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# M3 SHM for springs help Watch

1. In the simple harmonic motion for the springs in the edexcel m3 book (pg:57)

it states the acceleration to be away from the particle, how does that work.
you pull the particle and release it, it will move towards equilibrium, how can the acceleration be away from the particle. I don't really get that.

Thanks for helping
2. Hooke's law states that the restoring force on the particle is -kx. Our equation of motion is then ma = -kx. Thus acceleration is in the opposite direction of the motion. If I attach a bungee cord to you and you run away from me you will feel a force acting against your motion and through Newton's second law the acceleration acts in a different direction.
3. but we are talking about the moment we release the particle so when i stop and release my self, obviosly the bungee cord will pull me and i will accelerate through the point where extension is zero. so how is the acceleration is away from centre of oscillation to start with.
btw -kx works fine but in the book, it goes a bit more fundemental rather than only algebraic

4. here is what actually boggled my mind...
after this it says, -T = mẍ and finally proves that it is shm by saying ẍ = -λ/ml x
5. (Original post by thephysicsguy)

here is what actually boggled my mind...
after this it says, -T = mẍ and finally proves that it is shm by saying ẍ = -λ/ml x
It simply means that this is the positive direction for the measurement of the acceleration, not that the acceleration is in this direction...that is why the negative sign comes about!!
6. (Original post by mikelbird)
It simply means that this is the positive direction for the measurement of the acceleration, not that the acceleration is in this direction...that is why the negative sign comes about!!
okay that makes it quite clear but one last question to clarify. When the string goes to the other side of centre of oscillation, now the thrust and the acceleration is in the same direction as we are always measuring acceleration in AP. This means that formula for thrust = -λx/l and for
tension =
λx/l (if that makes sense ) because only this makes the algebra work...
7. There is a further problem here....the text mentions a string or spring....it should only mention a spring because that can be compressed (and thus the thrust is in the opposite direction) with a string the particle will pass through the point of not stretching at all and the particle will not return (as no force except its weight will be acting. The text is definitely wrong at this point!!!
8. (Original post by mikelbird)
There is a further problem here....the text mentions a string or spring....it should only mention a spring because that can be compressed (and thus the thrust is in the opposite direction) with a string the particle will pass through the point of not stretching at all and the particle will not return (as no force except its weight will be acting. The text is definitely wrong at this point!!!
I actually solved this problem and understood it by realizing that in compression, the extension is obviously negative so the maths work. And you are quiet right there, but if it is a string, the particle actually goes to the other side of point A and starts stretching again. So no compression. It is a bit of strange questions these ones.
9. (Original post by thephysicsguy)
I actually solved this problem and understood it by realizing that in compression, the extension is obviously negative so the maths work. And you are quiet right there, but if it is a string, the particle actually goes to the other side of point A and starts stretching again. So no compression. It is a bit of strange questions these ones.
Yes you are quite right....was'nt thinking of the bigger picture!!

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