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 $y=Acos(Bt-C)$


$$ a\cos(Bt-C) = A\cos(Bt)\cos(C)+A\sin(Bt)\sin(C) $$ 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 '15 at 19:30
  • $\begingroup$ Not really sure. $\endgroup$ – Vladamir Paklov Feb 19 '15 at 19:44
  • $\begingroup$ I know you can :). it is a rearrangement of $A\cos C =-1/12$? $\endgroup$ – Chinny84 Feb 19 '15 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 '15 at 19:58
  • $\begingroup$ So ω< $3\sqrt(2)/2$ right? $\endgroup$ – Vladamir Paklov Feb 19 '15 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...? $\endgroup$ – Vladamir Paklov Feb 21 '15 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 '15 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.