# Answer to Question #84261 in Classical Mechanics for Robin

Question #84261

To a massless spring of length LO a mass m is hung, which causes a periodic change in the length L(t) of the spring.

(a) In case the suspended mass is released from rest with unstretched spring [L(t=0)=L0}, determine the change in the length L(t) as a function of time and set L(t) in a diagram.

(a) In case the suspended mass is released from rest with unstretched spring [L(t=0)=L0}, determine the change in the length L(t) as a function of time and set L(t) in a diagram.

Expert's answer

Since the mass is hung, the spring will be stretched as soon as the mass is set free, and according to Hooke's law:

@$ΔL=\frac {mg}{k}.@$

This is a magnitude of oscillations also. Thus, the function length from the ceiling to the mass vs. time is

@$L(t)=L_0+ΔL\cdot sin(ωt),@$

@$ω=\sqrt{\frac{k}{m}},@$

@$L(t)=L_0+\frac {mg}{k}\cdot sin(\sqrt{k/m}\cdot t).@$

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