I have the following mechanical vibrations problem.
A mass of $150$ g stretches a spring $3.828125$ cm. If the mass is set in motion from its equilibrium position with a downward velocity of $50$ cm/second, and if there is no damping, determine the position $u$ of the mass at any time $t$.
When does the mass first return to its equilibrium position?
I am not sure what the spring constant, $k$, is. I did $k=.98/0.03828125$, assuming $k=.98/L$, where $L$ was the length in meters. I ended up getting $u(t)= \cfrac{25\sqrt 6}{16}\sin\big(\frac{16 \sqrt 2t}{\sqrt 3}\big)$, which was incorrect.
For the second part, I assumed the period of $\sin(t)$ was $2\pi$, so $\sin\big(\frac{16 \sqrt 2t}{\sqrt 3}\big)$ would have a period of $\frac{\pi \sqrt6}{16}$, and that it would first return to its equilibrium position in half that time (going up), to make the answer $\frac{\pi \sqrt6}{32}$. However, this was also incorrect, probably because of the incorrect equation I did in part 1.
Any help is appreciated. Thank you
Best Answer
For the first part, we usually determine $k$ by using the equation $kL=mg$ (the spring force and the force of gravity cancel out during equilibrium). So we have:
$$kL=mg\implies k=(9.8)(150)(0.03828125)^{-1}=38400,$$
which gives the differential equation $150u''+38400u=0$.
You can now use a characteristic equation to determine the imaginary roots $r_1=16i$ and $r_2=-16i$.
The solutions are then of the form $u(t)=c_1\cos(16t)+c_2\sin(16t)$, which you can use the initial condition to solve for precisely.
I hope I wasn't too vague about the solution process after $k$. It appears that you already know how to solve this type of equation, so I mainly focused on finding $k$. If you need some clarification on the steps afterwards just leave a comment below.