7.3 Representing and Analyzing SHM

1. AIts velocity is zero, and its acceleration has its greatest magnitude, toward −x. Correct
   At a turning point the object is momentarily at rest. Its displacement has its greatest magnitude, so a = −ω²x has its greatest magnitude, Aω², and points toward equilibrium, the −x direction.
2. BIts velocity is zero, and so its acceleration is zero at that instant too.
   A student who thinks an object at rest has no acceleration picks this. The velocity is zero for an instant, but it is changing from +x to −x; the restoring force, and so the acceleration, is largest at x = +A.
3. CIts velocity is zero, and its acceleration is greatest in magnitude, toward +x.
   A student who thinks the acceleration points along the displacement picks this. The size is right, but a = −ω²x points toward equilibrium: toward −x when x = +A.
4. DIts velocity and its acceleration both have their greatest values, toward +x.
   A student who thinks x, v and a peak together picks this. At x = +A the velocity is zero, a quarter cycle out of step with x, and the acceleration points toward −x.

CED 7.3.A.1.i · Read this in Fix

1. A0.0070 s
   A student who takes the slope itself as ω picks this: T = 2π/(900) = 0.0070 s. The slope of an a–x graph is −ω², so ω is its square root, 30 rad/s.
2. B0.033 s
   A student who treats ω as a frequency in hertz picks this: T = 1/ω = 1/(30) = 0.033 s. The period is 2π/ω.
3. C0.21 s Correct
   The line through the origin with negative slope shows a = −ω²x, so the motion is SHM with ω² = (45 m/s²)/(0.050 m) = 900 s⁻². Then ω = 30 rad/s and T = 2π/ω = 0.21 s.
4. D4.8 s
   A student who mixes up period and frequency picks this: ω/(2π) = 4.8 is the frequency in hertz, not the period. The period is its reciprocal, 0.21 s.

Working The graph is a straight line through the origin with negative slope, so a = −ω²x, the equation of SHM. ω² = −slope = (45 m/s²)/(0.050 m) = 900 s⁻², so ω = 30 rad/s. T = 2π/ω = 2π/(30 rad/s) = 0.21 s.

CED 7.3.A.2 · Read this in Fix

1. AP > Q > R
   A student who thinks the acceleration is largest where the object moves fastest picks this. Speed ranks P > Q > R, but the acceleration depends on position: it is zero at P, where the speed is greatest.
2. BP = Q = R
   A student who thinks the acceleration has the same magnitude throughout SHM picks this. a = −ω²x varies with position: zero at equilibrium, greatest at the turning points.
3. CQ > P = R
   A student who thinks zero velocity means zero acceleration sets the acceleration to zero at R, where the object is momentarily at rest, as well as at P. At R the acceleration is in fact the largest of the three.
4. DR > Q > P Correct
   |a| = ω²|x|, so the acceleration is proportional to the distance from equilibrium: ω²A at R, ω²A/2 at Q and zero at P.

Working |a| = ω²|x|: at P, 0; at Q, ω²A/2; at R, ω²A. So R > Q > P.

CED 7.3.A.3.i · Read this in Fix

1. AThe natural frequency shifting until it matches the driving force's frequency
   A student who thinks a driving force changes the system's own frequency picks this. The natural frequency is set by the system (for a spring, k and m); resonance happens when the driving frequency matches it.
2. BA large increase in amplitude that occurs whenever the driving force is strong
   A student who thinks only the size of the driving force matters picks this. A strong force at a frequency far from the natural frequency gives only a small amplitude; the frequency match is what produces resonance.
3. CA large increase in amplitude when a periodic force acts at the natural frequency Correct
   Resonance is the growth of the amplitude to a large value when a periodic (sinusoidal) external force is exerted at the system's natural frequency.
4. DOscillation that keeps going at the natural frequency after the driving force stops
   A student who uses 'resonate' in its everyday sense of continuing to ring picks this. Free oscillation after the force stops is not resonance, which is the large amplitude produced while the force acts.

CED 7.3.A.4 · Read this in Fix

1. A4.4 s
   A student who thinks the block takes twice as long because it travels twice as far picks this. It also moves twice as fast, because the restoring force at corresponding points of the cycle (for example, at the turning points) is doubled.
2. B2.2 s Correct
   The period of an object on a spring, T_s = 2π√(m/k), does not depend on the amplitude. With the same block and spring, the period stays 2.2 s.
3. C1.1 s
   A student who thinks the larger restoring force returns the block sooner picks this. The larger force is matched by the larger distance, so the period is unchanged.
4. D3.1 s
   A student who treats the motion as constant acceleration of the same size as before, t = √(2d/a), picks this: √2 × 2.2 s. In SHM the acceleration is not constant; it scales with the amplitude, so the time does not change.

Working For SHM the period, T_s = 2π√(m/k), does not depend on the amplitude. Doubling the amplitude from 3.0 cm to 6.0 cm leaves T = 2.2 s.

CED 7.3.A.5 · Read this in Fix

1. Av_X = 2v_Y
   A student who compares amplitudes only picks this: X's amplitude is twice Y's. Y completes four cycles while X completes one, so Y moves much faster.
2. Bv_Y = 4v_X
   A student who thinks the amplitude does not affect the maximum speed compares frequencies only: Y's is four times X's. Y's amplitude is half of X's, which halves the factor of 4.
3. Cv_Y = 8v_X
   A student who uses Aω² for the maximum speed picks this: (2.0)(4)² ÷ (4.0)(1)² = 8 in ratio. Aω² is the maximum acceleration; the maximum speed is Aω, giving a ratio of 2.
4. Dv_Y = 2v_X Correct
   v_max = Aω = 2πA/T. X: 2π(4.0 cm)/(2.0 s) = 4π cm/s. Y: 2π(2.0 cm)/(0.50 s) = 8π cm/s. Y has half the amplitude but one-fourth the period, so v_Y = 2v_X.

Working X: A = 4.0 cm, T = 2.0 s, v_X = 2πA/T = 2π(4.0 cm)/(2.0 s) = 4π cm/s. Y: A = 2.0 cm, T = 0.50 s, v_Y = 2π(2.0 cm)/(0.50 s) = 8π cm/s. v_Y = 2v_X.

CED 7.3.A.6 · Read this in Fix

1. A0.052 m
   A student who uses x = A sin(2πft) for every oscillator picks this: (0.060 m)sin(π/3) = 0.052 m. The sine form gives x = 0 at t = 0, but this object starts at x = +0.060 m, so the cosine form applies.
2. B0.020 m
   A student who assumes the object moves at constant speed picks this: 0.20 s is two-thirds of the quarter period (0.30 s), so they place it two-thirds of the way to equilibrium. The object starts slowly from rest, so it is still at 0.030 m.
3. C0.059 m
   A student who puts f in place of 2πf picks this: cos(ft) = cos(0.17 rad) ≈ 0.99, giving 0.059 m. The argument of the cosine is 2πft, which here is π/3 rad.
4. D0.030 m Correct
   The object starts at x = +A from rest, so x = A cos(2πft). With f = 1/(1.2 s), the phase at 0.20 s is 2π(0.20/1.2) = π/3 rad, and x = (0.060 m)cos(π/3) = 0.030 m.

Working Released from rest at maximum displacement, so x = A cos(2πft) with A = 0.060 m and f = 1/T = 1/(1.2 s). At t = 0.20 s: 2πft = 2π(0.20 s)/(1.2 s) = π/3 rad. x = (0.060 m)cos(π/3) = (0.060 m)(0.50) = 0.030 m.

CED 7.3.A.1 · Read this in Fix

1. A3T/8
   A student who assumes constant speed picks this: −A/2 is three-quarters of the way from +A to −A, a trip that takes T/2, so (3/4)(T/2) = 3T/8. The object moves fastest near equilibrium, so it reaches −A/2 sooner.
2. B7T/12
   A student who writes x = A sin(2πt/T) picks this: sin(2πt/T) = −1/2 first at 7π/6, giving 7T/12. The sine form has the object at x = 0 at t = 0, but this object starts at +A.
3. CT/3 Correct
   The object starts at +A from rest, so x = A cos(2πt/T). Then cos(2πt/T) = −1/2 first at 2πt/T = 2π/3, giving t = T/3, between T/4 (at x = 0) and T/2 (at −A).
4. D(2π/3)T
   A student who uses 1/T (the frequency) in place of the angular frequency picks this: cos(t/T) = −1/2 gives t/T = 2π/3. The argument must be 2πt/T, which gives T/3.

Working x = A cos(2πt/T). Setting x = −A/2: cos(2πt/T) = −1/2. The first positive solution is 2πt/T = 2π/3, so t = T/3. Check: T/3 lies between T/4 (x = 0) and T/2 (x = −A), as it must.

CED 7.3.A.1 · Read this in Fix

1. At = 0.20 s
   A student who thinks the acceleration points along the displacement picks this, the instant when x = +A. There the acceleration does have its greatest magnitude, but it points toward −x.
2. Bt = 0.60 s Correct
   a = −ω²x, so the acceleration is in the +x direction where x is negative and is greatest where x is most negative. The graph reaches x = −A at t = 0.6 s, where the object is momentarily at rest.
3. Ct = 0.40 s
   A student who thinks the restoring force acts like drag, opposite to the velocity and largest at the greatest speed, picks this: the object passes x = 0 moving in −x at its greatest speed. At x = 0 the acceleration is zero.
4. Dt = 0.80 s
   A student who thinks the acceleration is largest where the object moves fastest, in its direction of motion, picks this: the object passes x = 0 moving in +x. At x = 0 the acceleration is zero.

Working a = −ω²x: the acceleration is in +x and greatest where x is most negative, x = −A, which the graph shows at t = 0.6 s.

CED 7.3.A.1.ii · Read this in Fix

1. AIt is located at x = −A, and its acceleration is directed toward −x.
   A student who thinks the acceleration points along the displacement reads the negative slope as a < 0 and places the object at −A. The acceleration points toward equilibrium, so a < 0 means x > 0.
2. BIt is at x = 0, the equilibrium position, and its acceleration is zero.
   A student who thinks zero velocity means being at equilibrium picks this. The object is at rest only at a turning point; at x = 0 it would be moving at its greatest speed, 0.30 m/s.
3. CIt is at x = +A, and its acceleration is zero, since v = 0 there.
   A student who thinks zero velocity means zero acceleration picks this. The slope of the v–t graph at t = 1.0 s is steep and negative, so the acceleration there is large and directed toward −x.
4. DIt is at x = +A, and its acceleration is directed toward the −x side. Correct
   At t = 1.0 s the graph crosses v = 0, so the object is at a turning point, and the slope of the v–t graph, the acceleration, is negative there. Since a = −ω²x, a negative acceleration means x is positive: the object is at x = +A.

Working At t = 1.0 s, v = 0, so the object is at a turning point. The slope of the v–t graph there is negative, so a < 0. Since a = −ω²x, x is positive: x = +A. (Consistently, v > 0 from 0 to 1.0 s, so the object has been moving in +x for half a cycle.)

CED 7.3.A.1.ii · Read this in Fix

1. A2π√(m/(2β)) Correct
   F = −dU/dx = −2βx. Newton's second law gives d²x/dt² = −(2β/m)x, so ω = √(2β/m) and T = 2π/ω = 2π√(m/(2β)).
2. B2π√(m/β)
   A student who reads the coefficient β of x² as the spring constant picks this: T = 2π√(m/β). The force is −dU/dx = −2βx, so the effective spring constant is 2β, as U_s = (1/2)kx² also shows.
3. Cπm/β
   A student who takes the coefficient 2β/m in d²x/dt² = −(2β/m)x as ω picks this: T = 2π/ω = 2π/(2β/m) = πm/β. That coefficient is ω², so ω = √(2β/m).
4. D√(m/(2β))
   A student who finds ω = √(2β/m) correctly but treats it as a frequency in hertz picks this: T = 1/ω = √(m/(2β)). ω is in rad/s; one cycle is 2π rad, so T = 2π/ω.

Working F_x = −dU/dx = −2βx. Newton's second law: m d²x/dt² = −2βx, so d²x/dt² = −(2β/m)x. Comparing with d²x/dt² = −ω²x: ω² = 2β/m, ω = √(2β/m). Period: T = 2π/ω = 2π√(m/(2β)). Units: β in J/m² = N/m, so √(m/β) is in s.

CED 7.3.A.2 · Read this in Fix

1. A+Aω sin(ωt + φ)
   A student who takes the derivative of cos θ to be +sin θ picks this. Check: for φ = 0 the object starts at +A and must move toward −x, so v must be negative just after t = 0.
2. B−Aω sin(ωt + φ) Correct
   v = dx/dt. The derivative of cos θ is −sin θ, and the chain rule brings a factor ω from θ = ωt + φ, so v = −Aω sin(ωt + φ).
3. C+Aω cos(ωt + φ)
   A student who thinks the velocity rises and falls in step with the displacement picks this. The velocity is zero where x is greatest; it is a quarter cycle out of step with x.
4. D−Aω cos(ωt + φ)
   A student who thinks the velocity always points toward equilibrium, opposite to x, picks this. It is the acceleration, not the velocity, that is opposite to x; the velocity is a quarter cycle out of step.

Working v = dx/dt = A · (−sin(ωt + φ)) · ω = −Aω sin(ωt + φ) (chain rule: the derivative of ωt + φ is ω).

CED 7.3.A.3 · Read this in Fix

1. A−π/3 Correct
   cos φ = 1/2 gives φ = ±π/3. The velocity at t = 0 is v = −Aω sin φ, which is positive only if sin φ is negative, so φ = −π/3.
2. B+π/3
   A student who takes the calculator value of cos⁻¹(1/2) picks this. With φ = +π/3, v(0) = −Aω sin(π/3) is negative: that object would be moving in −x.
3. C+π/6
   A student who solves for φ with the sine form they habitually use for SHM (sin φ = 1/2 with motion in +x) picks this. That is the phase for x = A sin(ωt + φ); with the cosine form given here, cos φ = 1/2 and v(0) > 0 give −π/3.
4. D−π/4
   A student who thinks the object moves at constant speed puts x = A/2 halfway in phase between x = 0 (φ = −π/2) and x = A (φ = 0). Position is a cosine of the phase, not proportional to it, so cos φ = 1/2 gives −π/3.

Working x(0) = A cos φ = A/2, so cos φ = 1/2 and φ = ±π/3. v(0) = −Aω sin φ must be positive (motion in +x), so sin φ < 0 and φ = −π/3.

CED 7.3.A.3 · Read this in Fix

1. A0.25 m/s²
   A student who uses a = −ωx instead of a = −ω²x picks this: (12.6)(0.020) = 0.25. The acceleration carries ω², as d²x/dt² = −ω²x shows; ωx does not even have units of acceleration.
2. B13 m/s²
   A student who uses a_max = Aω² at every point picks this: (0.080 m)(12.6 rad/s)² = 13 m/s². The acceleration is proportional to the displacement; at P, x is only a quarter of A.
3. C3.2 m/s² Correct
   The period is 0.50 s (peak to peak), so ω = 2π/0.50 s = 12.6 rad/s. At P, x = 0.020 m, so |a| = ω²x = (12.6 rad/s)²(0.020 m) = 3.2 m/s².
4. D0.080 m/s²
   A student who uses the frequency f = 1/T in place of ω picks this: f²x = (2.0 Hz)²(0.020 m) = 0.080 m/s². The relation uses ω = 2πf.

Working From the graph: A = 0.080 m and T = 0.50 s, so ω = 2π/T = 12.6 rad/s. At P, x = 0.020 m. |a| = ω²x = (12.6 rad/s)²(0.020 m) = 3.2 m/s².

CED 7.3.A.3.i · Read this in Fix

1. A0.012 m/s
   A student who uses the frequency 1/T in place of ω picks this: A/T = (0.060 m)/(5.0 s) = 0.012 m/s. v_max = Aω with ω = 2π/T.
2. B0.075 m/s Correct
   The amplitude is 0.060 m and one cycle (peak to peak) takes 5.0 s, so ω = 2π/(5.0 s) = 1.26 rad/s and v_max = Aω = 0.075 m/s.
3. C0.048 m/s
   A student who thinks the object moves at a constant speed picks this: 4A per period, (0.24 m)/(5.0 s) = 0.048 m/s, is the average speed. The speed is greatest at x = 0, where it exceeds this average.
4. D0.095 m/s
   A student who uses Aω², the expression for the maximum acceleration, picks this: (0.060 m)(1.26 rad/s)² = 0.095. The maximum speed is Aω.

Working From the graph: A = 0.060 m, T = 5.0 s (peak to peak). ω = 2π/T = 1.26 rad/s. v_max = Aω = (0.060 m)(1.26 rad/s) = 0.075 m/s.

CED 7.3.A.3.ii · Read this in Fix

1. Ad√(k/m) Correct
   Newton's second law gives ω = √(k/m), and an object released from rest at x = d has amplitude d, so v_max = Aω = d√(k/m).
2. B2πd√(m/k)
   A student who uses the period expression 2π√(m/k) as the angular frequency picks this. ω = √(k/m); the expression 2π√(m/k) is T_s, and d·T_s has units of m·s, not m/s.
3. Cd√(k/m)/(2π)
   A student who uses the frequency f = (1/2π)√(k/m) in place of ω picks this. v_max = Aω, and ω = 2πf.
4. Ddk/m
   A student who uses Aω², the expression for the maximum acceleration, picks this: d(k/m). That is a_max; the maximum speed is Aω = d√(k/m).

Working Newton's second law: m d²x/dt² = −kx, so ω = √(k/m). Released from rest at x = d, so A = d. v_max = Aω = d√(k/m). (Check by energy: (1/2)kd² = (1/2)m v_max².)

CED 7.3.A.3.ii · Read this in Fix

1. A1.1 m/s
   A student who thinks ω is inversely proportional to the mass picks this: 4.4/4 = 1.1 m/s. ω = √(k/m), so four times the mass divides ω by 2, not by 4.
2. B8.8 m/s
   A student who takes ω = √(m/k), the ratio in the period expression, picks this: four times the mass doubles ω. In fact the heavier block oscillates more slowly.
3. C4.4 m/s
   A student who thinks the mass does not affect the motion, as for a pendulum bob, picks this. The spring's force does not depend on the block's mass, so a heavier block accelerates less and ω = √(k/m) is smaller.
4. D2.2 m/s Correct
   v_max = Aω and ω = √(k/m). With four times the mass on the same spring, ω falls by a factor √4 = 2; the amplitude is unchanged, so v_max = 2.2 m/s.

Working v_max = Aω with ω = √(k/m). Same k, four times the mass: ω is halved. Same A, so v_max = (4.4 m/s)/√4 = 2.2 m/s.

CED 7.3.A.3.ii · Read this in Fix

1. A8.0 Hz
   A student who thinks the natural frequency is inversely proportional to the mass picks this: 4 × 2.0 Hz. The frequency depends on 1/√m, so the factor is √4 = 2.
2. B4.0 Hz Correct
   Resonance occurs at the natural frequency, which the graph shows is 2.0 Hz. Since f₀ = (1/2π)√(k/m), one-fourth the mass doubles f₀ to 4.0 Hz, where the new amplitude peak occurs.
3. C1.0 Hz
   A student who uses the ratio m/k, as in the period expression, for the frequency picks this. A smaller mass makes the period shorter and the natural frequency higher, not lower.
4. D2.0 Hz
   A student who thinks the mass does not affect the oscillation picks this. The spring exerts the same force on a lighter block, which accelerates more, so the natural frequency rises.

Working The peak at 2.0 Hz is the natural frequency f₀ = (1/2π)√(k/m). With m/4: f₀ becomes (1/2π)√(4k/m) = 2 × 2.0 Hz = 4.0 Hz. Resonance occurs at the new natural frequency.

CED 7.3.A.4.i · Read this in Fix

1. AIt increases steadily, as a higher driving frequency delivers more energy.
   A student who links a higher frequency with more energy picks this. Above f₀ the force is out of step with the motion, and the amplitude decreases.
2. BIt stays the same, since the maximum driving force does not change.
   A student who thinks only the size of the driving force sets the amplitude picks this. The same force gives a much larger amplitude near f₀ than away from it.
3. CIt increases to a maximum when the driving frequency is near f₀, then decreases. Correct
   Resonance occurs when the driving force is exerted at the natural frequency: the amplitude rises to a peak as the driving frequency approaches f₀ and falls again above it.
4. DIt decreases steadily, as the force pushes for less time in each cycle.
   A student who thinks a longer push gives a larger amplitude picks this. Below f₀ the pushes are not in step with the motion; the amplitude grows as the driving frequency approaches f₀.

CED 7.3.A.4.ii · Read this in Fix

1. A16 Hz Correct
   Once the driving force is removed, only the spring force acts, so the block oscillates at its natural frequency: f = (1/2π)√(k/m) = (1/2π)√(1000/0.10) = 16 Hz.
2. B0.50 Hz
   A student who thinks the block keeps the driving frequency picks this. Without the driver, Newton's second law with only the spring force gives the natural frequency, set by k and m.
3. C100 Hz
   A student who quotes ω = √(k/m) = 100 rad/s as the frequency picks this. The frequency in hertz is ω/(2π) = 16 Hz.
4. D0.063 Hz
   A student who uses the period expression 2π√(m/k) = 0.063 s and quotes it as a frequency picks this. The frequency is the reciprocal of the period: 1/(0.063 s) = 16 Hz.

Working With the driver removed, only the spring force acts, so the block oscillates at its natural frequency: ω = √(k/m) = √(1000/0.10) = 100 rad/s; f = ω/(2π) = 100/(2π) = 16 Hz.

CED 7.3.A.4.iii · Read this in Fix

1. A0.11√(k/m)
   A student who uses the series rule for parallel springs takes k_eq = k/2 and picks this: f = √(k/(2m))/(2π) = 0.11√(k/m). Parallel springs share the stretch and both pull, so k_eq = 2k.
2. B1.41√(k/m)
   A student who gives the angular frequency as the natural frequency picks this: ω = √(2k/m) = 1.41√(k/m). That is ω in rad/s; the frequency in hertz is ω/(2π).
3. C4.44√(m/k)
   A student who gives the period expression as the frequency picks this: 2π√(m/k_eq) = 2π√(m/(2k)) = 4.44√(m/k). That is the period, in seconds; the frequency is its reciprocal.
4. D0.23√(k/m) Correct
   In parallel the spring constants add, k_eq = 2k. Newton's second law gives ω = √(2k/m), and the natural frequency is f = ω/(2π) = (√2/(2π))√(k/m) = 0.23√(k/m).

Working Parallel springs: k_eq = k + k = 2k. Newton's second law: m d²x/dt² = −2kx, so ω = √(2k/m) and f = ω/(2π) = (√2/(2π))√(k/m) = 0.23√(k/m). Checked with sympy. Distractors: series rule, k_eq = k/2 → f = √(k/(2m))/(2π) = 0.11√(k/m); ω given as the frequency → √(2k/m) = 1.41√(k/m); period given as the frequency → 2π√(m/(2k)) = 4.44√(m/k) (units s, not s⁻¹).

CED 7.3.A.4.iii · Read this in Fix

1. ATrial 2 has the longer period, and the maximum speeds are equal.
   A student who thinks a larger amplitude means a longer trip and so a longer period picks this. The curves repeat together every 0.8 s, and trial 2's steeper zero crossings show a greater maximum speed.
2. BTrial 2 has the shorter period and the greater maximum speed.
   A student who thinks the larger restoring force brings the block back sooner picks this. Trial 2 does have the greater maximum speed, but the graph shows both trials reaching x = 0 and their peaks at the same times.
3. CThe periods are equal, and trial 2 has the greater maximum speed. Correct
   Both curves peak at t = 0, 0.8 s and 1.6 s, so the periods are equal. Trial 2's curve has twice the amplitude and is twice as steep where it crosses x = 0, so its maximum speed is greater: v_max = Aω with the same ω.
4. DThe periods are equal, and the two maximum speeds are equal as well.
   A student who extends 'amplitude does not affect the period' to the speed picks this. The slopes at the zero crossings differ: trial 2 covers twice the distance in the same time, so it moves faster.

Working Both curves reach their maxima at the same times (0, 0.8 s, 1.6 s), so the periods are equal, 0.8 s. Trial 2 has twice the amplitude, and its curve is twice as steep where it crosses x = 0, so its maximum speed is twice as large (v_max = Aω, same ω).

CED 7.3.A.5 · Read this in Fix

1. AT = 4v_max/a_max
   A student who treats the quarter cycle from a turning point to equilibrium as constant acceleration a_max finds that it takes v_max/a_max to reach v_max, and so a period of 4v_max/a_max. The acceleration falls to zero as the object reaches equilibrium (a = −ω²x), so that quarter cycle takes longer, (π/2)v_max/a_max.
2. BT = v_max/a_max
   A student who finds ω = a_max/v_max correctly but then sets 1/T equal to ω gets v_max/a_max. The angular frequency is ω = 2π/T, so T = 2π/ω.
3. CT = 2πv_max/a_max Correct
   With x = A cos(ωt + φ), v_max = Aω and a_max = Aω². Dividing the second by the first gives ω = a_max/v_max, so T = 2π/ω = 2πv_max/a_max.
4. DT = a_max/(2πv_max)
   A student who finds ω = a_max/v_max and gives the frequency, ω/2π, as the period picks this. That expression is f, in hertz; the period is its reciprocal, 2πv_max/a_max.

Working v_max = Aω and a_max = Aω². Dividing: a_max/v_max = ω, so T = 2π/ω = 2πv_max/a_max (units: (m/s)/(m/s²) = s). Checked with sympy. Distractors: T = 1/ω → v_max/a_max (s); ω/2π given as the period → a_max/(2πv_max) (s⁻¹); constant acceleration a_max from a turning point to equilibrium, v_max = a_max(T/4) → 4v_max/a_max (s).

CED 7.3.A.3.ii · Read this in Fix