differentiating energy...Watch

#1
I am having a little trouble diff the following eqn for energy:

E=(1/2)mv2 +λ(1-cosµx)

v is velocity ie. dx/dt and x is distance

i want to diff with respect to t.

i think my confusion is all the variables. i would be grateful if someone could give some suggestions as to the techniques i use etc...
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13 years ago
#2
x and v are functions of t, and lambda and mu are constants. I'm assuming m is also a constant, since that clearly isn't a relativistic equation. Write v as v(t) and x as x(t), and then when you work out dE/dt, use usual methods on v(t)2 and cos[mu x(t)].
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#3
(Original post by AlphaNumeric)
x and v are functions of t, and lambda and mu are constants. I'm assuming m is also a constant, since that clearly isn't a relativistic equation. Write v as v(t) and x as x(t), and then when you work out dE/dt, use usual methods on v(t)2 and cos[mu x(t)].
thanks an,

i get

dE/dt=mvv'+λt+λ(uv)sin(ux)

is this right??

xx
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13 years ago
#4
Yep
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#5
(Original post by AlphaNumeric)
Yep
i want to write it in the form of an eqn of motion. i assume if i equate to 0(energy is constant) and rearrange for mv' i will have what i want?
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13 years ago
#6
Yep
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13 years ago
#7
Hang on a sec, I didn't see you had the λt term. That shouldn't be there. You should have

dE/dt=mvv'+λ(uv)sin(ux)

dE/dt = 0 so

0=mvv'+λ(uv)sin(ux)
-ma = λusin(ux) (assuming v isn't 0)
a = -(λu/m)sin(ux)
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#8
yep just ignore me i integrated the λ by mistake! thanks xxx
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#9
(Original post by realicetic)
yep just ignore me i integrated the λ by mistake! thanks xxx
now for eqm points on x axis. i take it these occur at -λµsin(µx)=0?

but that means eqm points are at periods of 2π/µ on the sin graph? ie. x=0,2π/u,...etc

is i right??

im quite enjoying this question!
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#10
am i?
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13 years ago
#11
Yup.
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#12
(Original post by Dimez)
Yup.
thankyou i like your little aviator thing
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13 years ago
#13
*avatar
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