# Undamped spring mass system

I have this study guide for an upcoming test for DE class I'm trying to figure out.

A mass of 400 grams stretches a spring by 5 centimeters.
(a) Find the spring constant k, the angular frequency ω, as well as the period T and frequency f of free undamped motion for this spring-mass system.
(b) Find the general solution of the DE for the free spring-mass system.
(c) Suppose that an exterior force of F(t) = 27sin(13t) Newtons

acts on the spring-mass system. Find the equation of motion of the system if the mass initially is at rest in its equilibrium position.

I know K is 784 (or do I need to convert to 5 centimeters to 0.05 meters?) and w is sqrt(k/m), but I'm not sure what I need to find T and F. I can find the general solution, but then I have no clue on what to do with part c.

You don't need to convert because your units of mass are grams, so you are using CGS system of units. For a: $\omega=2 \pi/T$ For c: You have to add the new force to Newton's law.
before you can set the equation you need $k,$ the spring constant. we will do all this in metric system. $k = \frac{5/100}{400 \times 9.8/1000} = 0.0127\,N/m, \omega^2 = \frac{k}{m} = (0.1785)^2sec^{-2}, T = \frac{2\pi}{\omega} = 35.185 \, sec$
equation of motion is $$m\frac{d^2x}{dt^2} + kx = 0 \to \frac{d^2x}{dt^2} + \omega^2 x = 0 \text{ where x is deviation from equilibrium.}$$ the general solution is $$x = A\cos(\omega t - \phi).$$
the equation of motion for a forced system is $$m\frac{d^2x}{dt^2} + kx = 27 \sin 13 t \to \frac{d^2x}{dt^2} + \omega^2 x = \frac{27 \sin 13 t}m = 67.5 \sin 13 t.$$
• @user3032755, the unit of $kx$ that is $unit \,of\, k \times m = Newton$ – abel Mar 9 '15 at 1:07
• In your calculations you have the units of $k$ be $\text{m/N}$. The answer should be 1 over that number – Dylan Mar 10 '15 at 2:41