I'm trying to create a physics simulation, and I need to be able to determine the amplitude of the oscillation of a mass-and-spring system given any position that the mass might be in and the velocity of the mass when it is in that position. For example, if the mass in in position $X$ (with the equilibrium being $0$) and the mass has no velocity, then we would know that the amplitude is simply the absolute value of $X$. But how would you find out the amplitude of the oscillation given $V$ and where $X$ is $0$? Or what about where both $X$ and $V$ are not zero? What is the general formula for the amplitude given some position $X$ and some velocity $V$ (i.e. if you throw the spring at velocity $V$ from position $X$ instead of just releasing it from position $X$ in an initial velcotity of zero)?

  • $\begingroup$ $v=\omega \sqrt{A^2-x^2}$ where $\omega$ is the angular frequency and $A$ is the amplitude. $\endgroup$ – harshit54 Jan 26 '19 at 18:11
  • $\begingroup$ @harshit54 Thanks, that's what I needed. $\endgroup$ – ElliotThomas Jan 26 '19 at 18:31

You'll need to know the mass and spring constant as well as the position and velocity to determine the amplitude.

$T = 2\pi\sqrt{\frac{m}{k}}$

$\omega = \frac{2\pi}{T} = \frac{2\pi}{2\pi\sqrt{\frac{m}{k}}} = \frac{1}{\sqrt{\frac{m}{k}}} = \frac{1}{\frac{\sqrt{m}}{\sqrt{k}}} = \frac{\sqrt{k}}{\sqrt{m}} = \sqrt{\frac{k}{m}}$ where $k$ is the spring constant and $m$ is the mass of the mass.

$v = \omega\sqrt{A^2 - x^2}$

$\frac{v}{\omega} = \sqrt{A^2 - x^2}$

$\frac{v^2}{\omega^2} = A^2 - x^2$

$\frac{v^2}{\omega^2} = A^2$

$+\sqrt{\frac{v^2}{\omega^2} + x^2} = A$ (Since we know that the amplitude is positive, we don't need to bother with the plus or minus square root)

$A = \sqrt{\frac{v^2}{\frac{k}{m}} + x^2} = \sqrt{\frac{v^2m}{k} + x^2}$

So that is the formula for the amplitude of a spring given position and velocity.

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  • $\begingroup$ Have you done a dimensional check on these equations? Try it--always a good check. $\endgroup$ – user45664 Jan 27 '19 at 16:53
  • $\begingroup$ @user45664 Thanks. I forgot to square the ω in the denominator of the fraction in the square root. $\endgroup$ – ElliotThomas Jan 27 '19 at 21:04

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