Question

Potential energy ($U$) of a body of unit mass moving in a one-dimensional conservative force field is given by, $U=\left( {{x}^{2}}-4x+3 \right)$. All units in S.I.
$(1)$ Find the equilibrium position of the body.
$(2)$ Show that oscillations of the body about this equilibrium position is simple harmonic motion and find its time period.
$(3)$ Find the amplitude of oscillations if the speed of the body at equilibrium position is $2\sqrt{6}m/s$.

Answer 1

We will find the equilibrium position by finding the expression for force from the given expression for energy and equating it with zero. The force acting on a body at equilibrium will be zero. Then we will find whether the motion is harmonic or not by comparing the dependence of force on displacement. Then we will find the time period by finding $\omega$ from the acceleration of the body when it is moved from the equilibrium position. We will find the amplitude of this using the expression for energy at equilibrium position. We must be aware that at equilibrium position, kinetic energy will be maximum and potential energy will be minimum. Total energy will be maximum when amplitude of the motion is at maximum.

Formula used:

& F=-\dfrac{\delta U}{\delta x} \\\ & T=\dfrac{2\pi }{\omega } \\\ & {{U}_{\max }}=K.{{E}_{\max }}+P.{{E}_{\min }} \\\ \end{aligned}$$ **Complete step by step answer:** $$(1)$$ We will differentiate the expression for energy with respect to distance and find the expression for force. $$F=-\dfrac{\delta U}{\delta x}=-\dfrac{\delta \left( {{x}^{2}}-4x+3 \right)}{\delta x}=-\left( 2x-4 \right)$$ We know, at equilibrium position, $$F=0$$ . $$\Rightarrow 0=-\left( 2x-4 \right)$$ $$\begin{aligned} & 2x=4 \\\ & x=2m \\\ \end{aligned}$$ Therefore, the equilibrium position is at $$x=2m$$ . $$(2)$$ Now we will prove that this body is in harmonic motion by comparing the relation between force and distance. Then we will deduce the expression for time period from the acceleration by finding $$\omega $$ . We have, $$F=-\left( 2x-4 \right)$$ $$\begin{aligned} & \Rightarrow F\propto -x \\\ & \Rightarrow a\propto -x \\\ \end{aligned}$$ This proves the body is in SHM because force increases when we move the object from equilibrium position. Now, to find the time period, we will find $$\omega $$ . Let us displace the body from equilibrium position by $$\Delta x$$ . $$\Rightarrow F=-\left( 2(x+\Delta x)-4 \right)$$ We know $$x=2m$$ and the mass of the body is unity. $$\begin{aligned} & \Rightarrow F=-\left( 2(2+\Delta x)-4 \right) \\\ & \Rightarrow a=\dfrac{-2\Delta x}{m}=-2\Delta x \\\ \end{aligned}$$ We know, acceleration in an SHM is given as, $$a=-{{\omega }^{2}}\Delta x$$ . Now, we will equate both equations to get $$\omega $$ . $$\Rightarrow -{{\omega }^{2}}\Delta x=-2\Delta x$$ $$\begin{aligned} & {{\omega }^{2}}=\dfrac{2\Delta x}{\Delta x}=2 \\\ & \omega =\sqrt{2} \\\ \end{aligned}$$ Now, time period is given by, $$T=\dfrac{2\pi }{\omega }$$ $$\Rightarrow T=\dfrac{2\pi }{\sqrt{2}}=\sqrt{2}\pi \operatorname{s}$$ Therefore, the time period of this SHM by the body is $$\sqrt{2}\pi $$ seconds. $$(3)$$ We will use the expression for total energy at equilibrium position to find amplitude. $${{U}_{\max }}=K.{{E}_{\max }}+P.{{E}_{\min }}$$ $${{U}_{\max }}=\dfrac{1}{2}m{{v}^{2}}+({{x}^{2}}-4x+3)$$ We know mass is 1kg and velocity at equilibrium position is given as $$2\sqrt{6}m/s$$ . At equilibrium position, $$x=2m$$ . $$\Rightarrow {{U}_{\max }}=\dfrac{1}{2}{{\left( 2\sqrt{6} \right)}^{2}}+({{2}^{2}}-4\times 2+3)=12-1=11J$$ But, total energy can also be expressed as the energy at maximum amplitude. i.e. $${{x}_{\max }}=\left( A+x \right)=\left( A+2 \right)$$ . Then energy will be, $${{U}_{\max }}=({{\left( A+2 \right)}^{2}}-4\left( A+2 \right)+3)=({{A}^{2}}+4+4A-4A-8+3)=\left( {{A}^{2}}-1 \right)$$ Let us equate this with the previously found energy. $$\begin{aligned} & \Rightarrow \left( {{A}^{2}}-1 \right)=11 \\\ & \Rightarrow {{A}^{2}}=12 \\\ & \Rightarrow A=\sqrt{12}=2\sqrt{3}m \\\ \end{aligned}$$ Therefore, the amplitude of the body is equal to $$2\sqrt{3}m$$ . **So we can conclude this answer as, $$(1)$$ The equilibrium position is at $$x=2m$$ . $$(2)$$ The body executes simple harmonic motion with a time period of $$\sqrt{2}\pi $$ seconds. $$(3)$$ Amplitude of oscillations if speed of the body at equilibrium position is $$2\sqrt{6}m/s$$ will be $$2\sqrt{3}m$$ .** **Note:** In this question we can find the time period by another method. We can compare the force expression from this SHM to the force expression of a spring which is given as $$F=-kx$$ . Then we will find the spring constant. Time period is given by, $$T=2\pi \sqrt{\dfrac{\text{Inertia Factor}}{\text{Spring Factor}}}$$. Here, the inertia factor is just the mass of the body.