This diagnostic is for the algebra-based Physics 1 course, Unit 7 (Oscillations). It shows you which topics to revisit. Allow about 30 minutes.
These are original Marlbridge practice questions, not past exam questions. They have not been calibrated against real exam results, so your total does not predict an exam score. Use a scientific calculator, in radian mode whenever you work out sin(2πft) or cos(2πft), and g = 9.8 m/s². Springs and strings are ideal, surfaces are frictionless, pendulums swing through small angles, and air resistance is negligible unless stated.
Question 1 (multiple choice · 7.1)
Take +x to the right, with x = 0 at equilibrium. A 0.50 kg block on a level, frictionless surface sits between two springs, each fixed to a wall. The left spring has k = 30 N/m and the right spring has k = 20 N/m. Both springs are relaxed when the block is at x = 0. What is the block’s acceleration when it is at x = +2.0 cm?
- (A) −1.2 m/s²
- (B) −0.40 m/s²
- (C) −2.0 m/s²
- (D) +2.0 m/s²
Answer and explanation
Answer: (C). At x = +2.0 cm the left spring pulls left and the right spring pushes left, so F_net = −(30 + 20)x = −(50 N/m)x. Then a = −(50 ÷ 0.50) × 0.020 = −2.0 m/s².
- (A) uses the left spring only.
- (B) subtracts the springs. Both act toward equilibrium.
- (D) has the right size but points away from equilibrium. A restoring force points opposite to the displacement.
If you missed this: read “The condition for SHM” in the Topic 7.1 study guide.
Question 2 (multiple choice · 7.1)
A pendulum has a 0.40 kg bob on a string 0.50 m long. The string makes an angle of 0.10 rad with the vertical. Using the small-angle model, what is the size of the restoring torque about the pivot?
- (A) 2.0 N·m
- (B) 0.39 N·m
- (C) 0.0034 N·m
- (D) 0.20 N·m
Answer and explanation
Answer: (D). The weight has lever arm ℓ sin θ about the pivot, so τ = mgℓ sin θ ≈ mgℓθ = 0.40 × 9.8 × 0.50 × 0.10 = 0.196 N·m. The torque is proportional to θ, so the motion is SHM.
- (A) is mgℓ. It leaves out the angle, which sets the lever arm.
- (B) is mgθ, a force. It leaves out the length of the lever arm.
- (C) treats 0.10 as degrees. The small-angle model needs radians.
If you missed this: read “The pendulum: a restoring torque” in the Topic 7.1 study guide.
Question 3 (multiple choice · 7.2)
A block on one spring oscillates with a period of 0.60 s. It is replaced by a block with four times the mass, hung from two of the same springs side by side, which act like one spring with twice the spring constant. What is the new period?
- (A) 1.2 s
- (B) 0.85 s
- (C) 0.60 s
- (D) 0.42 s
Answer and explanation
Answer: (B). T = 2π√(m/k), so T depends on m/k. The ratio becomes 4m ÷ 2k = 2(m/k), so T is multiplied by √2: 0.60 × 1.41 = 0.85 s.
- (A) uses only the mass change (× √4 = 2), or forgets the square root.
- (C) assumes the two changes cancel. The mass rose more than the stiffness.
- (D) uses only the spring change (÷ √2).
If you missed this: read “Factor-of-change reasoning” in the Topic 7.2 study guide.
Question 4 (multiple choice · 7.2)
A museum wants a pendulum that swings with a frequency of 0.40 Hz. About how long must it be?
- (A) 0.040 m
- (B) 1.6 m
- (C) 0.62 m
- (D) 61 m
Answer and explanation
Answer: (B). First the period: T = 1/f = 2.5 s. Then square T = 2π√(ℓ/g) and rearrange: ℓ = gT² ÷ 4π² = 9.8 × 6.25 ÷ 39.5 = 1.6 m.
- (A) puts the frequency, 0.40, in place of the period.
- (C) forgets to square T.
- (D) leaves out the 4π².
If you missed this: read “Period and frequency” and “The simple pendulum” in the Topic 7.2 study guide.
Question 5 (multiple choice · 7.3)
Take +x to the right of equilibrium. A glider moves so that x = (0.050 m) sin(2πt ÷ 0.80 s). Which statement describes the glider at t = 0.50 s?
- (A) x = −0.035 m, moving left
- (B) x = +0.035 m, moving right
- (C) x = −0.035 m, moving right
- (D) x = +0.0034 m, moving right
Answer and explanation
Answer: (A). The angle is 2π × 0.50 ÷ 0.80 = 3.93 rad, and sin 3.93 = −0.707, so x = −0.035 m. The glider passes x = 0 moving left at T/2 = 0.40 s and reaches −A at 3T/4 = 0.60 s, so at 0.50 s it is moving left.
- (B) drops the sign of the sine.
- (C) reads the direction from the sign of x. Velocity is the slope of the x–t graph.
- (D) comes from a calculator in degree mode.
If you missed this: read “Displacement as a function of time” and Worked example 1 in the Topic 7.3 study guide.
Question 6 (multiple choice · 7.3)
Two identical spring–block systems sit side by side. Block P is released from rest 2.0 cm from equilibrium. At the same instant, block Q is released from rest 6.0 cm from equilibrium. What is true a quarter of a period later?
- (A) Only P has reached equilibrium, because it has less distance to travel.
- (B) Both blocks are at equilibrium, moving at the same speed.
- (C) Both blocks are at equilibrium, and Q is moving three times as fast as P.
- (D) Q has not yet reached equilibrium, because its period is √3 times longer.
Answer and explanation
Answer: (C). The period does not depend on the amplitude, so both blocks reach x = 0 together, after T/4. Q’s force and acceleration are three times as large at every matching point, so it reaches three times the speed.
- (A) and (D) assume a bigger amplitude takes longer. The larger force lets Q cover its longer path in the same time.
- (B) gets the timing right but forgets that Q covers three times the distance in the same time.
If you missed this: read “Amplitude does not change the period” in the Topic 7.3 study guide.
Question 7 (multiple choice · 7.4)
A pendulum bob passes through its lowest point at 0.70 m/s. Taking the bob–Earth system, how high above the lowest point does the bob rise?
- (A) 0.050 m
- (B) 0.036 m
- (C) 0.071 m
- (D) 0.025 m
Answer and explanation
Answer: (D). All kinetic energy at the bottom becomes gravitational potential energy at the top: ½mv² = mgh, so h = v² ÷ 2g = 0.49 ÷ 19.6 = 0.025 m. The mass cancels, which is why it is not given.
- (A) drops the ½ in the kinetic energy.
- (B) uses v instead of v².
- (C) uses v ÷ g, which mixes up the units.
If you missed this: read Worked example 2 (a pendulum) in the Topic 7.4 study guide.
Question 8 (multiple choice · 7.4)
A block oscillates on a spring with amplitude A. The spring is replaced by one with four times the spring constant, and the block is set oscillating with the same amplitude. Which row is correct?
- (A) Total energy × 4; maximum speed × 2; period × ½
- (B) Total energy × 4; maximum speed × 4; period unchanged
- (C) Total energy unchanged; maximum speed × 2; period × ½
- (D) Total energy × 2; maximum speed × 2; period × 2
Answer and explanation
Answer: (A). E = ½kA², so with the same A the energy is × 4. All of it becomes kinetic at equilibrium: ½mv_max² = ½kA², so v_max = A√(k/m), × √4 = × 2. The period T = 2π√(m/k) is × 1/√4 = × ½.
- (B) gives the speed the same factor as the energy. Kinetic energy goes with v², so the speed only doubles.
- (C) assumes the same amplitude means the same energy. The stiffer spring stores more energy at the same stretch.
- (D) takes a square root for the energy, and has the period growing. A stiffer spring gives a shorter period.
If you missed this: read “Trading energy back and forth” and “Quick checks with extreme cases” in the Topic 7.4 study guide.
Question 9 (short answer · 7.2)
On the fictional moon Teris, an astronaut builds a pendulum 0.50 m long and times 10 full swings in 31.4 s.
(a) Find the gravitational field strength on Teris. (b) A spring–block toy has a period of 0.60 s on Earth. Predict its period on Teris, and explain. (c) How long would a pendulum on Teris need to be to have the same period as a 0.50 m pendulum on Earth?
Worked answer
(a) T = 31.4 ÷ 10 = 3.14 s. Square T = 2π√(ℓ/g) and rearrange: g = 4π²ℓ ÷ T² = 39.5 × 0.50 ÷ 3.14² = 2.0 m/s².
(b) 0.60 s, unchanged. T = 2π√(m/k) contains no g. Weaker gravity only moves the equilibrium position of a hanging spring.
(c) On Earth, T = 2π√(0.50 ÷ 9.8) = 1.42 s. For the same period, ℓ ÷ g must be the same: ℓ = 0.50 × (2.0 ÷ 9.8) = 0.10 m.
Check yourself: 1 point each for dividing by 10 for T, g = 2.0 m/s², the unchanged spring period with the reason “no g in T_s”, and 0.10 m from keeping ℓ/g fixed.
If you missed this: read “The object–spring oscillator” and “The simple pendulum” in the Topic 7.2 study guide.
Question 10 (short answer · 7.4)
Take +x to the right of equilibrium. A 0.60 kg cart on a level, frictionless track is attached to a spring with k = 60 N/m. It is pulled to x = +0.10 m and released from rest at t = 0.
(a) Find the total energy of the cart–spring system and the cart’s maximum speed. (b) Find the cart’s speed and acceleration when it is at x = −0.060 m. (c) At which times in the first period is the kinetic energy at its largest? (d) A student says: “The system has the most energy at equilibrium, because that is where the cart moves fastest.” Evaluate this claim.
Worked answer
(a) At release K = 0, so E = ½kA² = ½ × 60 × 0.10² = 0.30 J. At equilibrium all of it is kinetic: v_max = √(2 × 0.30 ÷ 0.60) = 1.0 m/s.
(b) U = ½ × 60 × 0.060² = 0.108 J, so K = 0.30 − 0.108 = 0.192 J and v = √(2 × 0.192 ÷ 0.60) = 0.80 m/s. a_x = −(k/m)x = −100 × (−0.060) = +6.0 m/s², to the right, toward equilibrium.
(c) T = 2π√(0.60 ÷ 60) = 0.63 s. K is largest each time the cart passes x = 0: at T/4 = 0.16 s and 3T/4 = 0.47 s.
(d) Incorrect. With no friction the total energy is 0.30 J everywhere. Only the share between K and U changes.
Check yourself: 1 point each for 0.30 J and 1.0 m/s, 0.80 m/s from E − U, +6.0 m/s² with direction, both times in (c), and rejecting the claim because the total is constant.
If you missed this: read Worked example 1 and “Energy against time” in the Topic 7.4 study guide.
Your next step
Mark each question right or wrong. Count Questions 9 and 10 as right only if you earned every point.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 7.1 Defining simple harmonic motion | 1, 2 | Topic 7.1 study guide |
| 7.2 Frequency and period of SHM | 3, 4, 9 | Topic 7.2 study guide |
| 7.3 Representing and analyzing SHM | 5, 6 | Topic 7.3 study guide |
| 7.4 Energy of simple harmonic oscillators | 7, 8, 10 | Topic 7.4 study guide |
How to use your result
- Missed one in a topic? Re-read the section named in that answer, then try the topic’s practice set.
- Missed two or more in a topic? Restart that study guide, including its worked examples, before mixed practice.
- Guessed? Treat the question as missed.
- Look for patterns. Errors in Questions 3 and 8 usually point to the square-root rule; in 2 and 5, to degree mode; in 1 and 10(b), to signs.
- All correct? Move straight to the mixed unit review.