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=+Aandx=−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)=ωA2−x2
At x=±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 A2−x2 negative, which is impossible for a real velocity. This is why x=±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=21mω2A2(constant)
where K=21mω2(A2−x2) is kinetic energy and U=21mω2x2 is potential energy.
At a turning point (x=±A): K=0 and U=E. All the energy is potential; none is kinetic.
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=−ω2x) reach their maximum magnitude exactly at x=±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=±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.) …
KE = (1/2)k(A^2 - x^2): maximum at mean position (x=0) and zero at extreme positions (x=±A) — an inverted (downward) parabola in x. PE = (1/2)k x^2: zero at the mean position and maximum at the extremes — an upward parabola in x. The two curves cross at x = ±A/sqrt(2), and KE+PE = (1/2)kA^2 = constant total energy at every x.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
CBSE 2024Set ANNUAL2 marks
Q.Draw a graph of kinetic energy and potential energy of an oscillating particle with displacement.
›Reveal solutionSolution
[!TLDR]
KE = (1/2)k(A^2 - x^2): maximum at mean position (x=0) and zero at extreme positions (x=±A) — an inverted (downward) parabola in x. PE = (1/2)k x^2: zero at the mean position and maximum at the extremes — an upward parabola in x. The two curves cross at x = ±A/sqrt(2), and KE+PE = (1/2)kA^2 = constant total energy at every x.
Q.Where is the potential energy of a body maximum and minimum?
›Reveal solutionSolution
[!TLDR]
Potential energy is maximum at the extreme position (point of maximum displacement from the equilibrium/mean position, e.g. greatest height or maximum compression/extension) and minimum (often taken as zero) at the equilibrium/mean position or reference level.