Energy of Shm

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Step 1

Write the total mechanical energy of a spring oscillator in terms of k and amplitude A.

Hint

At the turning points (x = ±A) the mass is momentarily at rest, so all energy is potential: E = ½kA².

Step 2

At the equilibrium position x = 0, all energy is kinetic. Use energy conservation to find the maximum speed.

Step 3

Express the angular frequency ω in terms of k and m.

Solution walkthrough

In simple harmonic motion, energy sloshes continuously between kinetic and potential form but the total never changes. The formula E = ½kA² captures this elegantly: it doesn't matter where in the cycle you look, the total is always ½kA². At the amplitude A, the mass is at rest — zero kinetic, all potential. At x = 0 (equilibrium), the mass moves fastest — all kinetic, zero potential. Setting these equal: ½mv²_max = ½kA² → v_max = A√(k/m) = Aω. With k = 80 N/m and A = 0.15 m: E = ½ × 80 × 0.0225 = 0.9 J. Maximum speed: v_max = √(2 × 0.9 / 2) = √0.9 ≈ 0.949 m/s. Angular frequency: ω = √(80/2) = √40 ≈ 6.32 rad/s. The same E = ½kA² appears in any harmonic oscillator — an atomic bond vibrating in a crystal lattice stores the same kind of energy.

Step-by-step solution

Write the total mechanical energy of a spring oscillator in terms of k and amplitude A.

At the equilibrium position x = 0, all energy is kinetic. Use energy conservation to find the maximum speed.

Express the angular frequency ω in terms of k and m.

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