Question : A 3-kg block is suspended from a spring having a stiffness o...

A 3-kg block is suspended from a spring having a stiffness of k = 200 N/m. If the block is pushed 50 mm upward from its equilibrium position and then released from rest, determine the equation that describes the motion. What are the amplitude and the frequency of the vibration? Assume that positive displacement is downward.

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{ \omega }_{ n }=\sqrt { \frac { k }{ m } } =\sqrt { \frac { 200 }{ 3 } } =8.16 rad/s\\ x=A\sin { { \omega }_{ n }t } +B\cos { { \omega }_{ n }t }\\ x = -0.05 m when t = 0,
– 0.05 = 0 + B;\quad \quad B = -0.05\\ v=Ap\cos { { \omega }_{ n }t } -B{ \omega }_{ n }\sin { { \omega }_{ n }t }\\ v = 0 when t = 0,
0 = A(8.165) – 0;\quad \quad A = 0

Hence,

x = – 0.05\cos { (8.16t) }\\ C=\sqrt { { A }^{ 2 }+{ B }^{ 2 } } =\sqrt { ({ 0) }^{ 2 }+(-0.05) }= 0.05 m = 50 mm