This is a Physics problem linked to differential equations.
In order to solve it, we define our initial values:
[tex]m=0.1Kg\\x=0.05m\\g=9.8m/s^2[/tex]
We have given initial conditions,
The mass starts from a point of displacement at rest, that is,
[tex]U(0)=0[/tex]
However, it has a resting point speed equal to 30cm/s
[tex]U'(0) = 0.3m/s[/tex]
By definition we know that,
By Newton's law we know that
[tex]F=mg[/tex]
And by Hook's law we know that
[tex]F=kX[/tex]
where,
k elongation / elastic constant
X the change or distance traveled
Equating the expressions we have
[tex]mg=kX[/tex]
Solving for k,
[tex]k=\frac{mg}{x}[/tex]
Replacing,
[tex]k= \frac{(0.1)(9.8)}{0.05}[/tex]
[tex]k=19.6N/m[/tex]
The differential equation given for the movement is,
[tex]mu(t)''+ku(t)=0[/tex]
[tex]0.1 u(t)''+19.6u(t)=0[/tex]
[tex]u(t)''+196u(t)=0[/tex]
Being a differential equation we can obtain its auxiliary equation, given by
[tex]r^2+196=0[/tex]
[tex]r= \pm 14i[/tex]
By definition, this is a second order linear differential equation,
We know that every function of the form
[tex]y''+y=0,[/tex] its solution is
[tex]y(t)=c*cos(t)+d*sin(t)[/tex]
Applying the definition,
[tex]u(t)=c*cos(14t)+dsin(14t)[/tex]
And the first derivative would be
[tex]u'(t)=14c*cos(14t)-14dsin(14t)[/tex]
Applying our ideal conditions we can find d and c, so
For u(0) =0
[tex]u(0) = c*cos(14*0)+sin(14*0)[/tex]
[tex]c=0[/tex]
For u(0)=0.3
[tex]u'(0)=14c*cos(14*0)-14dsin(14*0)[/tex]
[tex]u'(0)=3/140[/tex]
We have know that our equation for the position of the mass at any time is,
[tex]u(t)=\frac{3}{140}sin(14t)[/tex]
Solving for the question: The time when the mass return to equilibrium we have
[tex]u(t)=0[/tex]
[tex]0=\frac{3}{140}sin(14t)[/tex]
[tex]\rightarrow sin(14t)=0[/tex]
[tex]t=\frac{\pi}{14}[/tex]
