- #1

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- Homework Statement:
- At t=0, a particle of mass m having velocity u starts moving through a liquid kept in a horizontal tube and experiences a drag force F = -kdx/dt. It covers a distance L before coming to rest. If the times taken to cover the distances L/2 and L/4 are t2 and t4 respectively, then the ratio t2/t4 (ignoring gravity) is ?

- Relevant Equations:
- F = -kdx/dt

Since given

so I equated

which gave

hence

and using

then I integrated

From this I got

Then I substituted L/2 = mu/k (1 - exp(-kt2/m) )

and L/4 = mu/k (1 - exp(-kt4/m) )

Now after this, I'm not getting how to find the ratio t2/t4 from the last two expressions of L/2 and L/4.

*F = -kdx/dt*so I equated

*mx'' = -kx'*which gave

*x(t) = A + B exp(-kt/m)*hence

*v(t) = (-kB/m) exp(-kt/m)*and using

*v(0) = u*,**v(t) = u exp(-kt/m)**then I integrated

*dx = v(t)dt*,*dx from 0 to L and v(t)dt from 0 to t*to find the distance covered L in terms of time taken t.From this I got

**L = mu/k (1 - exp(-kt/m) )**Then I substituted L/2 = mu/k (1 - exp(-kt2/m) )

and L/4 = mu/k (1 - exp(-kt4/m) )

Now after this, I'm not getting how to find the ratio t2/t4 from the last two expressions of L/2 and L/4.