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- anonymous

Simple harmonic motion question

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- anonymous

Simple harmonic motion question

- chestercat

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- anonymous

it said that the particles acceleration is proportional to displacement but sign is opposite? What does that mean?
I know that the displacement graph looks like this and the acceleration graph looks like this for SHM ....|dw:1441423084425:dw|
The textbook also said that acceleration magnitude is zero when displacement is zero (which for velocity is the max) and acceleration is max when particle at the crest or trough (for velocity is 0). Is that what it means by sign is opposite?

- anonymous

- zepdrix

|dw:1441436026318:dw|

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- zepdrix

Yes, it means opposite sign :)
In this example, I was trying to show that if you're displaced an amount \(\large\rm x\) from equilibrium,
then the acceleration is \(\large\rm -x\) at the same time.
I'm not so good with the physics, it would be a lot easier to explain this using calculus lol oh well!

- IrishBoy123

the key to shm is the restorative force. All proper shm starts with the equation F = -kx
the restorative force F is proportional to displacement x but acts in the other direction [hence - sign], ie back toward the origin/point of equilibrium
everything everything everything follows from that.

- Michele_Laino

the harmonic motion can be modeled by this function:
\[\Large x\left( t \right) = A\cos \left( {\omega t + \phi } \right)\]
we can set \( \Large \phi=0 \), without loss of generality, so we can write:
\[\Large x\left( t \right) = A\cos \left( {\omega t} \right) \qquad \qquad (1)\]
of course, the constant \( \large A \) is the amplitude of our motion.
Now taking the second derivative with respect to time of the last equation, we get:
\[\Large \ddot x\left( t \right) = - A{\omega ^2}\cos \left( {\omega t} \right) = - {\omega ^2}x\left( t \right) \qquad (2)\]
As you can see, your graph are the graphs of the equations \( (1), \; (2) \) respectively
@dareintheren

- Michele_Laino

neglecting the scale factor \( \Large \omega^2 \), of course

- IrishBoy123

and from
\[\Large \ddot x\left( t \right) = - A{\omega ^2}\cos \left( {\omega t} \right) = - {\omega ^2}x\left( t \right) \qquad (2)\]
note that
\[\Large m\ddot x\left( t \right) = F = - m{\omega ^2}x\]
the **restorative** force

- IrishBoy123

and just to be totally clear, the maths follows the idea of a restorative force, and not the other way round:p

- anonymous

Thank you guys for responding, more clear now, I can only medal the best one, but all was good!

- IrishBoy123

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