M4 - need help on harmonic motion questions Watch

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cktlee1
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Could someone teach me how to do question 4 and 5?
I have no problem have the second order differentiation, ie the pure maths.
but the mechanics side..my teacher did not teach any conditions for certain oscillation or limits or tending to values or graphs or whatever...

I hate my further maths teacher....useless! never teach the hard stuff, just teach the simple stuff like eaxmple 1 of each chapter, and go "u can think about it yourself" not explaining other stuff......

Please could someone answer question 5 with full method...and then teach me conditions that I meant to know?

Thanks for the help people!
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Squishy
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4) A force diagram may help, but the situation's quite simple. You have a 1kg particle, P, that's directed towards a point O. The force pushing P has magnitude 2x N and a resistive force in the opposite direction has magnitude 3v N, where v is the velocity of P.

By Newton's second law:

2x - 3v = - ma
m=1, so:
a - 3v + 2x = 0

where a (acceleration) is the second differential of x with respect to t (time) and v (velocity) is the first.

If you're comfortable solving this, you can use the boundary conditions to get x = 2e^t - e^2t

Then when t=0.1, x=0.99m

5) Not sure how I'd answer this one. I suppose the simplest way is just to sketch it. If you've never seen that kind of function drawn before, it looks something like the curve here:

http://www.ltcconline.net/greenl/cou...er/vibrat4.gif

Because it's a cos curve, it repeats with constant period. Also, a geometric progression has exponential growth/decay, which is quite obviously the case here because there's an e^(-t) term that's controlling the amplitude of the curve.

For c), just differentiate and equate. It's only pure maths.

Hope that's some use.
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dvs
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4.
Use F=ma. This gives you:
-2x - 3x' = x'' [since m=1]
x'' + 3x' + 2x = 0
The auxilary equation of this ODE is:
m^2 + 3m + 2 = 0
(m+1)(m+2) = 0
m=-1 or m=-2
So the general solution is:
x = Ae^(-t) + Be^(-2t)

P starts from rest when x=1 and t=0. So:
1 = A + B (1)
We als have:
x' = -Ae^(-t) - 2Be^(-2t) [by differentiating the general solution]
Substitute the values t=0 and v=x'=0 into this to get:
0 = -A - 2B (2)
Solve (1) & (2) simultaneously to get:
A=2 & B=-1

So:
x = Ae^(-t) + Be^(-2t)
x = 2e^(-t) - e^(-2t), which is x as a function of t.

To find t when x=1/10 we just substitute those values in x(t) to get:
1/10 = 2e^(-t) - e^(-2t)
e^(2t) - 20e^t + 10 = 0
e^t = (20 +/- sqrt[360])/2 = 10 +/- 3sqrt[10]
e^t = 10 + 3sqrt[10], since e^t is always positive
t = ln(10 + 3sqrt[10])

Read the book and try to do the examples, then attempt Q5. If you're still having problems, post back.
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dvs
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Too slow, as usual.
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Squishy
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(Original post by dvs)
Too slow, as usual.
Actually, we get different answers...you're probably right though, since it's been a while since I've done these things.
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