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.- Find the extension, xe, when the mass is in static equilibrium.
- Show that the net downward force acting on the mass, in any position, can be written in the form F=-kX where X≡x-xe is the displacement from the position of static equilibrium.
- Hence
show that the coordinate X obeys the equation
.
- Describe the motion implied by this equation.
Hint
Reveal
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Hint
Remember that gravity continues to act on the mass throughout...and must be included in the EOM.Solution
Reveal
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In general (that is when x is not necessarily equal to xe; see
figure (c)) the net downward force is
These two equations can be combined to give
is the displacement from the static equilibrium position.
But
since xe is a constant (independent of t).
The equation of motion for the mass can thus be written as:
which is of the general SHM form
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.

Solution
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:
![\[ 0= mg-kx_e \hspace{0.5cm}\mbox{\rm (equilibrium)}\hspace{0.5cm} \]](mastermathpng-1.png)
![\[ F= mg-kx \hspace{0.5cm}\mbox{\rm (general)}\hspace{0.5cm} \]](mastermathpng-2.png)
![\[ F= kx_e-kx = -kX \hspace{0.5cm}\mbox{\rm where}\hspace{0.5cm} X\equiv x-x_e \]](mastermathpng-3.png)
Application of Newton’s second law to the mass gives
![\[ m\ddot{x} = F = -kX \]](mastermathpng-4.png)
![\[ \ddot{x} = \frac{d^2}{dt^2} (X+x_e)= \frac{d^2X}{dt^2}=\ddot{X} \]](mastermathpng-5.png)
![\[ m\ddot{X} = -kX \]](mastermathpng-6.png)
![\[ \ddot{X} = -\omega^2 X \hspace{0.5cm}\mbox{\rm with}\hspace{0.5cm} \omega=\sqrt{k/m} \]](mastermathpng-7.png)