Oscillator Energy: Worked Solutions

Try the questions on the lesson page first. Then compare both the answer and the reasoning. The worked steps here are original.

Question 1: At x = A/2 in a horizontal ideal spring oscillator, what fraction of total energy is kinetic?

At x = A/2 in a horizontal ideal spring oscillator, what fraction of total energy is kinetic?

1. Use total energy E = ½kA² and U = ½kx². Why: These equations describe the horizontal ideal spring system.
2. At x = A/2, U/E = (x/A)² = 1/4. Why: Potential energy depends on squared displacement, not displacement itself.
3. Use K = E − U, giving K/E = 3/4. Why: The remaining conserved mechanical energy is kinetic.

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Question 2: Amplitude doubles while the ideal spring and mass stay unchanged. Total energy…

Amplitude doubles while the ideal spring and mass stay unchanged. Total energy…

1. Write E = ½kA². Why: At a turning point all the oscillator’s energy is potential.
2. Replace A with 2A: E₂ = ½k(2A)² = 4E₁. Why: Squaring the amplitude doubles the factor twice.
3. Conclude that energy quadruples. Why: The unchanged mass and stiffness also mean the period stays the same.

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Question 3: In a vertical spring, x is displacement from the stretched equilibrium. The term ½kx² represents…

In a vertical spring, x is displacement from the stretched equilibrium. The term ½kx² represents…

1. Distinguish natural-length extension y from equilibrium displacement x = y − mg/k. Why: Gravity has already stretched the spring at equilibrium.
2. Include U = ½ky² − mgy for spring plus gravity. Why: The chosen spring–mass–Earth system includes both potential contributions.
3. Shift the potential zero to write U_relative = ½kx². Why: The equilibrium-centered expression represents their combined change, not spring energy alone.

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Question 4: equal kinetic and potential energies

For a horizontal ideal spring, find the positions where K = Uₛ. Then explain why there are four such moments per cycle.

1. Set K = U = E/2, where E = ½kA². Why: Equal energy shares must each be half of the fixed total.
2. Solve ½kx² = ¼kA² to get x = ±A/√2. Why: Both sides of equilibrium can have the same potential energy because it depends on x².
3. Each position is crossed in both travel directions, giving four moments per cycle; speed is v_max/√2. Why: The same energy state does not specify velocity direction. K = E/2 makes squared speed half its maximum.

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