June 04, 2020, 06:26:12 pm

### AuthorTopic: Spring and collision  (Read 5593 times) Tweet Share

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• Victorian
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##### Re: Spring and collision
« Reply #15 on: April 28, 2008, 11:03:45 pm »
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Also how did you get

$A = \frac{v_2T}{2\pi}$?

#### Neobeo

• Victorian
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##### Re: Spring and collision
« Reply #16 on: April 28, 2008, 11:51:49 pm »
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Also how did you get

$A = \frac{v_2T}{2\pi}$?

Well, by first principles you could derive for the spring:

$v(t) = A\frac{2\pi}{T} \cos(\frac{2\pi t}{T})$

and then $v_2 = v(0)$
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#### /0

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##### Re: Spring and collision
« Reply #17 on: May 01, 2008, 01:36:25 am »
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Sorry for replying so late, but I checked in the book and it says

$x(t) = x_m \cos{(\omega t + \phi)}$

$v(t) = -\omega x_m \sin{(\omega t + \phi)}$

I'm guessing this is if the block is being stretched at t = 0, is this right?

When the block starts at rest, does that mean

$x(t) = x_m \sin{(\omega t + \phi)}$ ?

Also, which direction do you usually choose to be 'negative'? The direction in the compression of the spring, or the direction stretching the spring?

Thanks again :p