Q.The potential energy of a particle executing linear simple harmonic motion is $V(x) = \tfrac{1}{2}kx^2$, where $k$ is the force constant; the graph of $V(x)$ against $x$ is an upward-opening parabola with its minimum ($V=0$) at $x=0$. Take $k = 0.5\ \text{N m}^{-1}$. The particle has total energy $E$ and turns back (momentarily comes to rest) when it reaches the extreme positions $x = \pm x_m$, where the horizontal line of constant energy $E$ meets the parabola. If $V$ and $K$ denote the potential energy and kinetic energy of the particle at $x = +x_m$, which of the following is correct?

Classical Turning Points

Picture a pendulum bob swung to one side, or a block on a spring pulled to its farthest stretch. At that extreme position, the object is momentarily still before it reverses direction. That instant — and that position — is what physicists call a classical turning point.

Where Turning Points Occur in SHM

For a particle executing simple harmonic motion (SHM) with amplitude $A$, the two turning points are the extreme positions:

These are the farthest points the particle reaches on either side of the mean (equilibrium) position, $x = 0$.

Why "Turning" — The Velocity Condition

For SHM, the velocity as a function of displacement is:

At $x = \pm A$, the term under the square root becomes zero, so $v = 0$. The particle cannot move past $x = A$ (or below $x = -A$) — doing so would make $A^2 - x^2$ negative, which is impossible for a real velocity. This is why $x = \pm A$ are hard boundaries for the motion.

The Energy Picture

Turning points are easiest to understand through energy. For SHM, total mechanical energy is conserved:

where $K = \frac{1}{2}m\omega^2(A^2 - x^2)$ is kinetic energy and $U = \frac{1}{2}m\omega^2 x^2$ is potential energy.

This is the opposite of what happens at the mean position ($x = 0$), where $K = E$ (maximum speed) and $U = 0$.

The Restoring Force Is Maximum Here

Even though velocity is zero at a turning point, the particle is not in equilibrium. The restoring force $F = -kx$ (and acceleration $a = -\omega^2 x$) reach their maximum magnitude exactly at $x = \pm A$, which is precisely why the particle doesn't stay there — it is pulled straight back toward the centre.

Why "Classical"?

The word "classical" distinguishes this boundary from quantum mechanics. In classical mechanics, a particle governed by SHM can never be found beyond $x = \pm A$, because that would require negative kinetic energy — physically impossible. The region beyond the turning points is called the "classically forbidden region." (In quantum mechanics a particle's wavefunction can extend slightly beyond this boundary — tunnelling — but that lies outside the Class 11 syllabus.) …