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 horizontal spring oscillates in SHM on a frictionless surface. Which of the following is the total mechanical energy of the block–spring system at any instant?
Answer and reasoning
AThe block's kinetic energy plus the spring's potential energy at that same instantCorrect Mechanical energy is the sum of the system's kinetic and potential energies at the same instant: Etotal = U + K, where K is the block's and U is the spring's.
BThe block's maximum kinetic energy plus the spring's maximum elastic potential energy A student who adds the two maxima picks this. They occur at different places and each equals the total energy, so their sum is twice the total.
CThe block's kinetic energy at that instant only, as only the block has mass A student who leaves the spring out of the system picks this. The stretched or compressed spring stores elastic potential energy, which is part of the system's mechanical energy.
DThe kinetic energy plus the potential energy plus the work done by the spring A student who counts the spring's work in addition to its potential energy picks this. The potential energy already accounts for the spring force inside the system; adding the work counts it twice.
A block on an ideal spring oscillates in SHM with angular frequency ω on a horizontal frictionless surface. When the block is a distance d from equilibrium, its speed is v. What is the block's maximum speed?
Answer and reasoning
Av + ω²d² A student who treats kinetic energy as proportional to speed writes (1/2)mvmax = (1/2)mv + (1/2)mω²d² and picks this. K = (1/2)mv², so the squares of the speeds add.
B√(v² + ω²d²)Correct Conservation of energy: (1/2)mv² + (1/2)kd² = (1/2)m vmax². With k = mω², this gives vmax² = v² + ω²d².
Cωd A student who takes the present position d as the amplitude, ignoring the block's kinetic energy there, picks this: vmax = Aω with A = d. The block is moving at x = d, so its amplitude is larger than d.
Dv A student who thinks the block moves at the same speed everywhere picks this. The speed increases as the block approaches equilibrium, where U is least and K is greatest.
Working Energy is conserved: (1/2)mv² + (1/2)kd² = (1/2)m vmax². With k = mω²: vmax² = v² + ω²d², vmax = √(v² + ω²d²).
The diagram shows a block on a horizontal frictionless surface attached to an ideal spring. The block oscillates in SHM between the positions marked −A and +A. Which ranks the block's kinetic energy at the positions marked P, Q and R?
Answer and reasoning
AQ = R > P A student who thinks K and U are largest at the same places picks this. U is larger at Q and R than at P, so K, which equals E − U, is smaller there.
BP = Q = R A student who thinks conservation keeps the kinetic energy constant picks this. Only K + U is constant; K is largest where U is smallest, at P.
CP > Q = RCorrect K = Etotal − (1/2)kx². At P, x = 0, so U = 0 and K = E. Q and R are the same distance A/2 from equilibrium, so U = E/4 and K = 3E/4 at both.
DR > P > Q A student who thinks a compressed spring has negative potential energy gives R (compressed side) the most kinetic energy. U = (1/2)k(Δx)² is the same for a compression or a stretch of A/2.
Working K = E − (1/2)kx². P (x = 0): K = E. Q (x = +A/2) and R (x = −A/2): U = E/4, K = 3E/4 each. So P > Q = R.
A block attached to an ideal spring oscillates in SHM on a horizontal frictionless surface with amplitude A. Take x to be positive when the spring is stretched. Where is the elastic potential energy of the system at its maximum?
Answer and reasoning
AAt x = +A only, since the spring has negative energy when compressed A student who thinks a compressed spring stores negative energy picks this. U = (1/2)k(Δx)² is the same for a compression or a stretch of the same size.
BAt x = +A and at x = −A, where the block's kinetic energy is zeroCorrect U = (1/2)kx² is largest where |x| is largest, at both turning points, where the block is momentarily at rest and K = 0.
CAt x = 0, where the kinetic energy of the block is also at its maximum A student who thinks K and U peak together picks this. At x = 0 the spring is relaxed, so U = 0, and all the energy is kinetic.
DAt every position equally, since mechanical energy is conserved A student who thinks conservation keeps each form constant picks this. Only U + K is constant; U changes with position as (1/2)kx².
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 system's kinetic and potential energies, Etotal = U + K. For a block on a horizontal ideal spring, K = (1/2)mv² is the block's kinetic energy and U = (1/2)kx² is the spring's elastic potential energy, with x measured from equilibrium. Unit: joule (J).
Elastic potential energy in SHM
For a spring–object system, Us = (1/2)k(Δx)², where Δx is the stretch or compression of the spring from its relaxed length (for a horizontal spring, the block's displacement x from equilibrium). It is never negative and is proportional to the square of the displacement. Unit: joule (J).
Students often think The total mechanical energy of an oscillator is the maximum kinetic energy plus the maximum potential energy. In fact No. The maximum kinetic energy and the maximum potential energy occur at different positions and are each equal to the total energy. Adding them gives twice the total energy.
Students often think The energy of a block–spring system is the block's kinetic energy, because the block is what moves; when the block is at rest the system has no energy. In fact No. The system includes the spring, which stores elastic potential energy (1/2)kx² whenever it is stretched or compressed. The total energy is U + K, and it is not zero even when the block is momentarily at rest.
7.4.A.2 Conservation of energy in SHM Fix
Conservation of energy in SHM
With no work done on the system by external forces and no dissipation, the total mechanical energy of an oscillating system stays constant: energy is transferred back and forth between kinetic and potential forms while U + K keeps the same value.
Students often think Because mechanical energy is conserved in SHM, each form of energy stays constant: the kinetic energy keeps its value, and so does the potential energy. In fact No. Only the total, U + K, is constant. The kinetic and potential energies each change throughout the motion, rising and falling in opposite senses so that their sum stays the same.
Students often think The kinetic energy of an object is proportional to its speed, so energies can be added or shared in the same proportions as speeds. In fact No. K = (1/2)mv², so the kinetic energy is proportional to the square of the speed: half the maximum speed means one-quarter of the maximum kinetic energy.
7.4.A.3 Kinetic energy at the equilibrium position Fix
Kinetic energy at the equilibrium position
At the equilibrium position the potential energy is at its minimum (zero for a horizontal spring–object system), so the kinetic energy is at its maximum and equals the total energy: (1/2)m vmax² = Etotal.
Students often think The speed of an object in SHM falls in proportion to its distance from equilibrium, so at half the amplitude it moves at half its maximum speed. In fact No. From energy conservation, v = vmax√(1 − x²/A²). At x = A/2 the speed is still about 0.87 vmax; it drops steeply only close to the turning points.
Students often think A compressed spring has negative elastic potential energy, because its displacement Δx is negative, so the system has less energy on the compressed side. In fact No. Us = (1/2)k(Δx)² depends on the square of the stretch or compression, so it is positive for both. A spring compressed by a distance d stores the same energy as one stretched by d.
7.4.A.4 Energy at the turning points Fix
Energy at the turning points
At x = ±A the object is momentarily at rest, so the kinetic energy has its minimum value, zero, and the potential energy has its maximum value, equal to the total energy.
Total energy and amplitude, Etotal = (1/2)kA²
For a spring–object system, the total energy equals the maximum potential energy at the turning points, Etotal = (1/2)kA². The energy is proportional to the square of the amplitude: doubling A multiplies Etotal by 4.
Students often think The kinetic and potential energies of an oscillator rise and fall together, so both are largest at the same positions and the object moves fastest where the spring is most deformed. In fact No. They change in opposite senses: the potential energy is largest at the turning points, where the kinetic energy is zero, and the kinetic energy is largest at equilibrium, where the potential energy is least.
Students often think The block's kinetic energy is zero at the equilibrium position, because equilibrium means being at rest. In fact No. At the equilibrium position the net force is zero, but the block moves through it at its maximum speed. It is momentarily at rest, with zero kinetic energy, only at the turning points.
7 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 7
The graph shows the elastic potential energy U of a block–spring system as a function of the block's position x. The horizontal line E shows the system's total mechanical energy. What is the block's kinetic energy when it is at x = 0.050 m?
Answer and reasoning
A0.20 J A student who takes U to be proportional to x picks this: halfway to 0.10 m, U would be 0.20 J and K = 0.20 J. U is proportional to x², so U = 0.10 J and K = 0.30 J.
B0.10 J A student who reads the height of the U curve at 0.050 m as the kinetic energy picks this. 0.10 J is the potential energy there; K is the gap up to the E line, 0.40 J − 0.10 J = 0.30 J.
C0.40 J A student who thinks conservation keeps the kinetic energy itself constant, equal to the total, picks this. Only K + U is constant; at x = 0.050 m some of the 0.40 J is potential energy.
D0.30 JCorrect The total energy is 0.40 J. U = (1/2)kx² rises to 0.40 J at x = 0.10 m, so at half that position it is (1/2)² × 0.40 J = 0.10 J. K = E − U = 0.30 J.
Working From the graph E = 0.40 J, and U = E at x = ±0.10 m, so A = 0.10 m. U ∝ x²: U(0.050 m) = (0.050/0.10)²(0.40 J) = 0.10 J. K = E − U = 0.40 J − 0.10 J = 0.30 J.
A block on an ideal spring oscillates in SHM with amplitude A. How far from the equilibrium position is the block when its speed is half its maximum speed?
Answer and reasoning
A0.87ACorrect Half the maximum speed means one-quarter of the maximum kinetic energy, so K = E/4 and U = 3E/4. Since U ∝ x², (x/A)² = 3/4 and x = (√3/2)A ≈ 0.87A.
B0.50A A student who thinks the speed falls linearly with distance from equilibrium picks this. From energy conservation, v depends on √(A² − x²), and at x = A/2 the speed is still 0.87 vmax.
C0.71A A student who takes K to be proportional to v sets K = E/2 and U = E/2, giving x = A/√2. Half the speed is one-quarter of the kinetic energy, so U = 3E/4.
D0.75A A student who takes U to be proportional to x correctly finds U = 3E/4 but then sets x = (3/4)A. U ∝ x², so x = √(3/4)A.
Working K ∝ v², so K = (1/2)² Kmax = E/4. Then U = E − K = 3E/4. U/E = (x/A)², so x = A√(3/4) = (√3/2)A ≈ 0.87A.
A block–spring system oscillates in SHM with total mechanical energy E. Which statement correctly describes the system at an instant when the block's kinetic energy has its minimum value?
Answer and reasoning
AThe kinetic energy is zero, and the potential energy is at its minimum too. A student who thinks K and U rise and fall together picks this. When K is least, U must be greatest, since U + K = E.
BThe kinetic energy is zero, and the block is at its equilibrium position. A student who thinks equilibrium means rest picks this. The block moves fastest at equilibrium; it is at rest only at x = ±A.
CThe kinetic energy is small but not zero, since the system still has energy E. A student who identifies the system's energy with the block's motion picks this. At a turning point the block is at rest and all of E is stored in the spring.
DThe kinetic energy is zero, and the potential energy then has the value E.Correct The kinetic energy is least at a turning point, where the block is momentarily at rest, so K = 0. Since U + K = E, the potential energy there is E, its maximum.
A block–spring system oscillating in SHM has a total mechanical energy of 1.1 J. The spring is replaced by one with half the spring constant, and the block is set oscillating with twice the original amplitude. What is the new total mechanical energy?
Answer and reasoning
A1.1 J A student who takes the energy to be proportional to the amplitude picks this: (1/2) × 2 = 1. The energy depends on A², so doubling A multiplies it by 4.
B4.4 J A student who thinks only the amplitude sets the energy picks this: × 4 for doubling A. Halving k also halves the energy.
C2.2 JCorrect Etotal = (1/2)kA². Halving k halves the energy and doubling A multiplies it by 4, so E = 1.1 J × (1/2) × 4 = 2.2 J.
D8.8 J A student who thinks a softer spring stores more energy multiplies by 2 for the new spring and by 4 for the amplitude. At a given amplitude the energy is proportional to k, so halving k halves it.
Working Etotal = (1/2)kA². New: (1/2)(k/2)(2A)² = 2 × (1/2)kA². E = 2 × 1.1 J = 2.2 J.
A 0.30 kg block on a horizontal frictionless surface oscillates on an ideal spring with an amplitude of 0.080 m. As the block passes through the equilibrium position, a 0.10 kg lump of putty is dropped vertically onto it and sticks. What is the amplitude of the oscillation afterward?
Answer and reasoning
A0.080 m A student who thinks kinetic energy is conserved when the putty sticks picks this. The collision is perfectly inelastic: momentum is conserved, but one-quarter of the kinetic energy becomes thermal energy.
B0.060 m A student who uses A = vmax/ω with the original ω picks this: 0.75 × 0.080 m. Adding mass lowers ω = √(k/m), which partly offsets the drop in speed.
C0.092 m A student who thinks the block keeps its speed, since the putty falls vertically, finds A′ = v/ω′ with a smaller ω′. The putty must be brought up to the block's horizontal speed, so the block slows to 0.75v.
D0.069 mCorrect Horizontal momentum is conserved: v′ = (0.30/0.40)v. The kinetic energy at equilibrium, which is the new total energy, is then (0.30/0.40) of the old one. With E = (1/2)kA², A′ = (0.080 m)√0.75 = 0.069 m.
Working Horizontal momentum: (0.30 kg)v = (0.40 kg)v′, v′ = 0.75v. Energy after: (1/2)(0.40 kg)v′² = (0.30/0.40)(1/2)(0.30 kg)v² = 0.75E. E = (1/2)kA², so A′ = A√0.75 = (0.080 m)(0.866) = 0.069 m.
A block of mass m is at rest at the equilibrium position on a horizontal frictionless surface, attached to an ideal spring of spring constant k. The block is given a speed v₀. What is the amplitude of the resulting oscillation?
Answer and reasoning
Av₀√k/(2π·√m) A student who uses the period expression 2π√(m/k) as ω picks this: A = v₀/ω. The angular frequency is √(k/m), so A = v₀/ω = v₀√(m/k).
Bv₀√(m/k)Correct All the energy is kinetic at equilibrium and all is elastic at a turning point: (1/2)mv₀² = (1/2)kA², so A = v₀√(m/k).
Cv₀√(m/(2k)) A student who takes the spring's energy at the turning point as kA², maximum force times stretch, picks this. The force grows from zero, so the stored energy is (1/2)kA².
D2πv₀√(m/k) A student who uses the frequency f in place of ω picks this: A = v₀/f. vmax = Aω with ω = 2πf, so A = v₀/ω.
Working Energy: (1/2)mv₀² = (1/2)kA², so A = v₀√(m/k). (Equivalently A = v₀/ω with ω = √(k/m).)
A block–spring system oscillates in SHM with amplitude A. A student draws the energy bar chart shown for the instant when the block is at x = +A/2. Which statement correctly evaluates the chart?
Answer and reasoning
AIt is wrong: U should be E/4 and K should be 3E/4, as U is proportional to x².Correct U = (1/2)kx² and E = (1/2)kA², so at x = A/2, U = E/4 = 1 unit and K = E − U = 3 units. The chart's equal bars of 2 units are wrong; E = 4 units is right.
BIt is correct: at x = A/2, U and K are each E/2, as the block is halfway out. A student who takes U to be proportional to x picks this. U depends on x², so at x = A/2 it is one-quarter of E, not one-half.
CIt is wrong: E should be under 4 units, as the spring is stretched less than at +A. A student who identifies the total energy with the spring's energy at the present position picks this. E stays the same throughout the motion; less stretch means less U and more K.
DIt is wrong: E should exceed 4 units, as the spring also does work on the block. A student who counts the spring's work on top of its potential energy picks this. The elastic potential energy already accounts for the spring's force, so E = K + U.
Working At x = A/2, U = (1/2)k(A/2)² = (1/4)(1/2)kA² = E/4 and K = 3E/4. The chart shows U = K = E/2, so it is wrong: U should be 1 unit and K 3 units, with E = 4 units.
Compiled from the AP Physics C: Mechanics 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
