Question

Show that the combined spring energy and gravitational energy for a mass $m$ hanging from a light spring of force constant $k$ can be expressed as ${U_0} + \dfrac{1}{2}k{y^2}$, where $y$ is the distance above or below the equilibrium position and ${U_0}$ is constant.

Answer 1

Hint The energy can be determined by the sum of the energy of the spring energy and the gravitational energy. And also by using the condition of the equilibrium position, then the combined spring energy and gravitational energy for a mass $m$ hanging from a light spring of force constant $k$ can be determined.

Complete step by step solution
When the mass of the object is in the equilibrium position, then the condition is given as,
$k{x_0} = mg\,..............\left( 1 \right)$
When the mass of the object is in displaced position, then
$U = \dfrac{1}{2}k{\left( {{x_0} + y} \right)^2} - mgy$
Now using the formula of ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, then the above equation is written as,
$U = \dfrac{1}{2}k\left( {{x_0}^2 + {y^2} + 2{x_0}y} \right) - mgy$
By multiplying the term inside the bracket, then the above equation is written as,
$U = \dfrac{1}{2}k{x_0}^2 + \dfrac{1}{2}k{y^2} + \dfrac{1}{2}k2{x_0}y - mgy$
By cancelling the terms in the above equation, then the above equation is written as,
$U = \dfrac{1}{2}k{x_0}^2 + \dfrac{1}{2}k{y^2} + k{x_0}y - mgy$
By substituting the equation (1) in the above equation, then the above equation is written as,
$U = \dfrac{1}{2}k{x_0}^2 + \dfrac{1}{2}k{y^2} + mgy - mgy$
By cancelling the same terms in the above equation, then the above equation is written as,
$U = \dfrac{1}{2}k{x_0}^2 + \dfrac{1}{2}k{y^2}$
From the above equation the term $\dfrac{1}{2}k{x_0}^2 = {U_0}$, so, the above equation is also written as,
$U = {U_0} + \dfrac{1}{2}k{y^2}$
Thus, the above equation shows the combined spring energy and gravitational energy for a mass $m$ hanging from a light spring of force constant $k$ can be expressed as ${U_0} + \dfrac{1}{2}k{y^2}$, where $y$ is the distance above or below the equilibrium position and ${U_0}$ is constant.

Note In the displacement equation, the terms ${x_0}$ and $y$ are added, because the term ${x_0}$ is the initial length of the mass when it is hung on the spring. Then the term $y$ is the distance between the original position to the displaced position, then both the distance is added.