3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A cart attached to a horizontal spring oscillates on a level track with negligible friction. Its position, measured from its equilibrium position, is given by x = A cos(2πft). Which of the following is true of the cart at t = 0?
Answer and reasoning
AIt is at equilibrium, x = 0. A student who takes cos 0 to be 0, or treats the cosine and sine forms as interchangeable, picks this. cos 0 = 1, so x = +A at t = 0; it is x = A sin(2πft) that starts the cart at x = 0.
BIt moves at its greatest speed. A student who thinks the cart moves fastest where it is farthest from equilibrium picks this. The cart is at x = +A at t = 0, a turning point, where it is momentarily at rest; it moves fastest at x = 0.
CIt is momentarily at rest.Correct At t = 0, x = A cos 0 = A: the cart is at a turning point, x = +A. At a turning point the cart reverses direction, so its velocity is zero for an instant.
DIts acceleration equals zero. A student who thinks zero velocity means zero acceleration picks this. At x = +A the cart is momentarily at rest, but the stretched spring exerts its largest force there, so the acceleration has its greatest magnitude, directed toward equilibrium.
A simple pendulum swings with a small angular amplitude. Which of the following changes would change its period?
Answer and reasoning
AMaking its bob twice as heavy A student who thinks heavier objects fall, and so swing, faster picks this. The mass does not appear in Tp = 2π√(ℓ/g): a heavier bob has a larger weight but also a larger mass to accelerate.
BReleasing it from 6° instead of 3° A student who thinks a wider swing takes longer picks this. For small angles the period does not depend on the amplitude: the bob travels farther but moves proportionally faster.
CMaking its string twice as longCorrect The period of a simple pendulum at small angles is Tp = 2π√(ℓ/g). Doubling the length multiplies the period by √2.
DGiving it a small push at release A student who thinks a faster start makes each swing quicker picks this. The push only gives the pendulum a larger (still small) amplitude, with higher speeds but a longer path, so the period is unchanged.
The graph shows the position x of a block attached to an ideal spring with spring constant 80 N/m, as a function of time t. The block oscillates on a horizontal surface with negligible friction. What is the mass of the block?
Answer and reasoning
A0.081 kg A student who reads the period as the time between neighboring crossings of x = 0 uses T = 0.20 s and gets (80)(0.20)²/(4π²) = 0.081 kg. Neighboring crossings are half a period apart; the period is 0.40 s.
B13 kg A student who substitutes the frequency for the period uses f = 1/0.40 s = 2.5 Hz in place of T and gets (80)(2.5)²/(4π²) = 13 kg. The equation needs the period, T = 0.40 s.
C5.1 kg A student who drops the square root, treating the period as proportional to m/k, gets m = kT/(2π) = (80)(0.40)/(2π) = 5.1 kg. Squaring both sides of T/(2π) = √(m/k) gives m = kT²/(4π²).
D0.32 kgCorrect From the graph, one cycle (maximum to maximum) takes T = 0.40 s. Rearranging Ts = 2π√(m/k) gives m = kT²/(4π²) = (80 N/m)(0.40 s)²/(4π²) = 0.32 kg.
Working Read the period: maxima at t = 0, 0.40 s and 0.80 s, so T = 0.40 s. Ts = 2π√(m/k) → m = kT²/(4π²) = 80 N/m × (0.40 s)² / 39.48 = 12.8/39.48 kg = 0.32 kg.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
7.3.A.1 Displacement equations for SHM Fix
Displacement equations for SHM
For an object in simple harmonic motion, the displacement from the equilibrium position can be written x = A cos(2πft) or x = A sin(2πft), where A is the amplitude, f the frequency and t the time. The cosine form describes an object that is at x = +A at t = 0 (for example, released from rest there); the sine form describes one that is at x = 0 and moving in the +x direction at t = 0. The quantity 2πft is an angle in radians, and x repeats each time t increases by one period, T = 1/f.
Amplitude, A
The magnitude of the maximum displacement of an oscillating object from its equilibrium position. The object moves between x = +A and x = −A, so the distance from one extreme to the other is 2A. SI unit: m.
Turning points (extremes) of SHM
The positions x = +A and x = −A, where the object reverses direction. There the displacement has its greatest magnitude, the velocity is zero for an instant, and the acceleration (and the restoring force) has its greatest magnitude, directed toward the equilibrium position.
Equilibrium position in SHM
The position x = 0, where the net force on the object is zero. There the displacement and acceleration are zero and the speed has its greatest value.
Acceleration and displacement in SHM
Because the restoring force is proportional to the displacement and opposite to it (max = −kΔx for a spring), the acceleration is always directed toward the equilibrium position and its magnitude is proportional to the distance from it: half the maximum at x = ±A/2, zero at x = 0. Unit of acceleration: m/s².
Speeding up and slowing down in SHM
An object in SHM speeds up while it moves from a turning point toward equilibrium (velocity and acceleration in the same direction) and slows down while it moves from equilibrium toward a turning point (velocity and acceleration in opposite directions). Locating the zeros and extremes of x, vx and ax in a cycle is enough to describe the motion qualitatively in each quarter of the cycle.
Students often think The cosine and sine forms both start the object at the equilibrium position at t = 0 (cos 0 is taken to be 0), so the choice between them does not matter for where the object starts. In fact At x = +A, because cos 0 = 1. It is the sine form, x = A sin(2πft), that places the object at x = 0 at t = 0, because sin 0 = 0.
Students often think The farther an oscillating object is from its equilibrium position, the faster it is moving, so its speed is greatest at the turning points and least at equilibrium. In fact No. At the turning points, x = ±A, the object is momentarily at rest; it moves fastest as it passes through the equilibrium position, x = 0.
7.3.A.2 Independence of period from amplitude Fix
Independence of period from amplitude
The period of a system in SHM does not depend on its amplitude. With a larger amplitude the object travels farther each cycle, but the restoring force, and so the acceleration and the speeds, are larger in the same proportion, so each cycle takes the same time. (For a pendulum this holds for small angular amplitudes.)
Students often think The period of an oscillator increases with its amplitude, because with a larger amplitude the object has farther to travel in each cycle (taking the period to grow in proportion to the amplitude). In fact No. With a larger amplitude the object travels farther each cycle, but the restoring force, the acceleration and the speeds are all larger in the same proportion, so the period is unchanged (for a pendulum, as long as the angle stays small).
Students often think The period of a block–spring oscillator is proportional to the mass (the square root in Ts = 2π√(m/k) is dropped), so multiplying the mass by 4 multiplies the period by 4, and m = kT/(2π). In fact No. The period depends on the square root of the mass, Ts = 2π√(m/k): four times the mass gives twice the period, and solving for the mass gives m = kT²/(4π²).
7.3.A.3 Reading an x–t graph of SHM Fix
Reading an x–t graph of SHM
On a graph of position against time for SHM, the amplitude is the greatest height of the curve above (or depth below) the equilibrium line, and the period is the time for one complete cycle, for example from one maximum to the next or between alternate crossings of x = 0. Successive crossings of x = 0 are only half a period apart.
Slopes of SHM graphs
The velocity at an instant is the slope of the x–t graph, so the object is at rest at the maxima and minima of x and moves fastest where the x–t graph crosses the time axis. The acceleration is the slope of the vx–t graph.
Students often think The period is the time between successive crossings of the equilibrium position (or the time for one swing from one side to the other), so it is read from a graph as the time between two neighboring zero crossings. In fact No. Successive crossings of equilibrium are half a period apart (one crossing in each direction). The period is the time for one complete cycle, such as from one maximum of the x–t graph to the next.
Students often think The period and the frequency are interchangeable in the SHM equations: once one of them is known, either value can be put into an equation such as Ts = 2π√(m/k), so the frequency can be substituted where the equation n… In fact No. The period T is the time per cycle (in s); the frequency f is the number of cycles per unit time (in Hz). They are reciprocals, T = 1/f; their numerical values match only for T = 1 s and f = 1 Hz.
5 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 5
An object undergoes simple harmonic motion about its equilibrium position. During which part of each cycle are the object's velocity and acceleration in the same direction?
Answer and reasoning
AWhile it moves away from equilibrium toward a turning point A student who thinks the acceleration points the same way as the displacement, away from equilibrium, picks this. The restoring force points toward equilibrium, so while the object moves away from equilibrium its acceleration opposes its velocity and it slows down.
BDuring the whole cycle, since acceleration is along the motion A student who thinks the acceleration of a moving object points in its direction of motion picks this. The acceleration points toward equilibrium at every instant, so on the way out to a turning point it is opposite to the velocity.
CDuring no part of it, as acceleration opposes the motion A student who thinks a restoring force always opposes motion, like friction, picks this. The restoring force opposes the displacement, not the velocity; on the way back to equilibrium it points along the velocity and speeds the object up.
DWhile it moves from either turning point back toward equilibriumCorrect The acceleration always points toward equilibrium. While the object moves from a turning point toward equilibrium, its velocity also points toward equilibrium, so the two are in the same direction and the object speeds up.
The graph shows the position x of an object in simple harmonic motion as a function of time t. Points P, Q and R are marked on the graph. Let aP, aQ and aR be the magnitudes of the object's acceleration at P, Q and R. Which of the following ranks them?
Answer and reasoning
AaR > aQ > aP A student who thinks the acceleration is greatest where the object moves fastest picks this. The speed is greatest at R and zero at P, but the acceleration follows the displacement: zero at R and greatest at P.
BaP > aQ > aRCorrect In SHM the acceleration is proportional to the displacement from equilibrium (max = −kΔx). P is at x = +A, Q at x = +A/2 and R at x = 0, so aP is the greatest, aQ is half of aP, and aR is zero.
CaP = aQ = aR A student who thinks the acceleration in SHM is constant picks this. The restoring force, and so the acceleration, is proportional to the displacement, so it differs at the three points.
DaQ > aP = aR A student who thinks an object at rest for an instant has zero acceleration sets aP = 0 at the turning point; with aR = 0 at equilibrium, Q is left as the only point with an acceleration. At P the velocity is zero but changing, and the acceleration is greatest there.
Working |a| ∝ |x| (max = −kΔx). From the graph: P at +A, Q at +A/2, R at 0. So aP = amax, aQ = amax/2, aR = 0: aP > aQ > aR.
A block attached to an ideal spring oscillates on a horizontal surface with negligible friction, with period T. The block is replaced by a block with 4 times the mass, and the new block is set oscillating with 3 times the original amplitude. The new period is how many times T?
Answer and reasoning
A2Correct Ts = 2π√(m/k) does not contain the amplitude, so tripling the amplitude has no effect. Multiplying the mass by 4 multiplies the period by √4 = 2.
B4 A student who drops the square root and takes the period to be proportional to the mass picks this. Ts = 2π√(m/k), so 4 times the mass gives √4 = 2 times the period.
C6 A student who thinks the period grows in proportion to the amplitude multiplies the correct mass factor, 2, by 3. The amplitude does not appear in Ts = 2π√(m/k): a larger amplitude means a longer path but proportionally higher speeds.
D1 A student who carries over the pendulum result that mass does not affect the period picks this. For a block on a spring the mass appears in Ts = 2π√(m/k); only the amplitude change has no effect.
Working Ts = 2π√(m/k); amplitude does not appear. New period = 2π√(4m/k) = √4 × 2π√(m/k) = 2T.
A student sets the same block–spring system oscillating twice, with different amplitudes. The graph shows the block's position x as a function of time t for trial 1 (solid line) and trial 2 (dashed line). Which claim is supported by the graph?
Answer and reasoning
ATrial 1 has the longer period, since its block travels farther from equilibrium. A student who thinks a larger amplitude means a longer period picks this. Trial 1's block does travel farther, but the graph shows it completing each cycle in the same 0.8 s as trial 2: it moves proportionally faster.
BBoth trials have a period of 0.8 s, so the period does not depend on the amplitude.Correct Both curves reach their maxima at the same times, 0, 0.8 s and 1.6 s, so both have a period of 0.8 s, although trial 1 has an amplitude of 0.10 m and trial 2 an amplitude of 0.05 m. Halving the amplitude left the period unchanged.
CTrial 2 has the longer period, since its block moves more slowly than in trial 1. A student who judges the time for a cycle from speed alone picks this. Trial 2's block is slower (its x–t graph is less steep), but it also travels only half as far, and the graph shows the same 0.8 s period for both trials.
DBoth trials have a period of 0.4 s, the time between two successive crossings of x = 0. A student who takes the time between neighboring crossings of equilibrium as the period picks this. Neighboring crossings (at 0.2 s and 0.6 s, for example) are half a cycle apart; a full cycle, maximum to maximum, takes 0.8 s.
Working Both curves have maxima at t = 0, 0.8 s and 1.6 s (and minima at 0.4 s and 1.2 s), so T = 0.8 s for both, while the amplitudes are 0.10 m (trial 1) and 0.05 m (trial 2). Same period, different amplitude: period independent of amplitude. Neighboring zero crossings (0.2, 0.6 s) are T/2 = 0.4 s apart.
An object in simple harmonic motion passes the point x = +A/2 twice during each cycle: once moving away from equilibrium (outward) and once moving back toward it (inward). How do the object's velocities at these two instants compare?
Answer and reasoning
AThey have equal magnitudes and point in opposite directions.Correct The return trip from the turning point retraces the outward trip in reverse: the object slows from x = +A/2 to rest at x = +A and then speeds up again over the same distance with the same acceleration at each position. So it passes x = +A/2 with the same speed both times, but moving in opposite directions.
BThey are equal, since the object is at the same position. A student who treats velocity like speed, as fixed by position, picks this. The speeds are equal, but the object moves away from equilibrium on one pass and toward it on the other, so the velocities are opposite.
CThe inward velocity is greater, since the object is speeding up. A student who confuses speeding up with having the greater speed picks this. Speeding up describes how the speed is changing, not its size; at x = +A/2 the speeds on the two passes are equal.
DThe outward velocity is greater, as it carries its top speed outward. A student who thinks a moving object carries forward the speed it was given picks this. On the way out the restoring force points back toward equilibrium and slows the object from the moment it passes x = 0; it returns to x = +A/2 with the same speed.
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
