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Simple Harmonic Motion - Acceleration and Time Period

  • Thread starter ProPM
  • Start date

I am a bit confused regarding the relation between the acceleration and time period in simple harmonic motion.

This questions asks what would be one change to the motion of a car's wing mirror in case the glass was switched for a heavier one. I don't understand for example, how would the period change. I can think that the same force will have to move a greater mass over the same distance but, wouldn't the amplitude decrease if the mass increases?



Homework Helper
the oscillation is going to occur around the _equilibrium_ location
(more massive glass will have a lower equilibrium);
the oscillation Energy depends on Amplitude (squared, times ½k)
but the frequency depends on the spring "stiffness" compared to the inertia
(inside a square root). If you have the same Force, but a bigger mass, what can you tell about the acceleration at these turn-around points?
(that is where it spends most of its time, since its speed is slow there).
Sorry, I couldn't follow very well the beginning. But, for the same force with a bigger mass the acceleration will decrease.

So, from the equation, either omega must decrease (frequency decreases/time period increases) or the displacement will decrease...

How do I know which of these will happen?


Homework Helper
IF the oscillation has the same (total) Energy, it has the same maximum PE (spring),
so it will have the same amplitude (maximum displacement from equilibrium).
. . . . if the mass is doubled . . .
acceleration will be halved, in x = ½at^2 , (here, t is roughly time for ¼ oscillation)
but it has to get just as far, so t^2 is doubled . . . so t is increased by sqrt(2)
that's why the omega formula is squareroot(k/m).

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