This review is for the algebra-based Physics 1 course, Unit 7 (Oscillations). It connects the topics. Do the unit diagnostic first if you have not yet done it.
Big ideas of the unit
- SHM is defined by its force. Only a restoring force proportional to the displacement from equilibrium, F_net = −kx, gives SHM (Topic 7.1).
- Measure displacement from equilibrium, where the net force is zero. For a hanging spring that is mg/k below the relaxed length, and the net force about it is still −kx (Topic 7.1).
- A pendulum is SHM only at small angles, where the restoring torque mgℓ sin θ is close to mgℓθ, with θ in radians (Topic 7.1).
- The period depends on inertia against stiffness: T = 2π√(m/k) or T = 2π√(ℓ/g). Square roots mean ×4 in m or ℓ gives ×2 in T (Topic 7.2).
- Plot T² to get a straight line. T² against m has slope 4π²/k; T² against ℓ has slope 4π²/g (Topic 7.2).
- Choose cos or sin from how the clock starts. x = A cos(2πft) for release from x = +A; x = A sin(2πft) for passing equilibrium in +x (Topic 7.3).
- At a turning point v = 0 and |a| is largest; at equilibrium a = 0 and the speed is largest. Acceleration always points opposite to displacement (Topic 7.3).
- Amplitude does not change the period. A bigger amplitude brings a proportionally bigger restoring force (Topic 7.3).
- Energy explains the speeds. K + U is constant; K is zero at the turning points and largest at equilibrium. E = ½kA², so amplitude sets the energy (Topic 7.4).
Key relationships and methods
| Idea | Relationship or method | When it applies |
|---|---|---|
| SHM condition | F_net = −kx, so a = −(k/m)x | x from equilibrium; any SHM system |
| Pendulum torque | τ = −mgℓ sin θ ≈ −mgℓθ | small angles, θ in radians |
| Period and frequency | T = 1/f | any periodic motion |
| Spring period | T = 2π√(m/k); so k/m = 4π²/T² | ideal spring, horizontal or vertical |
| Pendulum period | T = 2π√(ℓ/g) | small angles; ℓ to the centre of the bob |
| Position | x = A cos(2πft) or x = A sin(2πft) | radian mode; pick from the start state |
| Largest acceleration | a_max = kA/m | at x = ±A |
| Energy | E = K + U = ½kA² (spring); mgh_max (pendulum) | no friction or outside work |
| Speed at x | ½mv² + ½kx² = ½kA²; v_max = A√(k/m) | spring–object system |
Practice questions
These are original Marlbridge practice questions, not past exam questions. The rubric tables are a suggested Marlbridge rubric, not official scoring. Use g = 9.8 m/s² and radian mode. Springs and strings are ideal, pendulums swing through small angles, and friction and air resistance are negligible unless stated.
Question 1 (multiple choice · mixed)
Take +x to the right of equilibrium. A glider on a horizontal spring moves so that x = (0.060 m) cos(2π × 2.0 Hz × t). What is the size of its largest acceleration?
- (A) 0.75 m/s²
- (B) 0.24 m/s²
- (C) 9.5 m/s²
- (D) 4.7 m/s²
Answer and explanation
Answer: (C). From the equation, A = 0.060 m and f = 2.0 Hz, so T = 0.50 s. Squaring T = 2π√(m/k) gives k/m = 4π²/T² = 4π²f² = 158 s⁻². The largest acceleration is at x = ±A: a_max = (k/m)A = 158 × 0.060 = 9.5 m/s².
- (A) is 2πfA, the maximum speed in m/s.
- (B) uses f²A, leaving out the 4π².
- (D) uses f instead of f².
Question 2 (multiple choice · mixed)
A pendulum is released from rest at 4° to the vertical. It is then released from rest at 8°. Which statement is correct about the second release, compared with the first?
- (A) The period is the same, and the bob–Earth system has about 4 times the energy.
- (B) The period doubles, and the system has twice the energy.
- (C) The period is the same, and the system has twice the energy.
- (D) The period increases by √2, and the system has about 4 times the energy.
Answer and explanation
Answer: (A). Both angles are small, so the motion is SHM and T = 2π√(ℓ/g) has no amplitude in it. The release height is h = ℓ(1 − cos θ). For small angles this grows with the square of the angle: (1 − cos 8°) ÷ (1 − cos 4°) = 4.0. So mgh, the total energy, is about 4 times larger, just as E = ½kA² predicts for a spring.
- (B) assumes a bigger swing takes longer, and that energy goes with amplitude, not its square.
- (C) gets the period right but not the energy.
- (D) applies a square-root rule to the amplitude, which is not in the period equation.
Question 3 (multiple choice · mixed)
A spring–object system oscillates with a period of 0.80 s. How much time passes between one moment when its kinetic energy is largest and the next?
- (A) 0.80 s
- (B) 0.40 s
- (C) 0.20 s
- (D) 1.6 s
Answer and explanation
Answer: (B). K is largest each time the object passes through equilibrium. That happens twice per cycle, once moving each way, so the gap is T/2 = 0.40 s.
- (A) is the period of the motion. K repeats twice as often, because it depends on v², not on the direction of v.
- (C) is the time from a maximum of K to the next maximum of U.
- (D) doubles the period instead of halving it.
Question 4 (constructed response · mixed)
Take +x to the right of equilibrium. An inertial balance is a 0.50 kg tray on a horizontal spring, on a frictionless track. With the tray empty, its period is 0.40 s. An unknown object is clamped to the tray, which is pulled 0.050 m from equilibrium and released. Ten full oscillations now take 6.0 s.
(a) Explain, using the period equation, why the balance would give the same reading on the Moon. (b) Find the spring constant. Then derive an expression for the object’s mass in terms of the tray mass m_t, the empty period T₀ and the loaded period T, and evaluate it. (c) Find the maximum speed and the maximum acceleration of the loaded tray. (d) A student says: “Pull the tray further out next time. A bigger swing takes longer, so the mass reading will change.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) T = 2π√(m/k) depends only on mass (inertia) and stiffness. Weight, mg, does not appear, so weaker gravity changes nothing.
(b) k = 4π²m_t/T₀² = 39.48 × 0.50 ÷ 0.40² = 120 N/m (123 N/m). Loaded, T = 6.0 ÷ 10 = 0.60 s, and the oscillating mass is tray plus object. Since m ∝ T² for the same spring: m_t + m = m_t(T/T₀)², so m = m_t[(T/T₀)² − 1] = 0.50 × (2.25 − 1) = 0.63 kg (0.625 kg).
(c) v_max = A√(k/m_total) = 2πA/T = 2π × 0.050 ÷ 0.60 = 0.52 m/s. a_max = kA/m_total = 123.4 × 0.050 ÷ 1.125 = 5.5 m/s².
(d) Incorrect. The period does not depend on the amplitude. A larger pull gives a proportionally larger restoring force, so the tray covers the longer path in the same time. The reading stays the same; only the energy (½kA²) and the maximum speed change.
| Point | What earns it |
|---|---|
| 1 | T depends on m and k, not g |
| 1 | k ≈ 120 N/m from the empty period |
| 1 | T = 0.60 s, with tray plus object as the oscillating mass |
| 1 | m = m_t[(T/T₀)² − 1], or equivalent |
| 1 | 0.63 kg |
| 1 | v_max ≈ 0.52 m/s and a_max ≈ 5.5 m/s² |
| 1 | Rejects the claim because the period is independent of amplitude |
Total: 7 points.
Question 5 (constructed response · mixed)
Take +x to the right. A motion sensor records a pendulum bob’s horizontal displacement from its lowest point: x = (0.040 m) cos(2πt ÷ 1.6 s).
(a) Find the length of the pendulum. (b) Find the angular amplitude, and justify modelling the motion as SHM. (c) Find x at t = 0.30 s and at t = 1.1 s. State which way the bob moves at each time. (d) At what time in the first period is the bob’s acceleration largest and pointing to the right? (e) Find the bob’s speed at the lowest point using energy, and check it against v_max = 2πA/T.
Worked solution and suggested Marlbridge rubric
(a) T = 1.6 s. Rearranging T = 2π√(ℓ/g): ℓ = gT²/4π² = 9.8 × 1.6² ÷ 39.48 = 0.64 m (0.635 m).
(b) sin θ = 0.040 ÷ 0.635, so θ = 0.063 rad = 3.6°. At this angle sin θ and θ differ by less than 0.1%, so the restoring torque, mgℓ sin θ, is proportional to θ. That is the condition for SHM.
(c) t = 0.30 s: angle 1.18 rad, x = 0.040 cos 1.18 = +0.015 m. The bob started at +A and has not yet reached x = 0 (at T/4 = 0.40 s), so it moves left. t = 1.1 s: angle 4.32 rad, x = −0.015 m. It is between −A (0.80 s) and x = 0 (1.2 s), so it moves right, speeding up.
(d) a_x = −(g/ℓ)x, so a is largest and positive where x = −A, at T/2 = 0.80 s.
(e) h = ℓ(1 − cos θ) = 0.635 × (1 − cos 3.6°) = 1.26 × 10⁻³ m. ½mv² = mgh gives v = √(2gh) = 0.16 m/s. Check: 2πA/T = 2π × 0.040 ÷ 1.6 = 0.157 m/s. ✓ (For a spring, v_max = A√(k/m) = 2πA/T; here g/ℓ plays the part of k/m.)
| Point | What earns it |
|---|---|
| 1 | ℓ ≈ 0.64 m from the period |
| 1 | θ ≈ 3.6° (0.063 rad) |
| 1 | Torque proportional to θ because sin θ ≈ θ at this angle |
| 1 | Both x values, using radians |
| 1 | Both directions, with a reason from the time in the cycle |
| 1 | t = 0.80 s, where x = −A |
| 1 | v ≈ 0.16 m/s from mgh, agreeing with 2πA/T |
Total: 7 points.
Question 6 (constructed response · mixed)
Take +x to the right of equilibrium. A 0.20 kg glider on a level air track is attached to a spring. Its motion is x = (0.060 m) sin(2πt ÷ 0.50 s).
(a) Describe how the glider is moving at t = 0, and find the spring constant. (b) Find the total energy of the glider–spring system. (c) Find U, K and the glider’s speed at t = 0.10 s, and state its direction of motion. (d) Sketch K and U against time for one period on the same axes, labelling the times of their maxima. (e) A student claims: “Kinetic and potential energy are equal halfway between equilibrium and the turning point, at x = A/2, which happens at t = T/8.” Evaluate both parts of the claim.
Worked solution and suggested Marlbridge rubric
(a) At t = 0, x = 0 and x then increases, so the glider passes through equilibrium moving right at its maximum speed. From T = 2π√(m/k): k = 4π²m/T² = 39.48 × 0.20 ÷ 0.25 = 32 N/m (31.6 N/m).
(b) E = ½kA² = ½ × 31.6 × 0.060² = 0.057 J.
(c) x = 0.060 sin(2π × 0.10 ÷ 0.50) = 0.060 sin(1.257) = 0.0571 m. U = ½ × 31.6 × 0.0571² = 0.051 J; K = 0.0568 − 0.0514 = 0.0054 J; v = √(2 × 0.0054 ÷ 0.20) = 0.23 m/s. It is before T/4 = 0.125 s, so the glider is still moving right.
(d) K is largest (0.057 J) at 0, 0.25 s and 0.50 s, and zero at 0.125 s and 0.375 s. U is the mirror image. Their sum is a flat line at 0.057 J.
(e) Partly correct. At t = T/8 = 0.0625 s, x = A sin(π/4) = 0.71A, and U = ½k(0.71A)² = ½E, so K = U there. But at x = A/2, U = ¼E and K = ¾E, so the energies are not equal at A/2. The glider reaches A/2 at T/12, earlier than T/8, because it moves fastest near equilibrium.
| Point | What earns it |
|---|---|
| 1 | Starts at equilibrium moving right; k ≈ 32 N/m |
| 1 | E ≈ 0.057 J |
| 1 | U and K at 0.10 s, using radians |
| 1 | v ≈ 0.23 m/s, moving right with a reason |
| 1 | Sketch: K and U each peak twice per period, at the correct times, with constant sum |
| 1 | Shows K = U at T/8 (x = 0.71A) |
| 1 | Shows U = ¼E at A/2, so that part is wrong |
Total: 7 points.
Question 7 (constructed response · mixed)
Take +x in the direction the clay moves. A 0.60 kg block rests at equilibrium on a frictionless surface. It is attached to a horizontal spring with k = 80 N/m, fixed to a wall on the far side, so the clay pushes the block toward the wall. A 0.20 kg lump of clay moving at 2.0 m/s hits the block and sticks to it. Time t = 0 is the moment of impact.
(a) Find the speed of the block and clay just after impact. (b) Find the amplitude and the period of the oscillation that follows. (c) Write x as a function of t, and find when the spring is first most compressed. (d) A student says: “The oscillation’s energy equals the clay’s kinetic energy before impact, since no friction acts.” Evaluate this claim with numbers.
Worked solution and suggested Marlbridge rubric
(a) Momentum is conserved in the collision: 0.20 × 2.0 = 0.80v, so v = 0.50 m/s.
(b) The block starts at equilibrium, so all the energy is kinetic: E = ½ × 0.80 × 0.50² = 0.10 J. At the turning point ½kA² = 0.10 J, so A = √(2 × 0.10 ÷ 80) = 0.050 m. T = 2π√(0.80 ÷ 80) = 0.63 s, using the combined mass.
(c) It starts at x = 0 moving in +x, so x = (0.050 m) sin(2πt ÷ 0.63 s). The spring is most compressed at x = +A, a quarter period later: t = T/4 = 0.16 s.
(d) Incorrect. The clay had ½ × 0.20 × 2.0² = 0.40 J, but the oscillation has only 0.10 J, 25% of it. The collision is perfectly inelastic: 0.30 J became thermal energy as the clay deformed. “No friction” applies after the impact, not during it. Check: v_max = A√(k/m) = 0.050 × 10 = 0.50 m/s, matching (a).
| Point | What earns it |
|---|---|
| 1 | 0.50 m/s from momentum, not energy |
| 1 | E = 0.10 J and A = 0.050 m |
| 1 | T ≈ 0.63 s with the combined mass |
| 1 | Sine form with A and T, because motion starts at equilibrium |
| 1 | t ≈ 0.16 s (T/4) |
| 1 | Rejects the claim because kinetic energy is lost in the inelastic collision, with 0.40 J against 0.10 J |
Total: 6 points.
How did you do?
Questions 4–6 are worth 7 points and Question 7 is worth 6 on the suggested Marlbridge rubric. Use the points you lost to choose what to study.
- Question 5(b) (what makes it SHM): Topic 7.1 checklist.
- Questions 1, 4(a)–(b), 5(a) or 7(b) (period, frequency, rearranging): Topic 7.2 checklist.
- Questions 1, 4(d), 5(c)–(d), 6(a) or 7(c) (equations and graphs of motion, amplitude): Topic 7.3 checklist.
- Questions 2, 3, 4(c), 6 or 7(d) (energy): Topic 7.4 checklist.
- Quick check of every topic: retake the unit diagnostic.