Question

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

• A. $V=0,K=E$
• B. $V=E,K=0$
• C. $V<E,K=0$
• D. $V=0,K<E$

Answer 1

Answer: (B)

$V=E,K=0$

At any instant, the total energy of an oscillator is the sum of kinetic energy and potential energy.
Total energy, $E=K+V$
$E=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2$
At $x=+x_m$, the particle turns back, therefore its velocity at this point is zero, i.e. $v=0$
$\therefore K=0$
$\therefore E=\frac{1}{2}k{x}^2=V$.