4.9 H {a} The force exerted by a one—dimensional spring, ?xed at one end, is F = —k.r, where x is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is U = ital, if we choose U to be zero at the equilibrium position. {b} Suppose that this spring is hung vertically from the ceiling with a mass or suspended from the other end and constrained to move in the vertical direction only. Find the extension I.) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring Plus gravity} has the same form ilk}; if we use the coordinate y equal to the displacement measured from the new equilibrium position at x = x0 (and rede?ne our reference point so that U = i] at y = U).

I have answered the question fully step by step along with the formula used is mentioned so that you can use that for reference in future. 

Step-by-step explanation

Given : F = - kx it of tok ( one dimmsional spring] 7XF - 0 = J ( - Kx ) K - KX O = O FxF = ( conservative force ) . . F can be written as a scaloss potential u F = - Du = - dy, X -kx = - dy dx. du = [ kx dx - U = K / x. dx U = 1 kx' + a At x - 0 potential energy ( u) = 0 -) C = 0 u = 1 kx? Hence proved. 2 Expression for potential energy 6 Here Kx. = mg - ) / x0 = mg extension of new calm position. K Potential energy for this case Fg - Gravity force uly ) = - f ( fg + FB ) dy. FB - spring force.

= - mq y + Kxoy + 1 ky? =- may + K mo 4 + 1 ky ? (80 = mg = -may + may + } ky? 4(4)= 1 ky? B from
ARB=> Hence it is proved that total potential energy has the same form, if we use the coordinate
y equal to the displacement measured from new equilibrium position at X = Xd.