Prove that the motion of a mass $m$ on a linear spring with constant $k$, has the form $$y (t) = A \sin(wt+f),$$ where $t$ is the time and $A, w, f$ are constants. We know that for $t = 0, y(0)=y_{0}$ and $y'(0)=v_{0}$. If, in addition, the mass is subject to external force $F (t) = F_{0} \sin (w_{0}t)$, where $F_{0}$ the amplitude and $w_{0}$ the cyclic frequency, then calculate the amplitude of the motion.

When the mass is subject to external force $F (t) = F_{0} \sin (w_{0}t)$,we get this differential equation: $$y''+w^{2}y=\frac{F_{0}}{m} \sin(w_{0}t),$$ which has the solution: $$y(t)=c_{1} \cos(wt)+c_{2} \sin(wt)+\frac{F_{0}}{m(w-w_{0}^{2})} \sin(w_{0}t),$$ where $c_{1}=y_{0} $ and $ c_{2}=\frac{v_{0}}{w}-\frac{F_{0}w_{0}}{mw(w-w_{0}^{2})}$. Right? But how can I find the amplitude of the motion?

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    $\begingroup$ Try: $F=my''(t)$. $\endgroup$ – jinawee Feb 2 '14 at 23:04
  • $\begingroup$ You should have $w^2-w_0^2$ in your solution not $w-w_0^2$. $\endgroup$ – joshphysics Feb 3 '14 at 2:35
  • $\begingroup$ If $\omega\neq\omega_0$ then you will get interference in the output: $$y(t)=A\cos\left(\frac{\omega+{\omega}_0}{2}t+\phi\right)\cos\left(\frac{\omega-\omega_0}{2}t+\phi\right)$$ So in order for your initial equation to be true you would have to assume that $\omega=\omega_0$. $\endgroup$ – fibonatic Feb 3 '14 at 14:47

Amplitude is the maximum displacement at steady state. That is okay for a working definition. So, all you need to do is find out is the maximum value of $|y(t)|$ at steady state.

There some small algebraic errors in your solution. That said,

$y(t) = y_{cf}(t) + y_p(t)$

To illustrate my point better, let me use the following form for $y_{cf}(t)$

$ y_{cf}(t) = A e^{-i\lambda_1t} + Be^{-i\lambda_2t} $ with $ \lambda_1 , \lambda_2 > 0$

There is no difference between this and the form with sines and cosines apart from the values of the constants.

At steady state, that is, $\lim{t\to\infty}$ , $y_{cf}(t) = 0$

This means that the amplitude is simply, $|y_p(t)|= |\frac{F_0}{m(\omega^2 - \omega_0^2)}\sin (\omega_0 t)| = \frac{F_0}{|m(\omega^2 - \omega_0^2)|} $


Ampliyude $A$ will be determined by
$[\dfrac{v_0}{\omega ^2} + x_0^2]= A^2 $

To see how this equation is derived read this answer given by me:
i've posted here:
Using $\sin()$ or $\cos()$ for computing SHM?.

  • $\begingroup$ By the way anupam, I noticed your profile says "Please delete me" so if you wanted your account deleted, please see the relevant page in the help center. (If not, just ignore this.) $\endgroup$ – David Z Feb 3 '14 at 13:55
  • $\begingroup$ @DavidZ can i re-ask those questions which are deleted by the Community moderator? $\endgroup$ – user31782 Feb 3 '14 at 15:01
  • $\begingroup$ All those questions were highly downvoted and/or closed before being deleted, so no you shouldn't, not unless you have some way to improve them so that they don't get downvoted/closed again. $\endgroup$ – David Z Feb 3 '14 at 15:40
  • $\begingroup$ @DavidZ What is the current policy about that ? Should i ask on meta? $\endgroup$ – user31782 Feb 3 '14 at 15:49

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