Q7.9 A different mass-spring system? (H)

A mass m hangs from the bottom of a spring of spring constant k. Denote by x the extension of the spring with respect to its natural length.
  1. Find the extension, xe, when the mass is in static equilibrium.
  2. Show that the net downward force acting on the mass, in any position, can be written in the form F=-kX where Xx-xe is the displacement from the position of static equilibrium.
  3. Hence show that the coordinate X obeys the equation $m\ddot{X}=-kX$.
  4. Describe the motion implied by this equation.
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Hint

Remember that gravity continues to act on the mass throughout...and must be included in the EOM.
 

Solution

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Solution

Graphic - No title - verticalmassspring

Let’s think thorough this carefully, particularly in regard to signs.

Let x represent the downward displacement of the mass with respect to the position it would have if the spring were unstretched In the equilibrium position the displacement x has value x=xe (see figures (a), (b)), such that the net force (downward say) is zero:

\[ 0= mg-kx_e \hspace{0.5cm}\mbox{\rm (equilibrium)}\hspace{0.5cm} \]
In general (that is when x is not necessarily equal to xe; see figure (c)) the net downward force is
\[ F= mg-kx \hspace{0.5cm}\mbox{\rm (general)}\hspace{0.5cm} \]
These two equations can be combined to give
\[ F= kx_e-kx = -kX \hspace{0.5cm}\mbox{\rm where}\hspace{0.5cm} X\equiv x-x_e \]
is the displacement from the static equilibrium position.

Application of Newton’s second law to the mass gives

\[ m\ddot{x} = F = -kX \]
But
\[ \ddot{x} = \frac{d^2}{dt^2} (X+x_e)= \frac{d^2X}{dt^2}=\ddot{X} \]
since xe is a constant (independent of t). The equation of motion for the mass can thus be written as:
\[ m\ddot{X} = -kX \]
which is of the general SHM form
\[ \ddot{X} = -\omega^2 X \hspace{0.5cm}\mbox{\rm with}\hspace{0.5cm} \omega=\sqrt{k/m} \]
It follows that the coordinate X exhibits SHM at this angular frequency. The mass thus oscillates at the same frequency it would if it were on a surface, and attached to a horizontal spring; but it oscillates about the displaced equilibrium position x=xe.