A spring with a $3 - \mathrm { kg }$ mass is held stretched $0.9 \mathrm {~m}$ beyond its natural length by a force of $30 \mathrm {~N}$. If the spring begins at its equilibrium position but a push gives it an initial velocity of $1 \mathrm {~m} / \mathrm { s }$, find the position $x ( t )$ of the mass after $t$ seconds.
A) $x ( t ) = 0.3 \sin \left( \frac { 10 } { 3 } t \right)$
B) $x ( t ) = 0.4 \cos \left( \frac { 10 } { 3 } t \right)$
C) $x ( t ) = 1 \sin ( 7 t )$
D) $x ( t ) = 0.3 \sin \left( \frac { 10 } { 3 } t \right) + 0.4 \cos \left( \frac { 10 } { 3 } t \right)$
E) $x ( t ) = \frac { 10 } { 3 } \sin ( 7 t )$

Question

Quizplus · Accepted Answer

The equation of motion is $3 x ^ { \prime \prime } + 30 x = 0$. The characteristic equation is $3 r ^ { 2 } + 30 = 0$, which has roots $r = \pm \sqrt { 10 } / 3$. Therefore, the general solution is $x ( t ) = c _ { 1 } \sin \left( \frac { 10 } { 3 } t \right) + c _ { 2 } \cos \left( \frac { 10 } { 3 } t \right)$. Using the initial conditions, we get $x ( 0 ) = 0.9 = c _ { 2 }$ and $x ^ { \prime } ( 0 ) = 1 = \frac { 10 } { 3 } c _ { 1 }$. Therefore, $c _ { 1 } = 0.3$ and $c _ { 2 } = 0.9$. Thus, the position of the mass after $t$ seconds is $x ( t ) = 0.3 \sin \left( \frac { 10 } { 3 } t \right) + 0.9 \cos \left( \frac { 10 } { 3 } t \right)$.

A Spring with A 3−kg3 - \mathrm { kg }3−kg Mass Is Held Stretched

A Spring with A $3 - \mathrm { kg }$ Mass Is Held Stretched