- #106
Satvik Pandey
- 591
- 12
:Dehild said:I know that you won't. :D
ehild
Yes I will try to learn about them but not now.
:Dehild said:I know that you won't. :D
ehild
If you assume no elasticity, yes. If there is elasticity it becomes very difficult. There may be longitudinal oscillations.Satvik Pandey said:I have a confusion.
If a part of a chain is piled up on a table and rest part of it hangs from the table.And suppose we are asked to find the time in which whole chain falls off the table then will it's answer be the solution of the last equation in post #107?
Thank you!haruspex said:If you assume no elasticity, yes. If there is elasticity it becomes very difficult. There may be longitudinal oscillations.
Apart from the additional /2 in the final integral, which I assume is a typo, it is fine.Satvik Pandey said:Thank you!
Is solution in #post107 correct?
Thank you!Orodruin said:Apart from the additional /2 in the final integral, which I assume is a typo, it is fine.
ehild said:Do not run. Wait till you learn about them in school.
ehild
ehild said:It is correct, Satvik. Good job!
That was the hard way to do it.
The easy method is when one has studied differential equations, and recognises that it is a second order linear DE with constant coefficients.
One knows that the solutions are of the form x=e^{λt}.
Replacing back the trial solution into the DE, you get an equation for the constant λ.
λ^{2}=g/l; λ1=√(g/l), λ2=-√(g/l). Let be k= √(g/l). There are two independent solutions: y1=e^{kt} and y2=e^{-kt}.
Again, when you studied differential equations, you know, that any linear combinations of the independent solutions are a solution again.
So you can write up the general solution as x(t)=c_{1}e^{kt}+c_{2}e^{-kt}.
Fit the constants to the initial conditions: at t=0 x(0)=x_{0} and v(0)=0---> c_{1}+c_{2}=x_{0}.
v(t)=c_{1}ke^{kt}-c_{2}ke^{-kt}
v(0)=0=k(c_{1}-c_{2}).
The constants are c_{1}=c_{2}=x_{0}/2 . The solution is x(t)=(x_{0}/2) ( e^{kt}+e^{-kt}). You need to solve it for t when x=l.ehild