Simple Harmonic motion of spring Mass system when spring is horizontal :
If x is a small extension or compression in the spring from the equilibrium state the restoring force produced is given by
F = − k x
Where k is called force constant or spring factor.
∴ Equation of motion of the mass M is given by
$ \displaystyle M\frac{d^2 x}{dt^2} = -kx $
$ \displaystyle \frac{d^2 x}{dt^2} = -\frac{k}{M}x $
This represents an S.H.M. its angular frequency ω is
$ \displaystyle \omega = \sqrt{\frac{k}{M}} $
and the period of oscillation is
$ \displaystyle T = 2\pi \sqrt{\frac{M}{k}} $
Exercise : A spring is cut into two equal parts. What will be the difference in time period of the spring pendulum thus formed from the original spring pendulum ?
SHM of Spring Mass System (spring is vertical)
Simple Harmonic motion of Spring Mass System spring is vertical :
The weight Mg of the body produces an initial elongation, such that Mg − k yo = 0.
If y is the displacement from this equilibrium position the total restoring force will be
Mg − k(yo + y) = − ky
Again we get,
$ \displaystyle T = 2\pi \sqrt{\frac{M}{k}} $
Note that the gravity has no effect on the time period of oscillations.
In general , time period of a spring mass system depends only on spring and mass and it is independent of external forces provided that external forces are constant and acts throughout the motion.