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October 24, 2014

October 24, 2014

Posted by **Amanda** on Wednesday, July 14, 2010 at 10:28am.

A 40.0 N force stretches a vertical spring 0.250 m.

(A) What mass must be suspended from the spring so that the system will oscillate with a period of 1.0 s?

(B) If the amplitude of the motion is 0.050 m and the period is as in part (a), where is the object and in what direction is it moving 0.35 s after it has passed the equilibrium position moving downwards?

(C) What force (magnitude and direction) does the spring exert on the object when it is 0.030 m below the equilibrium position, moving upward?

- Physics -
**drwls**, Wednesday, July 14, 2010 at 11:53amFrom the force and deflecion information, the spring constant is k = F/X = 160 N/m

(A) Use the equation

P (period) = 2 pi sqrt(M/k)

(Solve for M)

(B) The total energy (kinetic + potential) = (1/2) k (Amplitude)^2

If t=0 is the equilibrium position,

X = -A sin (2 pi t/P)

A = 0.050 m

Substitute t = -0.35 s

That is more than 1/4 period past the t=0 postion. It should be at negatime deflection but on the way back.

(C) X = -0.030 m. Multiply than by k for the force. The force wil be up.

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