4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A block attached to an ideal spring oscillates on a horizontal surface with negligible friction. The graph shows the spring potential energy Us of the block–spring system as a function of the block's position x, measured from equilibrium; the curve is a parabola, and the dashed line shows the system's total energy. What is the kinetic energy of the block when it is at x = 0.050 m?
Answer and reasoning
A0.30 JCorrect The dashed line shows the total energy, 0.40 J, which is all spring energy at the turning points, x = ±0.10 m. Us = (1/2)kx² grows with x², so at x = 0.050 m, half the amplitude, Us = (1/2)² × 0.40 J = 0.10 J and K = 0.40 J − 0.10 J = 0.30 J.
B0.10 J A student who takes the height of the curve at x = 0.050 m, a quarter of the way up to the total-energy line, as the kinetic energy picks this. The curve shows the potential energy, 0.10 J; the kinetic energy is the gap between the curve and the total-energy line.
C0.20 J A student who assumes the potential energy grows in proportion to x takes Us at half the amplitude to be half the total, 0.20 J, leaving K = 0.20 J. The curve is a parabola, Us = (1/2)kx²: at half the amplitude Us is one quarter of the total, 0.10 J.
D0.40 J A student who thinks conservation of energy keeps the kinetic energy itself constant gives the block the full 0.40 J. It is the total Us + K that is constant; at x = 0.050 m, 0.10 J of it is stored in the spring.
Working Etotal = 0.40 J (dashed line; the curve reaches it at the turning points, x = ±0.10 m). Us = (1/2)kx² ∝ x², so at x = 0.050 m = A/2, Us = 0.40 J × (0.050/0.10)² = 0.10 J. K = Etotal − Us = 0.40 J − 0.10 J = 0.30 J.
A block of mass m attached to an ideal spring of spring constant k oscillates with amplitude A on a horizontal surface with negligible friction. Which expression gives the speed of the block when it is at x = A/2, halfway between the equilibrium position and a turning point?
Answer and reasoning
A0.71A√(k/m) A student who thinks the spring potential energy is proportional to the stretch gives the spring half the energy at half the amplitude, so K = (1/2)E and v = A√(k/2m) ≈ 0.71A√(k/m). The potential energy depends on x²: at x = A/2 it is only one-quarter of the total, which leaves three-quarters as kinetic energy.
B0.87A√(k/m)Correct The total energy, (1/2)kA², is the same at every point. At x = A/2 the spring stores (1/2)k(A/2)² = (1/4) of it, so the kinetic energy is (3/4)(1/2)kA². Then (1/2)mv² = (3/8)kA², v² = (3/4)(k/m)A² and v = (√3/2)A√(k/m) ≈ 0.87A√(k/m).
C0.75A√(k/m) A student who finds correctly that K is three-quarters of its maximum, but then takes the speed to be three-quarters of its maximum, A√(k/m), picks this. Kinetic energy depends on v², so K = (3/4)Kmax gives v = √(3/4)·vmax ≈ 0.87A√(k/m).
D0.61A√(k/m) A student who writes the kinetic energy as mv² while keeping (1/2)kx² for the spring gets mv² = (3/8)kA² and v = √(3/8)·A√(k/m) ≈ 0.61A√(k/m). With K = (1/2)mv² the factors of 1/2 cancel, giving v = (√3/2)A√(k/m).
Working The total energy is constant. At a turning point it is all spring potential energy: Etotal = (1/2)kA². At x = A/2: (1/2)kA² = (1/2)k(A/2)² + (1/2)mv² → mv² = kA² − kA²/4 = (3/4)kA² → v = √(3/4)·A√(k/m) = (√3/2)A√(k/m) ≈ 0.87A√(k/m).
A 0.50 kg block attached to an ideal spring with spring constant 200 N/m oscillates on a horizontal surface with negligible friction. The graph shows the block's position x as a function of time t. What is the maximum speed of the block?
Answer and reasoning
A2.0 m/sCorrect From the graph the amplitude is 0.10 m. The kinetic energy is greatest where the potential energy is least, at x = 0, where all of the energy (1/2)kA² is kinetic: (1/2)mv² = (1/2)kA², so v = A√(k/m) = (0.10 m)√(200/0.50) = 2.0 m/s.
B4.0 m/s A student who takes the amplitude to be the full distance between the extremes, 0.20 m, gets (0.20)√(200/0.50) = 4.0 m/s. The amplitude is measured from equilibrium to one extreme: 0.10 m.
C6.3 m/s A student who writes the spring energy as (1/2)kA, without squaring A, gets v² = kA/m = (200)(0.10)/0.50 = 40 m²/s², so v = 6.3 m/s. With Us = (1/2)kx², v² = kA²/m = 4.0 m²/s².
D1.4 m/s A student who writes the kinetic energy as mv², dropping the 1/2, gets v = A√(k/(2m)) = (0.10)√200 = 1.4 m/s. With K = (1/2)mv² the halves cancel, giving 2.0 m/s.
Working Amplitude from graph: A = 0.10 m. At x = 0, Us = 0 and K = Etotal: (1/2)mvmax² = (1/2)kA² → vmax = A√(k/m) = 0.10 m × √(200 N/m ÷ 0.50 kg) = 0.10 m × 20 s⁻¹ = 2.0 m/s.
A block attached to a horizontal ideal spring oscillates on a surface with negligible friction. At an instant when the block's kinetic energy has its minimum value, which of the following is true of the spring potential energy Us of the block–spring system?
Answer and reasoning
AIt is zero, because the block is momentarily at rest. A student who thinks a system whose object is at rest has no energy picks this. The block is at rest for an instant, but the spring is stretched or compressed by the amplitude, so the system stores its greatest potential energy then.
BIt has its maximum value, equal to the system's total energy.Correct The total energy Us + K is constant, so when K is at its minimum (zero, at a turning point) Us must be at its maximum. At that instant all of the system's energy is stored in the spring: Us = Etotal = (1/2)kA².
CIt has its minimum value, as the system then has the least energy. A student who thinks the system's energy is greatest when the block moves fastest and least when it is at rest picks this. The total energy does not change; the instant when K is least is exactly when Us is greatest.
DIt has the same value as at every other instant of the motion. A student who thinks conservation of energy keeps each form of energy constant picks this. Only the total is constant; Us changes from zero at equilibrium to its maximum at the turning points.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.4.A.1 Total mechanical energy of an oscillating system, EtotalFix
Total mechanical energy of an oscillating system, Etotal
The sum of the kinetic energy and the potential energy of the system, Etotal = U + K. For a block–spring system on a horizontal surface U is the spring potential energy Us; for a pendulum (bob–Earth system) it is the gravitational potential energy Ug; for a block on a vertical spring (block–spring–Earth system) it includes both. SI unit: J.
Kinetic energy, K
The energy an object has because of its motion, K = (1/2)mv². It depends on the square of the speed, so doubling the speed multiplies K by 4. SI unit: J.
Spring potential energy, Us
The energy stored in a stretched or compressed ideal spring, Us = (1/2)k(Δx)², where Δx is the stretch or compression from the spring's natural length. It is the same for a stretch and a compression of the same size and grows with the square of Δx. SI unit: J.
Students often think The energy asked about can be read directly as the height of whatever curve is plotted, so the height of a U–x curve at a position is taken to be the kinetic energy there. In fact No. The curve gives the potential energy U at each position. The kinetic energy there is the vertical gap between the total-energy line and the curve, K = Etotal − U.
Students often think Energy means motion: a system has the most energy when its objects move fastest and the least when they are at rest, so the total energy of an oscillator rises and falls with its kinetic energy, and the kinetic energy a… In fact No. The total energy of the system is the same at every instant. When the object moves fastest its kinetic energy is greatest, but the potential energy is then at its least.
7.4.A.2 Conservation of energy in SHM Fix
Conservation of energy in SHM
When no energy is transferred into or out of the oscillating system (no friction, air resistance or external work), its total energy U + K is constant. Energy is transferred back and forth between kinetic and potential forms, but the sum does not change.
Students often think Conservation of energy means each form of energy stays the same, so an oscillating system's kinetic energy (or potential energy) keeps one value throughout, and a change in either one shows that energy is not conserved. In fact No. Only the total, U + K, is constant. Energy is continually transferred between the two forms: K is zero at the turning points and greatest at equilibrium, while U does the opposite.
Students often think 'Energy is conserved' means oscillating systems all have the same total energy, so separate systems with different springs and amplitudes are equal in energy. In fact No. Conservation of energy says the total energy of one system stays constant over time when no energy is transferred in or out. Different systems can have very different energies, such as (1/2)kA² with different k and A.
7.4.A.3 Exchange between kinetic and potential energy Fix
Exchange between kinetic and potential energy
Because U + K is constant, the kinetic energy is greatest where the potential energy is least, and the potential energy is greatest where the kinetic energy is least. For a spring oscillator K is greatest at the equilibrium position and least at the turning points.
Vertical spring oscillator
For a block oscillating on a vertical spring, the block–spring–Earth system has potential energy Us + Ug. The sum is least at the equilibrium position, where the spring force balances the weight, so the block's kinetic energy is greatest there, not where the spring is unstretched and not at the lowest point.
Students often think An oscillating object moves fastest where it is farthest from equilibrium and slowest, or at rest, at the equilibrium position. In fact No. It is momentarily at rest at the turning points, where Us is greatest, and moves fastest at equilibrium, where Us is least.
Students often think Only the spring potential energy matters for a vertical spring oscillator, so the block moves fastest where the spring is least stretched. In fact No. For the block–spring–Earth system the potential energy is Us + Ug. K is greatest where this sum is least: at the equilibrium position, where the spring force balances the weight.
7.4.A.4 Energy at the turning points Fix
Energy at the turning points
At the turning points, x = ±A, the object is momentarily at rest, so its kinetic energy is at its minimum and the potential energy of the system is at its maximum, equal to the total energy.
Minimum kinetic energy in SHM
The kinetic energy of an object in SHM falls to zero at each turning point, twice per cycle; its minimum value is zero. (For a spring oscillator the kinetic energy therefore varies with half the period of the motion.)
Total energy and amplitude
For a spring–object system the total energy equals the maximum spring potential energy, Etotal = (1/2)kA². Changing the amplitude changes the maximum potential energy and so the total energy: doubling A multiplies Etotal by 4, although the period is unchanged.
Maximum speed of a spring oscillator
At the equilibrium position all of the energy is kinetic, so (1/2)mvmax² = (1/2)kA², giving vmax = A√(k/m). The maximum speed is proportional to the amplitude. Unit: m/s.
Students often think The spring potential energy is proportional to the stretch, so at half the amplitude the spring holds half the total energy, and doubling the amplitude doubles the total energy (the square in (1/2)kx² is dropped). In fact No. Us = (1/2)kx² grows with the square of the stretch: at half the amplitude Us is one quarter of its maximum, and doubling the amplitude makes Etotal = (1/2)kA² four times as large.
Students often think When an object is at rest the system has no energy at all (its potential energy is zero too), because energy goes with motion. In fact No. At a turning point the block is at rest, but the spring is stretched or compressed by the amplitude, so the system stores its greatest potential energy then.
6 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 6
A block of mass m is attached to an ideal spring of spring constant k and oscillates on a horizontal surface with negligible friction. At one instant the block is at position x, measured from the equilibrium position, and has speed v. Which expression gives the amplitude A of the oscillation?
Answer and reasoning
Av·√(m/k) A student who counts only the kinetic energy as the system's energy sets (1/2)kA² = (1/2)mv², giving A = v·√(m/k). At position x the spring already stores (1/2)kx², which is part of the total: Etotal = Us + K.
B√(x²+2mv²/k) A student who writes the kinetic energy as mv², leaving out the 1/2, gets (1/2)kA² = (1/2)kx² + mv² and picks this. With K = (1/2)mv² the halves cancel throughout, giving A² = x² + mv²/k.
Cx+v·√(m/k) A student who takes the square root of each term separately, as if √(a² + b²) = a + b, picks this. Energies add, not distances: A = √(x² + mv²/k), which is smaller than x + v√(m/k) whenever both terms are nonzero.
D√(x²+mv²/k)Correct The total energy is the same at every point. At the instant described it is (1/2)kx² + (1/2)mv²; at a turning point it is all potential energy, (1/2)kA². Setting these equal gives A² = x² + mv²/k.
Working Etotal = Us + K = (1/2)kx² + (1/2)mv². At a turning point K = 0, so Etotal = (1/2)kA². (1/2)kA² = (1/2)kx² + (1/2)mv² → A² = x² + mv²/k → A = √(x² + mv²/k).
A cart attached to a horizontal ideal spring oscillates on a track with negligible friction. A student uses motion-sensor measurements to calculate the spring potential energy Us and the cart's kinetic energy K at five instants. The table shows the results. Which claim is supported by the data?
Answer and reasoning
AEnergy is not conserved, since Us and K each change as the cart oscillates. A student who thinks conservation of energy means each form of energy stays constant picks this. Us and K do change, but conservation applies to their sum, which the data show is 0.36 J throughout.
BUs + K equals 0.36 J at every instant, so the total energy is constant.Correct Adding the two columns gives 0.36 J at all five instants (for example 0.19 J + 0.17 J at 0.10 s). Energy moves between the spring and the cart, but the total stays the same within the precision of the data, as conservation of energy requires.
CThe cart is at its equilibrium position at t = 0, the instant when K is zero. A student who thinks the cart moves slowest near equilibrium and fastest far from it picks this. K = 0 at t = 0 means the cart is momentarily at rest, which happens at a turning point; Us is greatest then, so the spring is at its greatest stretch or compression.
DThe total energy is greatest at t = 0.20 s, the instant when K is greatest. A student who takes the kinetic energy to stand for the system's energy picks this. At t = 0.20 s the energy is all kinetic, but the total, Us + K = 0.36 J, is the same as at the other instants.
Working Us + K at each instant: 0.36 + 0.00 = 0.36 J; 0.19 + 0.17 = 0.36 J; 0.00 + 0.36 = 0.36 J; 0.17 + 0.19 = 0.36 J; 0.35 + 0.01 = 0.36 J. Constant total: supports conservation of energy.
A block hangs at rest from a vertical ideal spring. The block is pulled down and released, and it oscillates vertically; the spring stays stretched throughout the motion, and air resistance is negligible. Consider the block–spring–Earth system. At which point in the motion is the block's kinetic energy greatest?
Answer and reasoning
AAt the highest point, where the spring is stretched the least A student who counts only the spring's potential energy picks this, since Us is least where the stretch is least. The highest point is a turning point, where the block is momentarily at rest; Ug is greatest there, and Us + Ug is least at the equilibrium position.
BAt the lowest point, where the gravitational potential energy is least A student who thinks an object moves fastest at the lowest point of its motion, as a falling object or a pendulum bob does, picks this. The lowest point is a turning point: the spring is stretched most and the block is momentarily at rest.
CAt the equilibrium position, where the total potential energy is the leastCorrect The system's total energy, K + Us + Ug, is constant, so K is greatest where Us + Ug is least. That is the equilibrium position, where the upward spring force balances the weight: moving up from there raises Ug more than it lowers Us, and moving down raises Us more than it lowers Ug.
DAt every point, since the total energy of the system is constant A student who thinks conservation of energy keeps the kinetic energy constant picks this. The total is constant, but energy moves between K and the potential energies: K is zero at the turning points and greatest at equilibrium.
A block attached to a horizontal ideal spring is pulled to x = +A and released from rest. It then oscillates on a surface with negligible friction. Which of the following correctly gives the minimum kinetic energy of the block and where it occurs?
Answer and reasoning
AZero at x = +A; at x = −A the block is still moving A student who thinks the block carries some of its motion through the far end of its path picks this. With negligible friction the total energy stays (1/2)kA², so at x = −A the spring stores all of it and the block is momentarily at rest there too.
BZero, at x = 0, where the net force on it is zero A student who thinks an object with no net force exerted on it must be at rest picks this. At x = 0 the net force is zero, but the block moves at its greatest speed there; zero net force means zero acceleration, not zero velocity.
CEqual to its maximum value, since energy is conserved A student who thinks conservation of energy keeps the kinetic energy constant picks this. The kinetic energy changes throughout the motion; only the total Us + K is constant, and K is zero wherever Us is at its maximum.
DZero, at each turning point, x = +A and x = −ACorrect The block is momentarily at rest at each turning point, so its kinetic energy falls to zero twice every cycle: at x = +A, where it was released, and at x = −A, where the spring is compressed by the same amount and again stores all of the energy.
A block attached to an ideal spring oscillates on a horizontal surface with negligible friction. The block is stopped and then set oscillating again, with the same spring, at twice the original amplitude. The maximum speed of the block is multiplied by what factor?
Answer and reasoning
A4.0 A student who treats kinetic energy as proportional to speed multiplies the speed by the same factor of 4 as the energy. K = (1/2)mv², so 4 times the energy gives √4 = 2 times the speed.
B1.0 A student who extends 'amplitude does not change the period' to every property of the motion picks this. The period is unchanged, but the energy (1/2)kA² grows with the amplitude, and so does the maximum speed.
C2.0Correct Changing the amplitude changes the total energy: Etotal = (1/2)kA², so doubling A makes the energy 4 times as large. At equilibrium all of it is kinetic, (1/2)mvmax² = Etotal, so vmax² is 4 times as large and vmax is √4 = 2 times as large.
D1.4 A student who thinks the energy grows in proportion to the amplitude takes it to double, giving a speed factor of √2 = 1.4. The energy depends on A², so it becomes 4 times as large.
Working Etotal = (1/2)kA² → doubling A gives 4Etotal. (1/2)mvmax² = Etotal → vmax ∝ √Etotal ∝ A. Factor = √4 = 2.0.
Three block–spring systems oscillate on horizontal surfaces with negligible friction. System 1 has spring constant k and amplitude A. System 2 has spring constant 2k and amplitude A. System 3 has spring constant k/2 and amplitude 2A. Which of the following ranks the total energies E1, E2 and E3 of the three systems?
Answer and reasoning
AE2 > E1 = E3 A student who takes the energy to be proportional to A rather than A² compares kA: k·A for system 1, 2k·A for system 2 and (k/2)(2A) = kA for system 3, and picks this. With A squared, doubling the amplitude of system 3 multiplies its energy by 4.
BE2 > E1 > E3 A student who thinks the amplitude does not affect the energy, just as it does not affect the period, ranks the systems by spring constant alone: 2k, k, k/2. The energy is (1/2)kA², so the amplitude matters.
CE1 = E2 = E3 A student who applies 'energy is conserved' to compare different systems picks this. Conservation says the energy of each system stays constant over time; it does not make separate systems with different springs and amplitudes equal in energy.
DE2 = E3 > E1Correct Etotal = (1/2)kA² for each system. E1 = (1/2)kA²; E2 = (1/2)(2k)A² = kA²; E3 = (1/2)(k/2)(2A)² = (1/2)(k/2)(4A²) = kA². So E2 and E3 are equal, and each is twice E1.
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
