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m3 smh springs with upload (I hope) watch

1. I have a problem with understanding Hooke's Law and F = ma combined.

In the question in the attachment, the particle is to the left of the centre of oscillation and acceleration is always in the direction of x increasing and T is also in that direction?

for part a) Is the string taut at the stating position? If so why is it still -T? Is it anything to do with x now being -x?

If it's only taut at when the extension is +x the I would have had to assume shm to show shm.

I think I need some serious help in understanding.

Thanks
Attached Files
2. m3.doc (428.0 KB, 48 views)
3. (Original post by maggiehodgson)
I have a problem with understanding Hooke's Law and F = ma combined.

In the question in the attachment, the particle is to the left of the centre of oscillation and acceleration is always in the direction of x increasing and T is also in that direction?

for part a) Is the string taut at the stating position? If so why is it still -T? Is it anything to do with x now being -x?

If it's only taut at when the extension is +x the I would have had to assume shm to show shm.

I think I need some serious help in understanding.

Thanks
you define an origin O
then say x say to the right
the direction of x, x dot, x double dot is to the right

when the particle is to the right of O, the acceleration is proportional to minus x
as x is positive then acceleration is pointing left

when the particle is to the left of O, the acceleration is still proportional to minus xas x is now negative then acceleration is now positive (minus, minus) so the acceleration is pointing right

when the particle is to the right the acceleration is proportional to minus xas x is positive then acceleration is pointing left

when the particle is to the right the acceleration is proportional to minus xas x is positive then acceleration is pointing left
4. (Original post by TeeEm)
you define an origin O
then say x say to the right
the direction of x, x dot, x double dot is to the right

when the particle is to the right of O, the acceleration is proportional to minus x
as x is positive then acceleration is pointing left

when the particle is to the left of O, the acceleration is still proportional to minus xas x is now negative then acceleration is now positive (minus, minus) so the acceleration is pointing right

when the particle is to the right the acceleration is proportional to minus xas x is positive then acceleration is pointing left

when the particle is to the right the acceleration is proportional to minus xas x is positive then acceleration is pointing left
So, I've just misinterpreted the video. When it say acceleration is always to the right it means when there's an extension to the right? And the book says The acceleration is always in the direction of x increasing so it means mod x increasing?

Am I just taking everything too literally? (BTW I'm self studying so have no one other than TSR to mither about these things)

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Updated: January 13, 2016
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