Physics(Please respond)
posted by Hannah on .
A 0.33kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 to 0.26 m (relative to its unstrained length), the speed of the sphere decreases from 5.3 to 4.2 m/s. What is the spring constant of the spring?
I know that the formula is F=kx
but i am not sure what to plug in since there are two sets of numbers plus the mass.

x1=A•sin(ω•t1),
sin(ω•t1)=x1/A,
cos(ω•t1)=sqrt(1sin²(ω•t1)=
=sqrt (1(x1/A)²),
v1=A•ω•cos(ω•t1)=
=A •ω•sqrt (1(x1/A)²).
(v1/A•ω)²= 1(x1/A)²,
v1²/ω² =A²  x1²,
A² = x1² + v1²/ω²,
Similar to this
A² = x2² + v2²/ω²,
Therefore,
x1² + v1²/ω²= x2² + v2²/ω²,
x1² •ω² + v1²= x2²• ω² + v2²,
v1² v2² = (x2²x1² ) •ω²
ω² = (v1² v2²)/( x2²x1²),
ω² =k/m,
k =m•(v1² v2²)/( x2²x1²)=
0.33•(28.09—17.64)/(6.761.44) •10^2=
= 0.33•10.45•100/5.32=64.82 N/m. 
thank you!