Question

A 0.33-kg 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.

Answer 1

x1=A•sin(ω•t1),
sin(ω•t1)=x1/A,
cos(ω•t1)=sqrt(1-sin²(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 = (x2²-x1² ) •ω²
ω² = (v1²- v2²)/( x2²-x1²),
ω² =k/m,
k =m•(v1²- v2²)/( x2²-x1²)=
0.33•(28.09—17.64)/(6.76-1.44) •10^-2=
= 0.33•10.45•100/5.32=64.82 N/m.

Answer 2

thank you!