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I am a little bit confused about the implication from my computation. Must have done something wrong in the computation.

Assume we hold the top end of the mass-spring-mass system in the air, and the system is at equilibrium. Now release the top end mass to let the system have a free fall, what would be the tension in the spring (assume massless) ? I solve a simple ODE and it turns out the tension is given by a sinusoid with magnitude the initial tension.

However, this seems against intuition. I must have solved the wrong ODE...I could have done better with the drawing but I'm lazy.

I'm confused because some people told me accelerometer is based on this principle...what would the accelerometer read in this case... enter image description here

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  • $\begingroup$ What ODE did you use here? Could you also include a diagram of your mass-spring system? $\endgroup$ – Kyle Kanos Mar 10 '15 at 14:38
  • $\begingroup$ Your result sounds correct. You might want to add a diagram and your computation for future reference - it will make the question more valuable to future visitors. $\endgroup$ – Floris Mar 10 '15 at 14:56
  • $\begingroup$ A practical accelerometer is most likely critically damped - this will limit the oscillation. $\endgroup$ – Floris Mar 10 '15 at 15:26
  • $\begingroup$ @Floris thank you so much, now this makes sense, dominated by exponential decay... $\endgroup$ – Troy Woo Mar 10 '15 at 15:27
  • $\begingroup$ Incidentally for your system the frequency is given by $\omega=\sqrt{\frac{\mu}{k}}$ where the reduced mass $\mu=\frac{m_1\ m_2}{m_1+m_2}$ $\endgroup$ – Floris Mar 11 '15 at 1:40
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Of course, the solution is correct. Letting the system fall, the gravitation doesn't act on it anymore, and you have a mass-spring-mass system oscillating in absence of external forces.

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  • $\begingroup$ Succinct and correct. $\endgroup$ – Floris Mar 10 '15 at 14:55

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