Question

A block of mass 4 kg hangs from a spring of force constant k=400 N/m. The block is pulled down 15 cm below equilibrium and released. How long does it take the block to go from 12 cm below equilibrium (on the way up) to 9 cm above equilibrium?

Answer 1

$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{400}{4}}=10$ rad/s
Let $t_1$ be the time from $x=0cmtox=12cm$ and $t_2$ the time from$x=0cmtox=9cm.$ Then,
$12=15\sin \left(10t_1\right)$
or $t_1=0.093$ s
$9=15\sin \left(10t_2\right)$
or $t_2=0.064$ s
$\therefore$ Total time$=t_1+t_2=0.157$ s