The potential energy function for a particle executing linear SHM… - Vidyadip

MCQ

The potential energy function for a particle executing linear $\text{SHM}$ is given by $\text{v(x)}=\frac{1}{2}\text{kx}^2$ where $k$ is the force constant of the oscillator. For $k = 0.5N/ m,$ the graph of $V(x)$ versus $x$ is shown in the figure. A particle of total energy $E$ turns back when it reaches $\text{x}=\pm\text{x}_\text{m}.$ If $V$ and $K$ indicate the $P.E$. and $K.E.,$ respectively of the particle at $x = +xm,$ then which of the following is correct?

A
$V = 0, K = E$
✓
$V = E, K = 0$
C
$V < E, K = 0$
D
$V = 0, K < E$

✓

Answer

Correct option: B.

$V = E, K = 0$

Key concept: Energy Graph for a Spring: If the mass attached with spring performs simple harmonic motion about its mean position then its potential energy at any position $(x)$ can be given by,

Now kinetic energy at any position $\text{K}=\text{E}-\text{U}=\frac{1}{2}\text{Ka}^2-\frac{1}{2}\text{Kx}^2$
$\text{K}=\frac{1}{2}(\text{a}^2-\text{x}^2)\ ....(\text{i})$
From the above formula, we can cheak that
$\text{U}_\text{max}=\frac{1}{2}\text{Ka}^2$
At extreme $x=\pm\text{a}\big]$ and $U_{min} = 0 [$At mean $x = 0]$
$\text{K}_\text{max}=\frac{1}{2}\text{Ka}^2 [$At mean $x = 0]$ and $K_{min} = 0 \big[$ At extreme $x=\pm\text{a}\big]$
$\text{E}=\frac{1}{2}\text{ka}^2=$ constant$($at all positions$)$
Because velocity of mass $= 0 [$at extreme position$]$
$\therefore\ \text{K}=\frac{1}{2}\text{mv}^2=0$
It means kinetic energy changes parabolically w.r.t. position but total energy remain always constant irrespective to the position of mass.
Total energy is $\text{E = PE + KE,}$ which remains constant.
According to the question, when paraticle is at $x = xm,$
i.e., at extreme position $\ce{x = xm \Rightarrow KE = 0}$
Hence, total energy will be
$\text{E}=\text{PE}+0=\text{PE}$
$\Rightarrow\text{V}\text{(x}_\text{m})=\frac{1}{2}\text{kx}^2_\text{m}$
Hence option $(b)$ is correct.
$\text{V = E, K = 0}$

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