Read Any SHM Graph in 5 Seconds
Graph-based questions are the highest-scoring opportunity in SHM. Most students get them wrong because they memorize shapes without understanding what each graph tells them. This page fixes that.
CBSE: sketch graphs of x-t, v-t, KE-t, PE-t with labels. Know shape and phase relationship. These appear in 3-mark questions.
NEET: "Which graph correctly represents...?" or "At which point does velocity equal maximum?" — all require instant graph reading. Practice elimination — 3 wrong options are always clearly wrong.
JEE Main: match-the-column with graphs. Given x(t), find which graph represents KE. Given a-x graph slope, find ω². Requires graph manipulation skill.
JEE Advanced: phase-space diagrams (v vs x plots), parametric graphs, energy diagrams where you identify turning points. These require deep conceptual understanding.
Interactive SHM Graph Visualizer
Adjust amplitude and frequency to see how x-t, v-t, a-t graphs change in real-time.
Understanding Each Graph Type
Displacement vs Time (x-t) Graph
Shape: Sinusoidal (cosine or sine)
Reading from x-t graph:
- Amplitude = peak value (max |x|)
- Time period T = distance between consecutive peaks
- Frequency f = 1/T
- At t = 0: initial position from y-intercept
- Initial phase φ₀ from initial conditions
Slope of x-t graph = velocity
- At x = A (peak): slope = 0 → v = 0
- At x = 0 (equilibrium): slope is steepest → v = max
- Positive slope → moving in +x direction
- Negative slope → moving in −x direction
Many students assume x = 0 always at t = 0. Wrong. Initial phase φ₀ determines starting position. If x = A at t = 0, the graph is cosine. If x = 0 and moving in +x at t = 0, the graph is sine.
Velocity vs Time (v-t) Graph
Shape: Sinusoidal, but shifted 90° (π/2) ahead of x-t
v(t) = −Aω sin(ωt) [if x = A cosωt]
x = A cos(ωt) → v = dx/dt = −Aω sin(ωt)
Since −sin(ωt) = cos(ωt + π/2), velocity leads displacement by π/2 in phase.
This means: when x reaches its maximum (+A), v is crossing zero. When x is zero (equilibrium), v is at its maximum (±Aω). They are always 90° apart.
In exams: given x(t) = 4 sin(πt), find v(t). Simply differentiate: v(t) = 4π cos(πt). The amplitude of v = Aω = 4π. No need to think further.
Acceleration vs Time (a-t) Graph
a(t) = −Aω² cos(ωt) = −ω²x(t)
Shape: Same shape as x-t but inverted (180° phase difference)
Since a = −ω²x, plotting a vs x gives a straight line through origin with slope = −ω². This is one of the most tested graphs in JEE!
Trick: In JEE, if given a vs x graph as a straight line with negative slope, directly find ω² = |slope|, then T = 2π/ω.
KE & PE vs Time
Both KE and PE oscillate, but at twice the frequency (2f) of the SHM.
- KE = ½mω²A²cos²(ωt) — oscillates at frequency 2f
- PE = ½mω²A²sin²(ωt) — oscillates at frequency 2f
- KE + PE = ½mω²A² = constant (independent of t)
- Average KE = Average PE = E/2 (over one full period)
Many students say "KE and PE have the same frequency as SHM." Wrong. They have TWICE the frequency. If T = 1s for SHM, then KE and PE complete a cycle in 0.5s.
KE & PE vs Position (x)
KE = ½mω²(A²−x²) → parabola opening downward in x
PE = ½mω²x² → parabola opening upward in x
- KE and PE are both parabolic in x
- They intersect at x = ±A/√2 (KE = PE = E/2)
- At x = 0: KE = E (maximum), PE = 0
- At x = ±A: PE = E (maximum), KE = 0
Phase Space Diagram (v vs x) — JEE Advanced Special
Plot velocity v on y-axis and displacement x on x-axis. The resulting curve is an ellipse.
From: x = A cosθ and v = −Aω sinθ
→ (x/A)² + (v/Aω)² = cos²θ + sin²θ = 1
This is the equation of an ellipse with semi-axes A (x-direction) and Aω (v-direction). The particle moves anticlockwise on this ellipse.
- Rightmost/leftmost points: x = ±A, v = 0 (extreme positions)
- Top/bottom points: x = 0, v = ±Aω (equilibrium, max speed)
- Area of ellipse = πA(Aω) = πA²ω (related to action variable in classical mechanics)
- If ellipse is a circle: Aω = A → ω = 1 rad/s
Phase Difference — The Concept That Trips Students
Phase Difference Between x, v, and a
| Pair | Phase Difference | What This Means | Exam Relevance |
|---|---|---|---|
| v leads x | π/2 (90°) | When x is max, v = 0. When x = 0 (crossing), v is max. | NEET, CBSE, JEE Main |
| a leads v | π/2 (90°) | When v is max (at x=0), a = 0. When v = 0 (at extremes), a is max. | NEET, JEE Main |
| a vs x | π (180°) | Acceleration is always exactly opposite to displacement. This IS the definition of SHM. | All exams |
| KE vs PE (time) | π/2 (90°) | When KE is max, PE = 0 and vice versa. KE and PE are 90° out of phase. | JEE, NEET |
x → v → a each step adds π/2 phase. Equivalently: x and a are π apart. Think of it as: "x delays, a leads — they're always opposing each other."
How to Extract Values from SHM Graphs
5-Step Method to Read Any SHM Graph in an Exam
- Identify the type — Is it x-t, v-t, a-t, or energy graph? Read the axis labels.
- Read the amplitude — Maximum value on the y-axis.
- Read the time period T — Horizontal distance between two consecutive peaks (or zero crossings in the same direction).
- Calculate ω — ω = 2π/T. If the graph shows a-x: ω² = |slope|.
- Find what's asked — Use T, A, ω to compute frequency, energy, velocity, or acceleration at specific points.
In energy (KE or PE) vs time graphs, the apparent period is T/2, not T. Don't read T directly — multiply by 2. This is consistently used as a trap in JEE Main.