Calculating time by the force which depends on x

  • Thread starter Thread starter xserx
  • Start date Start date
  • Tags Tags
    Force Time
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
xserx
Messages
1
Reaction score
0

Homework Statement



An object with mass m start its journey on x-axis at point (-l,0) with velocity v.
The object feels force which depends on the object's distance to origin.
After T second it reaches the origin. What is T?

Homework Equations



x(t=0)=-l
v(t=0)=v

f(x)=-kx

x(t=T)=0

T=?

The Attempt at a Solution



1: dx/dt = v(t)

2: dt = dx/v(t) [from 1]

3: dv/dt = a(t) = (-k/m) * x(t)

4: dt = m * dv/(-k * x(t)) [from 3]

5: v(t) * dv = (-k/m) * x(t) * dx [put 2 to 3]

6: integrating both sides of 5 and putting the values x=0 and the velocity at the origin (which can be calculated by the energy change with integrating f(x) dx from -l to zero) gives the relation between x and v.

7: Putting x in terms of v (found in 6) to 4 and integrating right side from initial velocity to velocity at the origin gives a time (dt at the left side).

But i am not sure about the answer. Any ideas will be great.
Thank you.
 
Physics news on Phys.org
Steps 1 to 5 are correct, and the final step you've written (step 5) is saying that the change in KE of the object is equal to the change in PE of the object (since PE depends on its distance from the origin). This is all correct.

And you're right, this gives the relation between x and v. And then yes, you can calculate what the final speed should be. And then yes, you could put x in terms of v to equation 4, and integrate the right hand side from initial to final speed, and that gives the time on the left hand side.

So yes, it is all correct. But the integral that you must do to integrate the speed is a bit difficult. Have you tried it already? I reckon you'll need to do a sinusoidal substitution (or just look it up in an integral table if you're allowed to do that).

There are probably loads of other ways to do the problem, but I think your way is correct (although its a little difficult).