Posted by **Hannah** on Sunday, June 10, 2012 at 3:46pm.

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.

- Physics(Please respond) -
**Elena**, Sunday, June 10, 2012 at 5:15pm
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.

- Physics(Please respond) -
**Hannah**, Sunday, June 10, 2012 at 6:09pm
thank you!

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