3
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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)|} $

$\endgroup$
1
$\begingroup$

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?.

$\endgroup$
  • $\begingroup$ 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

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.