Asked by Ray

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.

Answers

Answered by Anonymous
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)
Answered by Anonymous
Actually for part a it's;

k=w^2*m
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