A ball of metal weighing 3 pounds stretches a spring by 3 inches. If the ball is pushed upward a distance of 1 inch, and then set into motion with a downward velocity of 2 ft/sec, neglecting air resistance, find the position of the ball at time, t.

$\frac{3}{32}y''+12y=0$, $y(0)=\frac{-1}{12} ft$, $y'(0)= 2 ft/s$

Characteristic equation: $\frac{3}{32}r^2+12=0$, which yields $r=±8i(2)^\frac{1}{2}$


Is this correct?

How do I write it in terms of just cosine or just sine? like $y=Acos(Bt-C)$


2 Answers 2


$$ a\cos(Bt-C) = A\cos(Bt)\cos(C)+A\sin(Bt)\sin(C) $$ Compare terms with your solution you find that $$ B = 8\sqrt{2} $$ And that $$ A\cos C = -\frac{1}{12}\\ A\sin C = \frac{\sqrt{2}}{8}. $$ So you can find $C$ by using the inverse tangent. As you can see this method doesn't yeild the coefficent $A$ but you can find it. Ps I have not checked your solution directly. But a quick check to see if a solution is correct for a given Ode is to put it back in..this is the last trick (but the easiest step) that people learn. So try it. :)

To answer the second question remember that adding a dampening term is $$ \frac{3}{32}r^2 -\omega r +12 = 0 $$ So now to have a periodic solution we require to have complex roots. What conditions on $\omega$ do you need to fulfil this criteria?

  • $\begingroup$ I do know how to do it. The elements are in the answer, in particular if you look at any of the equations I have $A$ present can you figure out given $C$ how we could get $A$? $\endgroup$
    – Chinny84
    Feb 19, 2015 at 19:30
  • $\begingroup$ Not really sure. $\endgroup$ Feb 19, 2015 at 19:44
  • $\begingroup$ I know you can :). it is a rearrangement of $A\cos C =-1/12$? $\endgroup$
    – Chinny84
    Feb 19, 2015 at 19:49
  • $\begingroup$ There you go. So you can now get $A$ right? I have also edited my answer to help with the second part. $\endgroup$
    – Chinny84
    Feb 19, 2015 at 19:58
  • $\begingroup$ So ω< $3\sqrt(2)/2$ right? $\endgroup$ Feb 19, 2015 at 20:04

For a harmonic motion with the equation

$$ y'' + \omega^2y = 0 $$

In physics, the general solution has the form:

$$ y = A\cos(\omega t + \phi) $$

In this case, $\omega = 8\sqrt{2}$ so $y = A\cos(8\sqrt{2}\, t + \phi)$

Substituting the initial values:

$$ y(0) = A\cos \phi = -\frac{1}{12} $$ $$ y'(0) = -8\sqrt{2}\, A \sin \phi = 2 $$

which yields $A =\frac{\sqrt{22}}{24}$ and $\tan\phi = \frac{3}{\sqrt{2}}, \,\pi < \phi < \frac{3\pi}{2}$.

You should stick to your original solution as neither of those numbers are "nice".

  • $\begingroup$ What is this...? $\endgroup$ Feb 21, 2015 at 23:04
  • $\begingroup$ @VladamirPaklov What is what? I'm using the solution form that you're requesting, $y = A\cos(Bt + C)$ $\endgroup$
    – Dylan
    Feb 24, 2015 at 22:03

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