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August 30, 2014

August 30, 2014

Posted by **sara** on Friday, March 23, 2012 at 9:21am.

(a) Determine the maximum distance that the spring is compressed.

I have this part..just need help with part b

(b) Find the total elapsed time until the mass returns to its starting point. (Hint: The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.)

- physics -
**bobpursley**, Friday, March 23, 2012 at 9:27amfigure the period of the spring (see the spring harmonic oscillator equations). Then, the time on the spring is one quarter cycle (compressing it), then the spring accelerates the mass to some velocity and as the spring slows, the mass leaves. Max velocity occurs in the spring at the unstretched point, so in my mind, 1/2 cycle total.

- physics -
**Elena**, Friday, March 23, 2012 at 1:50pmThe law of conservation of energy

KE (of the mass) = PE (of the spring)

mv^2/2 = kx^2/2

d=x = sqrt (mv^2/k)=sqrt(5•9/844)=0.23 m

T = 2•π•sqrt(m/k) =2•π•sqrt(5/844) = 0.484 s

The time of the mass motion with spring is T/2 =0.242 s.

The time of uniform motion before hitting the spring and after leaving thespring is

t =D/v =0.25/3=0.0833 s.

Therefore, the total time is 0.242 + 2•0.0833=0.409 s

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