AP Physics 1 Quiz: Energy Of Simple Harmonic Oscillators

What this quiz covers

This quiz focuses on Energy Of Simple Harmonic Oscillators, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $0\,\text{J}$
2. $10\,\text{J}$
3. $20\,\text{J}$ (correct answer)
4. Less than $20\,\text{J}$ because energy is lost whenever velocity is zero.

Explanation: This problem tests energy conservation in simple harmonic oscillators at extreme positions. In SHM on a frictionless surface, total mechanical energy remains constant throughout the motion. At equilibrium, the block has maximum speed with 20 J of kinetic energy and zero spring potential energy, establishing total energy as 20 J. At maximum displacement where the block is momentarily at rest, all kinetic energy transforms into spring potential energy, so the spring stores all 20 J. Choice D incorrectly suggests energy is lost when velocity is zero, but energy simply changes form rather than disappearing. When solving SHM problems, identify total energy at any point and apply conservation to find energies at other positions.

Question 2

1. Kinetic energy is zero because the cart is far from equilibrium.
2. Kinetic energy equals spring potential energy. (correct answer)
3. Total mechanical energy is greater than at $x=0$.
4. Potential energy is zero because the cart is moving.

Explanation: This question tests energy statements at a specific position in simple harmonic motion for a cart-spring system. Total mechanical energy E is constant, exchanging between K and U. At x = A/√2, U = E/2 and K = E/2, as (x/A)^2 = 1/2. Thus, kinetic equals potential energy there. Distractor C incorrectly suggests total E is greater than at x=0, but conservation keeps it constant. Calculate the position's (x/A)^2 fraction to determine energy equality or ratios in SHM problems.

Question 3

1. Kinetic energy is $0\,\text{J}$. (correct answer)
2. Kinetic energy is $6\,\text{J}$.
3. Total energy is $0\,\text{J}$ because both energies are zero there.
4. Total energy is less than $6\,\text{J}$ because the spring has stopped doing work.

Explanation: This problem tests understanding of energy states in simple harmonic oscillators at turning points. In undamped SHM, total mechanical energy remains constant as kinetic and potential energies continuously interchange. At equilibrium, the mass has 6 J of kinetic energy and zero spring potential energy. At a turning point, the mass is instantaneously at rest, meaning velocity and kinetic energy are both zero. All 6 J of total energy is stored as spring potential energy at this position. Choice C incorrectly claims total energy is zero when both energies are zero, not recognizing that potential energy is maximum when kinetic energy is zero. When analyzing SHM at turning points, remember that zero velocity means zero kinetic energy, not zero total energy.

Question 4

1. At maximum displacement, $K > U$ because the cart has moved farthest.
2. At maximum displacement, $K = U$ because energy is shared equally there.
3. At maximum displacement, $U > K$. (correct answer)
4. At maximum displacement, total energy is smaller than $14\,\text{J}$ because the cart stops.

Explanation: This question examines energy distribution in simple harmonic oscillators at different positions. In frictionless SHM, total mechanical energy (14 J) remains constant while kinetic and potential energies exchange. At equilibrium, all 14 J is kinetic energy (K = 14 J, U = 0 J). At maximum displacement where the cart is momentarily at rest, all energy converts to spring potential energy (K = 0 J, U = 14 J), making U > K at this position. Choice A incorrectly claims kinetic energy is greater at maximum displacement, when actually the cart is at rest there. To analyze SHM energy comparisons, identify where speed is maximum (equilibrium) versus zero (turning points).

Question 5

1. Kinetic energy is maximum and potential energy is minimum. (correct answer)
2. Potential energy is maximum because the spring force is greatest at equilibrium.
3. Total mechanical energy is increasing because the mass speeds up through equilibrium.
4. Kinetic energy is zero because the displacement is zero.

Explanation: This question assesses energy distribution at equilibrium in simple harmonic oscillators. In frictionless SHM, total mechanical energy stays constant, comprising kinetic and potential components that interchange. At equilibrium (x=0), the spring is unstretched, minimizing potential energy, while the mass's maximum speed maximizes kinetic energy. As it moves outward, kinetic energy decreases and potential increases, peaking at maximum displacement where kinetic is zero. Distractor B incorrectly asserts potential is maximum at equilibrium due to spring force, but force is zero there, and potential is actually minimum. A key strategy is to use conservation of energy to predict that maximum kinetic occurs where potential is minimum, and vice versa.

Question 6

1. $0\,\text{J}$
2. $9\,\text{J}$ (correct answer)
3. $18\,\text{J}$
4. Less than $9\,\text{J}$ because energy is lost at each turning point.

Explanation: This problem tests understanding of energy conservation in simple harmonic oscillators. In undamped SHM, total mechanical energy remains constant as kinetic and potential energies continuously exchange. At equilibrium, the system has 9 J of kinetic energy and zero spring potential energy, establishing total mechanical energy as 9 J. This total energy remains constant throughout the motion, including at maximum displacement where all energy is potential. Choice D incorrectly suggests energy is lost at turning points, but in SHM without damping, energy is conserved. When solving SHM problems, recognize that total energy can be calculated at any convenient point and remains the same everywhere.

Question 7

1. Kinetic energy is larger than potential energy because the mass moved farther.
2. Kinetic energy equals potential energy because energy is shared equally at the ends.
3. Potential energy is $9\,\text{J}$ and kinetic energy is $0\,\text{J}$. (correct answer)
4. Total energy is less than $9\,\text{J}$ because the mass stops at the turning point.

Explanation: This problem involves energy conservation in simple harmonic oscillators. For frictionless SHM, total mechanical energy remains constant at K + U = 9 J (from equilibrium values). At maximum displacement, the mass has zero speed, which means kinetic energy K = 0 J. By energy conservation, all 9 J must be stored as potential energy in the spring, so U = 9 J. Choice A incorrectly claims kinetic energy is larger at maximum displacement, when actually kinetic energy is zero there because the mass is momentarily stationary. The strategy is to recognize that at turning points (maximum displacement), all energy is potential since velocity equals zero.

Question 8

1. Kinetic energy is maximum at maximum displacement, so $K=2\,\text{J}$ there.
2. Total mechanical energy changes with time, so $K+U$ is not constant.
3. Potential energy equals $2\,\text{J}$ and kinetic energy equals $0\,\text{J}$. (correct answer)
4. Some energy is lost at the turning point, so $U<2\,\text{J}$.

Explanation: This problem tests understanding of energy conservation in simple harmonic oscillators. In frictionless SHM, total mechanical energy remains constant throughout oscillation. At equilibrium, K = 2 J and U = 0 J, giving a total of 2 J. At maximum displacement where the mass is momentarily at rest, velocity equals zero, so kinetic energy K = 0 J. By conservation of energy, all 2 J must be stored as spring potential energy, making U = 2 J. Choice A incorrectly claims kinetic energy is maximum at maximum displacement, when it's actually zero there since the mass stops momentarily. Remember that in SHM, energy continuously exchanges between kinetic and potential forms while maintaining a constant total.

Question 9

1. It is $0\,\text{J}$ because equilibrium is where potential energy is stored.
2. It is $7\,\text{J}$ because all energy is potential energy when speed is zero. (correct answer)
3. It is less than $7\,\text{J}$ because some energy is lost at each turning point.
4. It is greater than $7\,\text{J}$ because potential energy increases over time.

Explanation: This question examines energy conservation in simple harmonic oscillators. In undamped SHM, mechanical energy is conserved, so the total K + U remains constant at 7 J (from equilibrium). At a turning point where speed equals zero, kinetic energy must be 0 J (since K = ½mv²). Therefore, all 7 J of mechanical energy exists as spring potential energy at the turning point. Choice C incorrectly suggests energy is lost at turning points, but energy only transforms between forms without being destroyed in conservative systems. To solve SHM energy problems, identify total energy from any known state and apply conservation to find energy distribution at other positions.

Question 10

1. The kinetic energy is $K_0$ because speed is greatest at maximum displacement.
2. The spring potential energy is $K_0$ because all the mechanical energy is stored in the spring. (correct answer)
3. The total mechanical energy is less than $K_0$ because energy is lost at the turning point.
4. The total mechanical energy increases above $K_0$ because the spring does work on the block.

Explanation: This problem tests understanding of energy conservation in simple harmonic oscillators. In SHM without friction, total mechanical energy remains constant throughout the motion. At equilibrium, all energy is kinetic (K₀), while at maximum displacement, the block is momentarily at rest so all energy converts to spring potential energy. Since total energy is conserved, the spring potential energy at maximum displacement must equal K₀. Choice A incorrectly states the block has maximum speed at maximum displacement, when actually speed is zero there. When solving SHM energy problems, remember that total mechanical energy stays constant, with continuous exchange between kinetic and potential forms.

Question 11

1. Kinetic energy is half of its maximum value.
2. Potential energy is greater than kinetic energy. (correct answer)
3. Potential energy is zero because the block is moving.
4. Total mechanical energy is smaller than at equilibrium.

Explanation: This question evaluates energy relations when speed is half maximum in simple harmonic motion for a block-spring system. Total energy E is conserved, with K = (1/2)mv² and U = E - K. When v = v_max/2, K = (1/4)E, so U = (3/4)E, meaning U > K. This occurs at x = (√3/2)A, where potential dominates. Distractor A incorrectly states K is half maximum, but it's actually one-quarter due to the squared velocity term. Use the relation between speed and energy fractions to determine comparative values in similar SHM scenarios.

Question 12

1. The system has lost mechanical energy because the mass stops.
2. Kinetic energy is maximum because the displacement is maximum.
3. Potential energy is maximum. (correct answer)
4. Total mechanical energy is larger than at equilibrium because the spring is stretched.

Explanation: This question examines energy at maximum displacement in simple harmonic motion. In a frictionless mass-spring oscillator, total mechanical energy is constant, with potential energy maximum at maximum displacement where kinetic is zero due to zero speed. At equilibrium, this reverses, with kinetic maximum and potential minimum. Energy conservation ensures no loss during the oscillation. Distractor A claims energy loss because the mass stops, but stopping is momentary with full energy in potential form, not lost. A useful strategy is to use energy conservation to verify that total energy equals maximum potential or maximum kinetic at extreme positions.

Question 13

1. $K$ is greater than $U$. (correct answer)
2. $U$ is greater than $K$.
3. $K=U$ because equilibrium means forces balance.
4. Total energy is smaller at equilibrium than at the endpoints.

Explanation: This question tests comprehension of kinetic and potential energy comparison in simple harmonic motion. In a frictionless mass-spring system, total mechanical energy is constant, oscillating between kinetic and spring potential forms. At equilibrium, potential energy is minimal (zero for horizontal springs) since displacement is zero, and kinetic energy is maximal due to peak speed. Away from equilibrium, kinetic energy decreases as it's converted to potential energy, which maximizes at endpoints where speed is zero. Choice D incorrectly claims total energy is smaller at equilibrium, but energy conservation means it's the same everywhere in the oscillation. Remember, for SHM energy problems, use the fact that K + U = constant, and evaluate at key positions like equilibrium and amplitudes.

Question 14

1. Spring potential energy, because the spring force is nonzero there.
2. Kinetic energy. (correct answer)
3. They are equal because equilibrium means equal energy sharing.
4. Neither; total energy is changing too quickly to compare.

Explanation: This question tests identification of dominant energy at equilibrium in SHM. In a frictionless cart-spring system, total mechanical energy is conserved, fluctuating between kinetic and potential forms. At equilibrium, maximum speed yields maximum kinetic energy, while zero displacement means minimum (zero) potential energy. This kinetic energy transforms into potential as the cart approaches turning points, where kinetic is zero. Distractor A incorrectly favors potential due to nonzero spring force, but at equilibrium, force and potential are both zero. For similar questions, apply the principle that kinetic energy peaks where velocity does, offering a reliable way to compare energies.

Question 15

1. $K$ is maximum and $U$ is minimum.
2. $K=U$ because the mass is about to move downward.
3. $U$ is maximum and $K=0$. (correct answer)
4. Total mechanical energy decreases at the turning point due to reversal of motion.

Explanation: This question examines energy in a vertical mass-spring system undergoing simple harmonic motion. In SHM, total mechanical energy, including gravitational and spring potential, remains constant without dissipative forces. At the top turning point, the mass is at rest, so kinetic energy is zero, and the combined potential energy is at its maximum. As the mass descends toward equilibrium, potential energy converts to kinetic energy, which reaches its maximum at the equilibrium point where potential is minimal. Distractor A wrongly suggests kinetic is maximum at the top due to maximum displacement, but actually velocity is zero there, so kinetic is zero. A transferable strategy is to identify turning points as locations of zero kinetic energy and maximum potential in oscillatory systems.

Question 16

1. Spring potential energy is $8\,\text{J}$. (correct answer)
2. Kinetic energy is $8\,\text{J}$.
3. Total mechanical energy is less than $8\,\text{J}$ because the speed is zero.
4. Total mechanical energy is greater than $8\,\text{J}$ because gravity adds energy each cycle.

Explanation: This question examines energy conservation in a vertical spring-mass system undergoing simple harmonic motion. In undamped SHM, total mechanical energy remains constant as kinetic and potential energies continuously exchange. At equilibrium, all 8 J is kinetic energy (with spring potential measured from equilibrium as zero). At the turning point where the mass is momentarily at rest, all kinetic energy converts to spring potential energy, so the spring stores all 8 J. Choice C incorrectly assumes total energy decreases when speed is zero, but energy is conserved in SHM. To solve SHM energy problems, identify total energy at any convenient point and recognize it remains constant throughout the motion.

Question 17

1. Kinetic energy begins decreasing while potential energy begins increasing. (correct answer)
2. Kinetic energy remains maximum until the cart reaches the turning point.
3. Total mechanical energy increases because the cart is still speeding up.
4. Potential energy is zero everywhere because equilibrium is at $x=0$.

Explanation: This question assesses energy changes immediately after equilibrium in simple harmonic motion for a cart-spring system. In frictionless SHM, total energy is conserved, but K and U exchange as the cart oscillates. After passing equilibrium moving away, K decreases as speed slows, while U increases as the spring stretches or compresses. This continues until the turning point where K=0 and U maximum. Distractor B wrongly claims K stays maximum until the turning point, ignoring the continuous energy transfer. A transferable approach is to consider the direction of motion and position to predict whether K is increasing or decreasing in SHM.

Question 18

1. The kinetic energy is $5\,\text{J}$ because acceleration is zero there.
2. The spring potential energy is $5\,\text{J}$. (correct answer)
3. The total mechanical energy is $0\,\text{J}$ because the speed is zero.
4. The total mechanical energy decreases at the turning point due to the spring force.

Explanation: This question examines energy conservation in simple harmonic oscillators at different positions. In frictionless SHM, total mechanical energy remains constant as energy continuously transforms between kinetic and potential forms. At equilibrium, the system has 5 J of kinetic energy and zero spring potential energy, establishing total energy as 5 J. At the turning point where the mass is at rest, all kinetic energy converts to spring potential energy, so the spring stores all 5 J. Choice C incorrectly claims total energy is zero when speed is zero, confusing kinetic energy with total energy. To solve SHM energy problems, recognize that total energy equals the sum of kinetic and potential energies at any instant.

Question 19

1. $K$ decreases while $U$ increases. (correct answer)
2. $K$ increases while $U$ decreases because the spring pulls harder.
3. Both $K$ and $U$ increase because the mass is moving and the spring is stretching.
4. Total mechanical energy decreases as the mass approaches the turning point.

Explanation: This question evaluates energy changes during motion in simple harmonic oscillators. In SHM with negligible friction, total mechanical energy remains constant as kinetic energy converts to potential and back. From equilibrium to a turning point, the increasing displacement raises potential energy, while decreasing speed lowers kinetic energy. At the turning point, all energy is potential, with kinetic at zero. Choice C is a distractor, wrongly suggesting both increase, but they cannot both rise since total energy is fixed. A transferable approach is to analyze energy trends based on position and velocity changes along the path of motion.

Question 20

1. $0\,\text{J}$, because speed is zero so no energy remains.
2. $10\,\text{J}$, because all mechanical energy is spring potential energy. (correct answer)
3. Greater than $10\,\text{J}$, because potential energy is maximum at maximum displacement.
4. Less than $10\,\text{J}$, because energy is lost when the cart stops.

Explanation: This question examines energy transformation in simple harmonic oscillators. In frictionless SHM, mechanical energy is conserved, meaning the sum of kinetic and potential energy remains constant. At equilibrium, the system has K = 10 J and U = 0 J, establishing a total mechanical energy of 10 J. At the turning point where speed equals zero, all energy exists as spring potential energy, so U = 10 J and K = 0 J. Choice D incorrectly suggests energy is lost when the cart stops, but stopping simply means kinetic energy has fully converted to potential energy. Remember that in SHM, energy continuously exchanges between kinetic and potential forms while the total stays constant.