Accelerating Lift Homework: Equations of Motion

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Homework Help Overview

The problem involves a block attached to a spring in an accelerating lift, where the block is held stationary at the spring's equilibrium length. The context is within the framework of equations of motion, specifically focusing on the dynamics of the block as it transitions into simple harmonic motion (SHM) after being released.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are exploring the relationship between the acceleration of the lift and the forces acting on the block. There is confusion regarding the implications of the lift's acceleration on the equilibrium state of the spring and the block.

Discussion Status

Some participants have proposed equations of motion, while others are questioning the assumptions made about the system, particularly the acceleration values and their effects on the forces involved. There is an ongoing exploration of the correct formulation of the equations governing the motion.

Contextual Notes

Participants are grappling with the implications of the lift's acceleration on the forces acting on the block and the spring, as well as the definitions of the variables involved. There is a sense of urgency in resolving these conceptual challenges.

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Homework Statement



A light spring force const. k hands from ceiling of an accelerating lift with a block of mass m attached to its bottom end. Block is held stationary (in lift frame) with the string at its equilibrium length while the lift is accelerating at a rate Ag where A is between -1 and 1 (const.) At t=0 the block is released from rest allowing it to execute SHM. Denote X by distance of block from origin of intertial FoR and Xo its distance when the spring is at equil. length..

write down eqn of motion of block in interial FoR..

then more qs

Homework Equations





The Attempt at a Solution



Im quite confused.
So taken intertial FoR to be the ground. but if block is stationary when spring is at equilibrium length and lift accelerating at Ag then surely that means A = -1?

help? Thanks
 
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Thoughts? This is quite urgent sorry!
 
Is it k(Xo-X) - mg = mx''?
 
Actually I get (Xo-X)'' + k/m(Xo-X) = g

I know Xo'' = Ag

but how do i solve this eqn of motion?
 

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