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 is initially stretched to 1.2 m by a force of 19.6 N. The spring constant $k$ can be found using Hooke's law $F = kx$, where $x = 1.2 m - 0.8 m = 0.4 m$ is the extension from the natural length. Thus, $k = \frac{F}{x} = \frac{19.6}{0.4} = 49 \, \text{N/m}$. When the spring is compressed to 0.4 m (0.4 m shorter than its natural length) and released, it oscillates with a simple harmonic motion described by $x(t) = A\cos(\omega t + \phi)$, where $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase angle. Since it starts from the maximum compression, $\phi = 0$, and $A = 0.4$ m. The angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{49}{16}} = 1.75 \, \text{rad/s}$. Therefore, the position $x(t)$ is given by $x(t) = -0.4 \cos(1.75t)$, indicating the mass starts 0.4 m to the left of the equilibrium position and oscillates from there.

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