As seen above, the acceleration value for a particle in MHS is given by:
Then, by Newton's 2nd Law, we know that the resulting force on the system is given by the product of its mass and acceleration, thus:
Since mass and pulsation are constant values for a given MHS, we can replace the product. mω² by the constant k, called MHS force constant.
Thus we conclude that the algebraic value of the resulting force acting on a particle describing an MHS is proportional to the elongation, although they have opposite signs.
This is the fundamental characteristic that determines whether a body performs a simple harmonic motion.
It is called the force acting on a body that describes MHS from restorative forcebecause it acts to ensure that the oscillations continue, restoring the previous movement.
Whenever the particle passes through the central position, force has the effect of slowing it down and then bringing it back.
MHS breakeven point
At the midpoint of the path, the elongation is numerically zero (x = 0), consequently the resulting force acting at this moment is also null (F = 0).
This point where the force is nullified is called balance point of the movement.
Much of the practical usefulness of the MHS is related to the knowledge of its period (T), since it is experimentally easy to measure and from it it is possible to determine other quantities.
As we defined earlier:
k = mω²
From this we can get an equation for the MHS heartbeat:
But we know that:
Then we can get the expression:
As we know, the frequency is equal to the inverse of the period, so:
(1) A system is formed by a spring hanging vertically to a support at one end and a 10kg block of mass. When set in motion the system repeats its movements after every 6 seconds. What is the spring constant and the oscillation frequency?
For a system consisting of a mass and a spring, the constant k is equivalent to the elastic spring constant, thus: