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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.

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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.

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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 $$m\frac{d^2x}{dt^2} + kx = 27 \sin 13 t \to \frac{d^2x}{dt^2} + \omega^2 x = \frac{27 \sin 13 t}m = 67.5 \sin 13 t. $$

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  • $\begingroup$ Is k not = mg/s, as opposed to your s/(mg)? $\endgroup$ – user3032755 Mar 9 '15 at 1:06
  • $\begingroup$ @user3032755, the unit of $kx$ that is $unit \,of\, k \times m = Newton$ $\endgroup$ – abel Mar 9 '15 at 1:07
  • $\begingroup$ In your calculations you have the units of $k$ be $\text{m/N}$. The answer should be 1 over that number $\endgroup$ – Dylan Mar 10 '15 at 2:41

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