motion with frictional force but without driving force

LockeZz

hm.. i think i know where did i went wrong, i miss out the secant:

[tex]

x-x_0=-\frac{m}{\beta}(\ln(\sec(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1))-\ln(\sec(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)))
[/tex]

kuruman

Incorrect. I suggest you look it up on the web. Just google "Integral Tables" and take your pick.

LockeZz

i have check on the table.. is cosine..
[tex]


x-x_0=-\frac{m}{\beta}(\ln(\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1))-\ln(\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)))

[/tex]

kuruman

Can you make the expression more compact? What is the difference between two logarithms? Also set t₀=0 and see what you get.

LockeZz

ok.. the logarithm still can be simplify :

[tex]



x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\a lpha}})-\frac{\sqrt{\alpha\beta}}{m}t_0)})


[/tex]
so.. for t₀=0.. i will get :

[tex]



x-x_0=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{ \alpha}})-\frac{\sqrt{\alpha\beta}}{m}t_1)}{\cos(\arctan(v_0\sqrt{\frac{\beta}{\a lpha}}))})


[/tex]

kuruman

Good. I can see the light at the end of the tunnel. Can you? What do you think you should do next?

LockeZz

get rid of the x₀ ad substitute the t₁?

kuruman

Yes, you may assume that x₀=0. Substitute t₁ and also the time has come to let v₀ become very large.

LockeZz

after the substitution, i get :
[tex]
x=-\frac{m}{\beta}(\ln(\frac{\cos(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})-\frac{\pi}{2})}{\cos(\arctan(v_0 \sqrt{\frac{\beta}{\alpha}}))})
[/tex]

since
[tex]


t_1=m(\frac{\pi}{2})\sqrt{\frac{1}{\alpha\beta}}


[/tex]
should i bother about v₀?

kuruman

Not yet. Do you recognize the ratio in the argument of the log for what it is? Specifically, can you simplify the numerator some?

LockeZz

i think i can convert to tangent:

[tex]
x=-\frac{m}{\beta}(\ln(\tan(\arctan(v_0\sqrt{\frac{\beta}{\alpha}})))
[/tex]

then simplify again:

[tex]
x=-\frac{m}{\beta}(\ln(v_0\sqrt{\frac{\beta}{\alpha}}))
[/tex]

kuruman

Note that if you let v₀ become very large, you get an infinity. I hadn't appreciated that in the beginning. One thing that bothers me is the negative sign up front. Certainly x must be positive and the argument of the log is greater than 1 especially if v₀ becomes very large. Check your integral tables and don't forget that you are integrating

[tex]
\int Tan(\delta - t)dt
[/tex]

There is a negative sign in front of t.

Check that and you are done.

LockeZz

hm.. does that mean that the negative sign represent a mistake in the expression?

kuruman

All I am saying is that

[tex]

\int Tan(\delta - t)dt=Log[Cos(t-\delta)]

[/tex]

LockeZz

I shall check back from the beginning the see if any mistake again.. thx for the guidance by the way.. i really appreciate on your help.. Thank you very much :)