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Maximum Acceleration |
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| Nov13-12, 12:51 PM | #1 |
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Maximum Acceleration
Helping a friend out with some physics 1 mid-semester and It's been a little since I have had a physics course.... so the question is
"If a 30 g bird lands on a slender branch where it oscillates up and down with simple harmonic motion of amplitude 3.00 x 10^-2m and period 1.2 s"... trying to find an equation to solve for max acceleration, expressed as a fraction of the acceleration of Gravity as well as maximum speed... any help would really be appreciated and thanks a lot to anyone that reads this |
| Nov13-12, 12:53 PM | #2 |
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SHM is defined as
a= -ω2x Thus a will be max when x (displacement) is maximum. So at what value of x will the displacement be maximum? |
| Nov13-12, 12:54 PM | #3 |
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I don't have a key and I feel like I may be way off but here is what I have so far
f= 1/t so F=.83 Hz A= (2(3.14)f)^2 x so A=27x then for X, X=Acos2(3.14)f(t) |
| Nov13-12, 12:55 PM | #4 |
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Maximum Acceleration
wouldn't it be 3?
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| Nov13-12, 12:56 PM | #5 |
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or would it be 6? I am confused in what to do with amplitude, if amplitude is 3 cm, that would mean peak to trough is 6 cm? Correct? Is that what I would go with?
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| Nov13-12, 01:07 PM | #6 |
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Its asking for the max speed and acceleration of the bird, the thing that throws me off is I don't know if it is asking for the speed the bird landed on the branch or the speed which it actually oscillates
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| Nov13-12, 02:36 PM | #7 |
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| Nov13-12, 03:36 PM | #8 |
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Ok great, A as in amplitude or acceleration... sorry, thanks again
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| Nov13-12, 06:34 PM | #9 |
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| Nov14-12, 04:26 PM | #10 |
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So then, would speed maximum or minimum when the finch hits max acceleration?
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| Nov14-12, 06:26 PM | #11 |
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Remember the bird is landing from some height. So as it hits, it will be at maximum velocity. If we go back to your equation for displacement x=Asin(ωt), how would you find velocity from this? |
| Nov14-12, 06:48 PM | #12 |
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makes sense, since the bird has yet to transfer its energy to the branch... Plug in amplitude -sin and multiply with the angular force times time period...?
So to get this straight, and I really appreciate all of this, A as in amplitude A= (2(3.14)f)^2 x so A=27x I would plug A= 27x and then solve for X which would be my maximum acceleration? In this case 3=27x so x would be 9 cm/s Maximum Velocity would be the moment before impact? The last thing it asks |
| Nov14-12, 06:50 PM | #13 |
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is to express it as a fraction of the acceleration of gravity, so just 9cm/s^2?
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| Nov14-12, 06:59 PM | #14 |
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You can express the acceleration as a fraction of gravity as both are acceleration terms. If you want to express velocity in terms of gravity then your units will not make too much sense. |
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