At any point: KE = ½m·ω²(A²-x²), PE = ½mω²x². Total E = ½mω²A² = ½kA². Proof.

Show Total Energy in SHM is Constant — E = ½kA²

Question

A particle undergoes Simple Harmonic Motion with amplitude $A$ , angular frequency $\omega$ , and spring constant $k$ . Show that the total mechanical energy at any displacement $x$ is constant and equals:

E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2

Solution — Step by Step

Let the displacement at time $t$ be $x = A\sin(\omega t)$ .

The velocity at any instant follows from differentiating: $v = \frac{dx}{dt} = A\omega\cos(\omega t)$ .

We’ll express everything in terms of $x$ and $v$ , not $t$ , so the result is general.

KE = \frac{1}{2}mv^2

We need $v$ in terms of $x$ . From SHM kinematics, we know:

v^2 = \omega^2(A^2 - x^2)

This comes from $\sin^2 + \cos^2 = 1$ — substitute $\sin(\omega t) = x/A$ and $\cos(\omega t) = v/(A\omega)$ , and the identity gives you this directly.

\boxed{KE = \frac{1}{2}m\omega^2(A^2 - x^2)}

The restoring force in SHM is $F = -kx$ , where $k = m\omega^2$ .

Potential energy is the work done against this restoring force to bring the particle from equilibrium to displacement $x$ :

PE = \int_0^x kx\, dx = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2

\boxed{PE = \frac{1}{2}m\omega^2 x^2}

E = KE + PE = \frac{1}{2}m\omega^2(A^2 - x^2) + \frac{1}{2}m\omega^2 x^2

The $x^2$ terms cancel perfectly:

E = \frac{1}{2}m\omega^2 A^2 - \frac{1}{2}m\omega^2 x^2 + \frac{1}{2}m\omega^2 x^2

\boxed{E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}kA^2}

This is independent of $x$ — it doesn’t matter where the particle is. Total energy is constant.

Why This Works

The key insight is that KE and PE are complementary — as one rises, the other falls by exactly the same amount. At the mean position ( $x = 0$ ), all energy is kinetic ( $KE = \frac{1}{2}m\omega^2 A^2$ , $PE = 0$ ). At the extreme positions ( $x = \pm A$ ), all energy is potential ( $PE = \frac{1}{2}kA^2$ , $KE = 0$ ).

This is only possible because the restoring force is linear ( $F = -kx$ ). A linear force gives a quadratic PE, which exactly mirrors the quadratic KE from the velocity relation. If the force were non-linear (like a pendulum at large angles), this clean cancellation would break down.

The result $E = \frac{1}{2}kA^2$ tells us something practically useful: total energy depends only on the amplitude. Double the amplitude → four times the energy. This is why in JEE problems, when two SHM systems combine or a mass is added at mean position, you track how $A$ changes to find the new energy.

Alternative Method — Using the Time Equations Directly

Instead of using $v^2 = \omega^2(A^2 - x^2)$ , substitute the explicit time expressions:

KE = \frac{1}{2}m(A\omega\cos\omega t)^2 = \frac{1}{2}m\omega^2 A^2 \cos^2(\omega t)

PE = \frac{1}{2}k(A\sin\omega t)^2 = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t)

Adding both:

E = \frac{1}{2}m\omega^2 A^2 (\sin^2\omega t + \cos^2\omega t) = \frac{1}{2}m\omega^2 A^2

The identity $\sin^2\theta + \cos^2\theta = 1$ does the heavy lifting. This method is slightly more intuitive — the two energies oscillate in phase opposition, and their squares always sum to 1.

In JEE Main, they sometimes ask for the position where $KE = PE$ . Set $\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2 x^2$ , which gives $x = \frac{A}{\sqrt{2}}$ . At this point, each equals half the total energy. This appeared in JEE Main 2023 Session 2.

Common Mistake

Students often write $PE = \frac{1}{2}mv^2$ (confusing it with KE) or forget to use $k = m\omega^2$ when switching between forms. The potential energy in SHM is elastic PE stored in the equivalent spring — it’s $\frac{1}{2}kx^2$ , not $mgh$ . There’s no gravity here (or if there is, the equilibrium position absorbs it). Writing $PE = mgx$ is the most common incorrect step in CBSE board answers, and examiners deduct marks specifically for it.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Time Equations Directly

Common Mistake

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