0
$\begingroup$

A spring AB with constant k is hooked in the end A to the ceiling. At the end B of the undefomed spring is hooked a mass of weight 100N. At $t=0$ the mass is let free with no initial velocity. Not taking into account the mass of the spring, find the movement of the spring, its period and amplitude.

My suggested solution is as follows. By Hooke's law $F=k(x-x_0)$, where $x_0$ is the position of the end B at the beginning.

They say the force acting on the mass at a position $x$ is $100N-k(x-x_0)N$, and suggest the equation $m\ddot{x}=100g\ddot{x}=100N-k(x-x_0)N$, where g is the acceleration due to gravity.

I think this is wrong and suggest the equation $m\ddot{x}=100g\ddot{x}=k(x-x_0)$

Who is correct?

p.s. I am not interested in the solution of the equation.

$\endgroup$
6
  • $\begingroup$ You really should check units in both solution, it's not clear for me what you want to express. In The first solution,you subtract a force from a dimensionless quantity. In your solution I have no dea why you divide 100 by g. Should this be the mass? $\endgroup$
    – Noldig
    Commented Feb 21, 2013 at 11:52
  • $\begingroup$ @Noldig Yes, about the first remark, that is what got me stumbled. I believe mass is implicitly assumed equivalent to weight for these problems, since we assume the gravitational force to be constant. For the second one, well mass=weight/gravity, I have clearly stated what g is. Let me change up the question a little bit. $\endgroup$ Commented Feb 21, 2013 at 12:13
  • $\begingroup$ The truth is that I never attended the exercises for this class. I attended algebraic geometry instead. I'm having a hard time going through someone else’s notes. $\endgroup$ Commented Feb 21, 2013 at 12:14
  • $\begingroup$ mass = weight / gravity is true, but the 100 grams ARE A MASS, so the lefthand side of your eq. has units ma=N, the right hand side kg/a*a = kg has units of kg not N, thats wrong. $\endgroup$
    – Noldig
    Commented Feb 21, 2013 at 12:45
  • $\begingroup$ Yes, indeed. I changed it up a little. It is weight now and it's in Newtons $\endgroup$ Commented Feb 21, 2013 at 12:47

1 Answer 1

1
$\begingroup$

Afer the discussions in the comments this would be my solution:

The l.h.s in the equations of motion is always simply $m\ddot{x}$, the r.h.s contains the sum of all acting forces. So you have gravity and the force of the spring. thats it. $m\ddot{x}=mg-k(x-x_{0})$

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.