A particle of mass 0.2kg hangs from and ideal spring. In equilibrium, the spring is stretched by amount of 0.49cm. The spring is then suspended from the ceiling of a lift and hangs motionless relative to the lift as the lift descends with a constant velocity of 2.0m/s. The lift then suddenly stops.(adsbygoogle = window.adsbygoogle || []).push({});

a) With what amplitude will the particle oscillate?

b) What is the equation of the motion for the particles?

Take the upward directions to be positive and g= 9.8ms-2

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solution for question a)

I know that at the point of the elevator stops. Vmax= 2.0m/s

at equilibirium

Fg=-kx

0.2*9.8= -(k)(0.0049)

k=-400

w=-sqrt(400/0.2)= -44.72

using v(t)= -wA sin(wt + teta)

at equilibirium teta= 0, v(t)=Vmax

2=-44.72*A sin (wt)

A= -(2/(44.72*sin(wt))

I am stuck now...suddenly my brain jammed....what is the value of the t here?

t= 0 or t= 2?

or since it is at equilibrium when at the point release. therefore

Vm= wA

Which then,

A= -(2/44.72)= 0.044722719

??????

Please help...

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# Homework Help: Spring and Elevator Questions i need help am stuck

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