Question

Question 8
A spring is attached to the ceiling and pulled 8 cm down from equilibrium and released. The
damping factor for the spring is determined to be 0.4 and the spring oscillates 12 times each
second. Find an equation for the displacement, D(t), of the spring from equilibrium in terms of
seconds, t.
D(t) =

Can someone please help me ASAP?!!!!

Answer 1

Answer: D(t) = $8.e^{-0.4t}.cos(\frac{\pi }{6}.t )$

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = $a.sin(\omega.t)$ or y = $a.cos(\omega.t)$

where:

|a| is initil displacement

$\frac{2.\pi}{\omega}$ is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

$y=a.e^{-ct}.cos(\omega.t)$ or $y=a.e^{-ct}.sin(\omega.t)$

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

$y=a.e^{-ct}.cos(\omega.t)$

period = $\frac{2.\pi}{\omega}$

12 = $\frac{2.\pi}{\omega}$

ω = $\frac{\pi}{6}$

Replacing values:

$D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)$

The equation of displacement, D(t), of a spring with damping factor is $D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)$.