Question

A mass suspended on a spring will exhibit sinusoidal motion when it moves. If the mass on a spring is 91 cm off the ground at its highest position and 27 cm off the ground at its lowest position and takes 3.6 s to go from the top to the bottom and back again, determine an equation to model the data.

Answer 1

Assume that the mass start at a height of
`59\, \rm cm` (its equilibrium position.)

`\displaystyle h(t) = 64\, \sin \left((2\pi)/(3.6)\,t\right) + 59`,

where

• 
  `t` is in seconds, and
• 
  `h(t)` gives the height (in
  `\rm cm`) of the mass (above the ground.)

Explanation:

A mass on an ideal, vertical spring is in a simple harmonic motion. Its displacement from its equilibrium
`x` position at time
`t` can be modelled with a sine equation:

`\displaystyle x(t) = A \, \sin \left((2\pi)/(T)\, t + \phi \right)`,

where

• 
  `A` is the amplitude of the motion (the difference between the max height and min height.)
• 
  `T` is the time period of the motion.
• 
  `\phi` gives the initial (angular) position of the object.

For convenience, assume that
`\phi = 0`, such that the object starts at equilibrium position.

`A = \rm 91 - 27 = 64\; cm`.

`T = 3.6\; \rm s`.

The equation would become:

`\displaystyle x(t) = 64 \, \sin \left((2\pi)/(3.6)\, t \right)`,

The height of the object at equilibrium position is the average between its max and min height.
`\displaystyle h(0) = (91 + 27)/(2) = 59\; \rm cm`. To find the height of the object, add its height at equilibrium to the displacement from its equilibrium position.

Hence the equation:

`\displaystyle h(t) = x(t) + h(0) = 64 \, \sin \left((2\pi)/(3.6)\, t \right) + 59`.