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

Answer:

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

$\displaystyle h(t) = 64\, \sin \left(\frac{2\pi}{3.6}\,t\right) + 59$,

where

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

Step-by-step 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:

$\displaystyle x(t) = A \, \sin \left(\frac{2\pi}{T}\, t + \phi \right)$,

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:

$\displaystyle x(t) = 64 \, \sin \left(\frac{2\pi}{3.6}\, t \right)$,

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

Hence the equation:

$\displaystyle h(t) = x(t) + h(0) = 64 \, \sin \left(\frac{2\pi}{3.6}\, t \right) + 59$.