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Posted by Ray on Thursday, February 17, 2011 at 1:05pm.

A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the system is 2.00 J, find (a) The force constant of the spring and (b) The amplitude of the motion.

a) T = 2pi / w so w = 2 pi / T w = sqrt ( k / m ) so k = w² / m b) (1/2)* k * xmax ² = 2,00 J so, xmax = sqrt( 2,00 * 2 / k)

Actually for part a it's; k=w^2*m

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