SHM — JEE Diagram Reference

These are the canonical SHM diagrams that appear repeatedly in JEE problems. Each snapshot shows a physically precise moment.

1. Block at positive extreme

$x = +A$, $v = 0$, $a = -\omega^2 A$ (maximum, toward centre), $KE = 0$, $PE = E$

SpringMass amplitude 80 k 100 m 1
Snapshot at extreme
Show spring block wall equilibrium displacement-arrow acceleration-arrow
Hide graph-x energy readouts velocity-arrow

2. Block at centre — moving right

$x = 0$, $v = +A\omega$ (maximum), $a = 0$, $KE = E$, $PE = 0$

SpringMass amplitude 80 k 100 m 1
Snapshot at centre
Show spring block wall equilibrium velocity-arrow
Hide graph-x energy readouts displacement-arrow acceleration-arrow

3. Block at negative extreme

$x = -A$, $v = 0$, $a = +\omega^2 A$ (toward centre), $KE = 0$, $PE = E$

SpringMass amplitude 80 k 100 m 1
Snapshot at negative
Show spring block wall equilibrium displacement-arrow acceleration-arrow
Hide graph-x energy readouts velocity-arrow

4. KE = PE position

$x = \pm A/\sqrt{2}$, $KE = PE = E/2$

SpringMass amplitude 80 k 100 m 1
Snapshot at equal-energy
Show spring block wall equilibrium displacement-arrow energy KE PE E
Hide graph-x readouts velocity-arrow acceleration-arrow

5. Phase relationships — all three graphs

x(t) leads v(t) by $-90°$. a(t) is antiphase with x(t).

SpringMass amplitude 80 k 100 m 1
Show spring block wall graph-x graph-v graph-a
Hide energy readouts displacement-arrow velocity-arrow acceleration-arrow equilibrium

6. Energy vs displacement

PE parabola (orange), KE inverted parabola (green), total E (blue dashed). Marker shows current position.

SpringMass amplitude 80 k 100 m 1
Snapshot at equal-energy
Show spring block wall graph-energy KE PE E
Hide energy readouts displacement-arrow velocity-arrow acceleration-arrow equilibrium

7. Effect of amplitude on energy

$E \propto A^2$. Double amplitude → 4× energy. Same $k$, same $m$, same $T$.

Small amplitude ($A = 40$):

SpringMass amplitude 40 k 100 m 1
Snapshot at extreme
Show spring block wall displacement-arrow energy KE PE E
Hide graph-x readouts velocity-arrow acceleration-arrow equilibrium

Large amplitude ($A = 80$):

SpringMass amplitude 80 k 100 m 1
Snapshot at extreme
Show spring block wall displacement-arrow energy KE PE E
Hide graph-x readouts velocity-arrow acceleration-arrow equilibrium

PE bars confirm 4× ratio.

Quick reference

$x = A\cos(\omega t + \phi)$ $v = -A\omega\sin(\omega t + \phi) = \pm\omega\sqrt{A^2 - x^2}$ $a = -\omega^2 x$ $\omega = \sqrt{k/m}, \quad T = 2\pi\sqrt{m/k}$ $KE = \tfrac{1}{2}k(A^2-x^2), \quad PE = \tfrac{1}{2}kx^2, \quad E = \tfrac{1}{2}kA^2$