Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the p

Exercises · 5.3

Q. Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

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Concept understanding — Classical Turning Points

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:

x = +A \quad \text{and} \quad x = -A

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

Note

At a turning point, the particle's velocity is exactly zero, and it is about to reverse the direction of its motion.

Why "Turning" — The Velocity Condition

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

v(x) = \omega \sqrt{A^2 - x^2}

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:

E = K + U = \frac{1}{2}m\omega^2 A^2 \quad (\text{constant})

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.

x = \pm A

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.)

Watch out

Don't confuse the turning points with the mean position. At $x = 0$ : velocity is maximum, acceleration is zero. At $x = \pm A$ : velocity is zero, acceleration (and restoring force) is maximum. They are opposite extremes of the same oscillation.

Quick Example

A simple pendulum released from one side swings to the equal-and-opposite extreme angle on the other side before swinging back. Both extreme positions — where the bob is momentarily at rest — are its classical turning points, and by energy conservation both occur at the same height (ignoring friction), since the total energy is unchanged.

"Classical Turning Points important questions" is a common search among CBSE and competitive-exam aspirants alike, since Classical Turning Points builds on the Work, Energy and Power coverage of the NCERT Class 11 Physics curriculum and is more of an important topic for JEE Main, NEET and state CETs than a direct board-exam fixture. Pairing this explanation with NCERT Physics textbook practice and previous years' questions is the surest way to lock the concept in before an exam.

A particle is allowed only where $V(x)\le E$ (so that $K=E-V\ge 0$ ), and its least possible total energy equals the lowest point of the potential. Applying this rule to each graph gives the forbidden region(s), the minimum energy, and a simple everyday example each shape resembles.

Since $V(x)=0$ for $x<a$ and $V(x)=V_0>E$ for $x\ge a$ , the kinetic energy $K=E-V(x)$ would be negative for $x\ge a$ — impossible. So the particle is confined to $x<a$ and can never be found for $x\ge a$ . The lowest possible total energy for which the particle can exist at all is $E_{\min}=0$ (the lowest value $V(x)$ ever takes). This is the classic single-step potential: like a ball rolling on a flat frictionless track that meets a step/ledge higher than its kinetic energy can climb, so it is turned back at $x=a$ and can never cross over.
The marked energy $E$ lies below $V_0$ , which is itself the lowest step of the rising staircase — so $V(x)\ge V_0>E$ everywhere the staircase is defined, and $K=E-V(x)<0$ there. The particle cannot be found anywhere along the staircase at this energy. Since even the first, smallest step already exceeds $E$ , the particle's minimum energy to exist there at all is $E_{\min}=V_0$ . Physically this is like a ball sitting at the base of a long flight of ever-higher stairs — it doesn't even have enough energy to climb onto the very first step, let alone the higher ones beyond.
Here $V(x)=V_0>E$ for $x<a$ and $x>b$ (the two outer walls), but $V(x)=-V_1$ (below zero) in between — so the particle is confined entirely to the well, $a<x<b$ , and cannot be found outside it. Its minimum possible total energy is the floor of the well, $E_{\min}=-V_1$ . This is the standard rectangular potential well — the same idealised model used for a particle (e.g. a nucleon) trapped inside a finite, attractive potential region with walls too high for it to climb over classically, such as inside a nucleus or a simple finite square well.
The barriers on either side rise to a peak $V_0>E$ $V_{0} > E$ over $a/2<|x|<b/2$ $a /2 < ∣ x ∣ < b /2$ , so the particle cannot be found in those two barrier regions, $-b/2<x<-a/2$ $- b /2 < x < - a /2$ and $a/2<x<b/2$ $a /2 < x < b /2$ ; outside the barriers ( $|x|>b/2$ $∣ x ∣ > b /2$ ) and inside the central well ( $|x|<a/2$ $∣ x ∣ < a /2$ ) it is allowed, since $V(x)\le E$ $V (x) \leq E$ there. The lowest point of the whole curve is the central well's floor, so $E_{\min}=-V_1$ $E_{m i n} = - V_{1}$ . This double-hump shape is a simple model for a particle trapped in a central attractive well flanked by two barriers it cannot classically cross — e.g. a particle bound between two humps, as in a simplified model of a symmetric double-well/double-barrier system.
✓Final answer
1. forbidden for $x\ge a$ , $E_{\min}=0$ — like a ball unable to climb a single step higher than its energy.
2. forbidden for all $x$ at the marked energy, $E_{\min}=V_0$ — like a ball at the base of a staircase too tall to climb even the first step.
3. allowed only in $a<x<b$ , $E_{\min}=-V_1$ — the classic rectangular potential well (e.g. a particle trapped inside a finite well).
4. forbidden in $-b/2<x<-a/2$ and $a/2<x<b/2$ , $E_{\min}=-V_1$ — a particle trapped in a central well flanked by two barriers it cannot cross.

A particle can only be where its kinetic energy $K=E-V(x)\ge 0$ , i.e. where the total-energy level lies at or above the potential curve. Reading each graph: (a) forbidden for $x\ge a$ ; (b) forbidden everywhere for the marked $E$ ; (c) confined to $a<x<b$ ; (d) trapped in the central well and barred from the two barrier humps. The minimum total energy in each case equals the lowest point of that potential.

Concept

Total mechanical energy is constant, $E=K+V(x)$ , so $K=E-V(x)$ . Kinetic energy can never be negative, therefore the particle is allowed only where $V(x)\le E$ and forbidden wherever $V(x)>E$ . The least total energy a particle can have equals the minimum value of $V(x)$ (there $K=0$ ).

(a) Single upward step

For $x<a$ , $V=0<E$ , so the region is allowed. For $x\ge a$ , $V=V_0>E$ , giving $K<0$ , so the particle cannot be found for $x\ge a$ . The lowest potential is $0$ , so the minimum total energy is $0$ . Physical picture: a particle meeting a potential step, e.g. an electron approaching a metal boundary.

(b) Rising staircase

The marked energy lies below the whole curve ( $E<V_0\le V(x)$ for all $x$ ), so $K<0$ everywhere and the particle cannot be found anywhere with this energy. To be found even in its lowest broad region it needs $E\ge V_0$ , so the minimum total energy is $V_0$ . Physical picture: a charge driven through a succession of rising potential steps.

(c) Rectangular well

$V=V_0>E$ for $x<a$ and for $x>b$ , so those regions are forbidden; $V=-V_1<E$ for $a<x<b$ , which is allowed. The particle is confined to $a<x<b$ , and the minimum total energy is $-V_1$ (the floor of the well). Physical picture: a particle trapped in a box / finite square well, like a molecule bouncing between two rigid walls.

(d) Twin barriers with a central well

$V=-V_1<E$ in the central well $|x|<a/2$ (allowed) and $V=0<E$ for $|x|>b/2$ (allowed), but each hump rises to $V_0>E$ . The particle is forbidden where the humps rise above $E$ , i.e. in $-\tfrac{b}{2}<x<-\tfrac{a}{2}$ and $\tfrac{a}{2}<x<\tfrac{b}{2}$ ; a particle sitting in the central well is classically trapped. The minimum total energy is $-V_1$ . Physical picture: a particle bound in a well guarded by potential barriers, e.g. an $\alpha$ -particle held inside a nucleus.

✓Final answer

x\ge a

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Concept

(a) Single upward step

(b) Rising staircase

(c) Rectangular well

(d) Twin barriers with a central well

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