compressed by $(8.75-x)$
The compressed force is,
$F=k x$
$F=300 x=400(8.75-x)$
On solving the above equation,
$x=5 \,cm$
Spring $B$ is compressed by $(8.75-5)=3.75 cm$
Now,
$\frac{E_{A}}{E_{B}}=\frac{\frac{1}{2} k_{A} x_{A}^{2}}{\frac{1}{2} k_{B} x_{B}^{2}}$
$=\frac{\frac{1}{2} \times 300 \times 5^{2}}{\frac{1}{2} \times 400 \times 3.75^{2}}$
$=\frac{4}{3}$

$\vec r = (\sin \,t\,\hat i\, + \,\cos \,t\,\hat j\, + \,t\,\hat k)m$
Find time $'t'$ when position vector and acceleration vector are perpendicular to each other
