For the above system I have the following expressions for kinetic and potential energy:

$$ V = \frac{1}{2}\,k\,x^{2}+m\,g\,l\,(1-cos\,\theta)-m\,g\,x\\ T = \frac{1}{2}\,m\,\dot{x}^{2}+\frac{1}{2}\,m\,\vec{v}.\,\vec{v}\\ \vec{v}=l\,\dot{\theta}\,cos\,\theta\,\hat{\imath}\,+\,(\dot{x}-l\,\dot{\theta}\,sin\,\theta)\hat{\jmath}\\ T = m\,\dot{x}^{2}+\frac{1}{2}\,m\,l^{2}\,\dot{\theta}^{2}\,-\,m\,l\,\dot{x}\,\dot{\theta}\,sin\,\theta $$ For small $\theta$: $$ T = m\,\dot{x}^{2}+\frac{1}{2}\,m\,l^{2}\,\dot{\theta}^{2}\\ V = \frac{1}{2}\,k\,x^{2}+\frac{1}{2}m\,g\,l\,\theta^{2}-m\,g\,x $$ For mechanical systems with no external forces Lagrange's equations give: $$ \frac{\mathrm{d} }{\mathrm{d} t}\Big(\frac{\partial T}{\partial \dot{q_{j}}}\Big)-\frac{\partial T}{\partial q_{j}}+\frac{\partial V}{\partial q_{j}}=0;\qquad q_{j}=1,2 $$ For $q_{1}=x$ we have: $$ \frac{\mathrm{d} }{\mathrm{d} t}\Big(\frac{\partial T}{\partial \dot{x}}\Big)=2\,m\,\ddot{x}\\ -\frac{\partial T}{\partial x}=0\\ \frac{\partial V}{\partial x}=kx-mg $$ For $q_{2}=\theta$ we have: $$ \frac{\mathrm{d} }{\mathrm{d} t}\Big(\frac{\partial T}{\partial \dot{\theta}}\Big)=m\,l^{2}\,\ddot{\theta}\\ -\frac{\partial T}{\partial \theta}=0\\ \frac{\partial V}{\partial \theta}=m\,g\,l\,\theta $$ Therefore equations of motion are: $$2m\ddot{x}+kx-mg=0;\qquad m\,l^{2}\,\ddot{\theta}+m\,g\,l\,\theta \quad\text{or}\quad\ddot{\theta}+\frac{g}{l}\theta=0; $$

I chose ground as reference point $(V=0)$.

However in the solution given by the book, the term $-m\,g\,x$ in the potential energy expression is not considered. It is written that "Potential energy of the slider need not be considered if $x=0$ corresponds to static equilibrium position".

Final equations according to book: $$2m\ddot{x}+kx=0;\quad\ddot{\theta}+\frac{g}{l}\theta=0; $$

I don't understand why $-mgx$ is not considered. Kindly help me understand.

Source of the question: Mechanical Vibrations by Singiresu S. Rao Fifth Edition Chapter 6 Problem 6.39


closed as off-topic by Aaron Stevens, John Rennie, stafusa, ZeroTheHero, Kyle Kanos Feb 13 at 10:59

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  • $\begingroup$ Welcome to physics SE. Homework questions are not in general accepted. However, you showed some effort and you ask on the meaning of a part of the text.... Good luck $\endgroup$ – jaromrax Feb 12 at 15:25
  • $\begingroup$ Here, $x$ represents the elongation of the spring from the static equilibrium position obtained when the gravitational force is balanced by spring force. At the equilibrium position, $x$ is by definition zero. Realise that potential energy depends on the reference point. Considering the equilibrium position as the reference point, the term $mgx$ will become $mg\times 0 = 0$ and hence can be ignored. $\endgroup$ – exp ikx Feb 12 at 16:51
  • $\begingroup$ At its core, I'm not sure this counts as a homework-like question. It is a duplicate of Where does the loss in gravitational energy of the load go when a spring is pulled?, though. $\endgroup$ – Chris Feb 16 at 19:01

Your expression is correct. What the solution is telling is that it's possible to redefine $x$ such that the term $mgx$ doesn't appear (we can always redefine the reference point for the potential energy). This is easily achievable by completing the square on $x$ $$ \frac 1 2 kx^2 + mgx = \frac k 2 \left [ x^2 + \frac{2mg}{k} x + \left (\frac{mg}{k}\right)^2 - \left (\frac{mg}{k}\right)^2 \right ] = \frac 1 2 k \left ( x + \frac{mg}{k}\right )^2 - \frac 1 2 \frac{(mg)^2}{k}. $$ Note how the function above is minimized at $x = - mg/k$; that's the equilibrium position. By making the substitution $$ x \to x' = x + \frac{mg}{k}, $$ we just substitute $\dot x$ by $\dot x'$ and we may drop the constant term $(mg)²/(2k)$ since it doesn't change the dynamics. Now your potential is just an harmonic oscillator in $x'$ and a $\theta$-dependent term.


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