A spring with a $16 - \mathrm { kg }$ mass has natural length $0.8 \mathrm {~m}$ and is maintained stretched to a length of $1.2$ $\mathrm { m }$ by a force of $19.6 \mathrm {~N}$. If the spring is compressed to a length of $0.4 \mathrm {~m}$ and then released with zero velocity, find the position $x ( t )$ of the mass at any time $t$.
A) $x ( t ) = 0.8 \sin ( 1.75 t )$
B) $x ( t ) = - 0.4 \cos ( 1.75 t ) + 0.4 \sin ( 1.75 t )$
C) $x ( t ) = - 0.4 e ^ { - 4 t } \cos ( 0.4 t )$
D) $x ( t ) = - 0.4 \cos ( 1.75 t )$
E) $x ( t ) = - 0.4 e ^ { - 4 t } \cos ( 1.75 t )$

Question

Quizplus · Accepted Answer

The spring constant $k$ can be found using Hooke's Law: $F = kx$. Given $F = 19.6 \, \text{N}$ and $x = 0.4 \, \text{m}$, $k = \frac{19.6}{0.4} = 49 \, \text{N/m}$. The angular frequency $\omega$ is $\sqrt{\frac{k}{m}} = \sqrt{\frac{49}{16}} = 1.75 \, \text{rad/s}$. The initial displacement is $-0.4 \, \text{m}$, so the solution is $x(t) = -0.4 \cos(1.75t)$.

A Spring with A 16−kg16 - \mathrm { kg }16−kg Mass Has Natural Length

A Spring with A $16 - \mathrm { kg }$ Mass Has Natural Length