Conservation of energy in SHM — kinetic and potential energy at any displacement

Question

A particle of mass 0.5 kg executes SHM with amplitude 10 cm and angular frequency $\omega = 20$ rad/s. Find the kinetic energy and potential energy when the displacement is 6 cm from the mean position. Verify that total energy is conserved.

(JEE Main 2023, similar pattern)

Solution — Step by Step

For a particle executing SHM with amplitude $A$ and angular frequency $\omega$ :

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

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

E_{\text{total}} = \frac{1}{2}m\omega^2 A^2

where $x$ is the displacement from the mean position.

E_{\text{total}} = \frac{1}{2} \times 0.5 \times (20)^2 \times (0.10)^2

= \frac{1}{2} \times 0.5 \times 400 \times 0.01 = \mathbf{1.0 \text{ J}}

PE = \frac{1}{2} \times 0.5 \times 400 \times (0.06)^2 = \frac{1}{2} \times 0.5 \times 400 \times 0.0036 = \mathbf{0.36 \text{ J}}

KE = \frac{1}{2} \times 0.5 \times 400 \times [(0.10)^2 - (0.06)^2]

= \frac{1}{2} \times 0.5 \times 400 \times [0.01 - 0.0036] = \frac{1}{2} \times 0.5 \times 400 \times 0.0064 = \mathbf{0.64 \text{ J}}

Verification: $KE + PE = 0.64 + 0.36 = 1.0$ J $= E_{\text{total}}$ ✓

Why This Works

In SHM, the restoring force is conservative ( $F = -kx = -m\omega^2 x$ ), so total mechanical energy is conserved. Energy continuously converts between kinetic and potential forms:

At mean position ( $x = 0$ ): KE is maximum, PE = 0
At extreme position ( $x = \pm A$ ): PE is maximum, KE = 0
At any intermediate point: both KE and PE are non-zero, but their sum equals $E_{\text{total}}$

The KE and PE are complementary — they add up to a constant. If you plot both against displacement, KE is an inverted parabola and PE is an upright parabola. They intersect at $x = A/\sqrt{2}$ , where KE = PE = $E_{\text{total}}/2$ .

Alternative Method

Use the velocity formula directly:

v = \omega\sqrt{A^2 - x^2} = 20\sqrt{0.01 - 0.0036} = 20\sqrt{0.0064} = 20 \times 0.08 = 1.6 \text{ m/s}

KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times (1.6)^2 = 0.64 \text{ J}

PE = E_{\text{total}} - KE = 1.0 - 0.64 = 0.36 \text{ J}

Quick ratio trick: At displacement $x$ , the fraction of total energy that is PE = $(x/A)^2$ and the fraction that is KE = $1 - (x/A)^2$ . At $x = 6$ cm, $A = 10$ cm: PE fraction = $(6/10)^2 = 0.36$ , KE fraction = $0.64$ . No need to calculate absolute values if the question only asks for the ratio.

Common Mistake

Students often use $PE = mgh$ for SHM problems. That formula applies only to gravitational potential energy. In SHM, the potential energy is $\frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2$ — it comes from the elastic restoring force. Unless the SHM is specifically a vertical spring (where gravity also contributes), always use the spring PE formula, not the gravitational one.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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