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Energy graphs in simple harmonic motion are a favourite for one-mark and assertion-reason questions. The trouble is that there are two different graphs, energy against displacement and energy against time, and students mix up their shapes and periods. This post sets out both, works a spring example with numbers and lists the slips to watch for.
What do the syllabus and textbook expect?
The JEE (Main) 2026 syllabus lists "energy in S.H.M.: kinetic and potential energies" under Oscillations, and the CBSE Physics curriculum for 2026-27 has the same topic in Class XI. Section 13.7 of the NCERT Class 11 chapter on oscillations derives the results with x = A cos(ωt + φ):
K = ½kA² sin²(ωt + φ)
U = ½kx² = ½kA² cos²(ωt + φ)
E = K + U = ½kA², constant
NCERT also notes that since the sign of v does not matter in K, the period of K is T/2, and that both energies peak twice in each period.
What does energy against displacement look like?
Rewrite K using v² = ω²(A² − x²) and k = mω²: K = ½k(A² − x²).
U = ½kx² is an upward parabola, zero at the mean position, maximum at x = ±A.
K = ½k(A² − x²) is a downward parabola, maximum at x = 0, zero at x = ±A.
E is a horizontal line at ½kA². At every x, the two parabolas add up to it.
They cross where ½kx² = ½k(A² − x²), which gives x² = A²/2, so x = A/√2 ≈ 0.71A. At x = A/2, the potential energy is only a quarter of the total.
What does energy against time look like?
Against time, K and U are sin² and cos² curves. Each swings between 0 and ½kA² and never goes negative. Each repeats every T/2, so the energy frequency is twice the oscillation frequency. The two curves are mirror images about the line E/2, and the average of each over a cycle is E/2.
Worked example: a block on a spring
These are sample numbers. A 0.5 kg block on a smooth floor is attached to a spring with k = 200 N/m and oscillates with amplitude A = 0.1 m.
ω = √(k/m) = √(200/0.5) = 20 rad/s. T = 2π/ω ≈ 0.314 s.
Total energy E = ½kA² = ½ × 200 × 0.01 = 1 J. Check: maximum speed = ωA = 2 m/s, and ½ × 0.5 × 2² = 1 J.
At x = 0.05 m (A/2): U = ½ × 200 × 0.0025 = 0.25 J, so K = 0.75 J.
At x = 0.06 m: U = 0.36 J, K = 0.64 J, so v = √(2 × 0.64/0.5) = 1.6 m/s. Check with v = ω√(A² − x²) = 20 × 0.08 = 1.6 m/s.
K = U at x = 0.1/√2 ≈ 0.0707 m, where each is 0.5 J.
The energy graphs against time repeat every T/2 ≈ 0.157 s. Starting from the mean position, the block reaches x = A/√2 after one eighth of a period, about 0.039 s.
Which mistakes cost marks?
Equal energies at A/2. They are equal at A/√2.
Energy period equals T. It is T/2.
Negative kinetic energy on a sketch. Both curves stay at or above zero.
Reading the graph as a picture. The parabola U = ½kx² is not the path of the block, which moves in a straight line.
Forgetting that E depends on amplitude squared. Doubling A makes E four times larger, at any mass.
Reading a graph as a picture is a known error. In a study of 895 high school and college students, Beichner (1994) listed "misinterpreting graphs as pictures" and slope-height confusion among the common difficulties with kinematics graphs, and the same habits show up with energy graphs. For other SHM problem types, such as springs in series and floating blocks, see our post on simple harmonic motion problems.
To see how the two energies trade off as the block moves, Fermi's Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Sketch both graphs from memory for one oscillation, marking A, A/√2, T/2 and E/2.
Redo the worked example with amplitude 0.2 m and predict the new energies before calculating.
Answer three quick questions without notes: the fraction of energy that is kinetic at x = A/2, A/3 and A/√2. Testing yourself beats rereading for long-term memory, as Roediger and Karpicke (2006) found on delayed tests.
Then try NCERT Example 13.7 and check each energy against the graphs.
Right now, sketch U and K against x on one set of axes and mark where they cross.
