- #1

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I have big problem with solve this equation:

[tex]m\frac{d^{2}x}{dt^{2}}+ksinx=0[/tex]

I can't go ahead, because I don't know how solve this

[tex]\frac{dx}{\sqrt{cosx}}=\sqrt{\frac{2k}{m}}dt[/tex]

Phizyk

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- Thread starter Phizyk
- Start date

- #1

- 25

- 0

I have big problem with solve this equation:

[tex]m\frac{d^{2}x}{dt^{2}}+ksinx=0[/tex]

I can't go ahead, because I don't know how solve this

[tex]\frac{dx}{\sqrt{cosx}}=\sqrt{\frac{2k}{m}}dt[/tex]

Phizyk

- #2

HallsofIvy

Science Advisor

Homework Helper

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Let v= dx/dt. Then d

Your differential equation can be reduced to vdv/dx= -ksin(x) which is "separable":

mvdv= - k sin(x)dx. Integrating both sides, (m/2)v

v= dx/dt= sqrt((2k/m) cos(x)+ C').

I assume that, in order to get rid of that constant of integration, C', you must have some initial condition on dx/dt.

- #3

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Great. Thanks Hallsoflvy.

- #4

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But this equation [tex]\frac{dx}{dt}=\sqrt{\frac{2k}{m}cosx+C^{'}}[/tex] can I solve? Can I obtain x(t)? For t=0 x=0. It's a equation of motion.

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