{10}{m}} t\right) + {c}_{5} \sin \left(\sqrt{\frac{10}{m}} t\right) + m g$.
(If you want to know more about how this is derived. Look below to Heavy Math Part section.)

${c}_{4}$ and ${c}_{5}$ cannot be determined without more information, such as value of $x \left(0\right)$ when $t = 0$.

However, we can still use this equation to find the oscillation frequency. The only oscillatory components are ${c}_{4} \cos \left(\sqrt{\frac{10}{m}} t\right)$ and ${c}_{5} \sin \left(\sqrt{\frac{10}{m}} t\right)$.
For frequency to be $1$, $\sqrt{\frac{10}{m}} = 2 \pi$ which means $m = 0.253$.
Since frequency is not dependent on $g$, the mass to create frequency of $1 {s}^{-} 1$ or period of $1 s$ would be $0.253 k g$ on both Earth and Mars.

Heavy Math Part

So what we have now is a second-order nonhomogeneous differential equation.

We first solve the left-hand side. That is: $m \frac{{d}^{2} x}{\mathrm{dt}} ^ 2 + 10 x = 0$
Substituting $\frac{\mathrm{dx}}{\mathrm{dt}}$ with $s$ gives $m {s}^{2} + 10 = 0$.
Solving for $s$ gives $s = \sqrt{\frac{10}{m}} i , - \sqrt{\frac{10}{m}} i$.

The general solution becomes $x \left(t\right) = {c}_{1} {e}^{\sqrt{\frac{10}{m}} i} + {c}_{2} {e}^{- \sqrt{\frac{10}{m}} i} + {c}_{3}$.
Substituting Euler's formula ${e}^{i t} = \cos \left(t\right) + i \sin \left(t\right)$ gives $x \left(t\right) = {c}_{1} \left(\cos \left(\sqrt{\frac{10}{m}} t\right) + i \sin \left(\sqrt{\frac{10}{m}} t\right)\right) + {c}_{2} \left(\cos \left(\sqrt{\frac{10}{m}} t\right) - i \sin \left(\sqrt{\frac{10}{m}} t\right)\right) + {c}_{3}$.

To get rid of imaginary components, we assume ${c}_{1}$ and ${c}_{2}$ to be multiples of certain ${c}_{1}$ and ${c}_{2}$ values.
If ${c}_{1} = {c}_{2} = {c}_{4} / 2$, $x \left(t\right) = {c}_{4} \cos \left(\sqrt{\frac{10}{m}} t\right)$.
If ${c}_{1} = {c}_{5} / \left(2 i\right)$ and ${c}_{2} = - {c}_{5} / \left(2 i\right)$, $x \left(t\right) = {c}_{5} \sin \left(\sqrt{\frac{10}{m}} t\right)$

Now we have an equation without imaginary components:
$x \left(t\right) = {c}_{4} \cos \left(\sqrt{\frac{10}{m}} t\right) + {c}_{5} \sin \left(\sqrt{\frac{10}{m}} t\right) + {c}_{3}$.

We are done with the left side of the first equation. Now we find ${c}_{3}$ using right side of the first equation.

$\frac{{d}^{2} x}{\mathrm{dt}} ^ 2 = - 10 {c}_{4} / m \cos \left(\sqrt{\frac{10}{m}} t\right) - 10 {c}_{5} / m \sin \left(\sqrt{\frac{10}{m}} t\right)$
$m \frac{{d}^{2} x}{\mathrm{dt}} ^ 2 + 10 x = - 10 {c}_{4} \cos \left(\sqrt{\frac{10}{m}} t\right) - 10 {c}_{5} \sin \left(\sqrt{\frac{10}{m}} t\right) + 10 {c}_{4} \cos \left(\sqrt{\frac{10}{m}} t\right) + 10 {c}_{5} \sin \left(\sqrt{\frac{10}{m}} t\right) + {c}_{3} = {c}_{3} = m g$
Thus, $x \left(t\right) = {c}_{4} \cos \left(\sqrt{\frac{10}{m}} t\right) + {c}_{5} \sin \left(\sqrt{\frac{10}{m}} t\right) + m g$.