Vertical Spring Deviations

When a mass is hung from a vertical spring, the spring stretches because  F = 0 = mg - kx_o.   Solving for x_o.  

           mg  
x_o  =  ------ 
            k   

If the spring is now stretched a distance downward of x,   F = mg -k ( x + x_o ).     Substitute blue area above.  

  
                     
                               mg  
F  =  mg - k ( x  +  ----- )  =  mg  - kx  -  mg  =  - kx
                                 k   

  
This shows that Hooke's Law is valid for Vertical as well as Horizontal springs.  The displacement x must be measured from the new equilibrium position after the mass has been added.

   
Energy is also valid for vertical springs.  Measuring PE from the equilibrium position of the original spring

PE  =  kx_o²  -  mgx_o                Includes both elastic and gravitational PE.  

When the spring is stretched an additional x

PE  =  k( x  + x_o) ²  -  mg ( x  +  x_o )              Now find the difference between the two PE's.   

           mg  
x_o  =  ------    Remember to use this relationship and expand the polynomial.
            k       

k( x  + x_o) ²  -  mg ( x  +  x_o )  -   kx_o²  +  mgx_o   =  kx²  +  kxx_o  -  mgx

        
                                                                                                           mg
                                                                                  =  kx²  + kx ( ----- )  -  mgx
                                                                                                             k
    

                                                                                  =   kx²  

This the PE of an oscillating system can be measured from the new equilibrium position after the mass has been added.