Unit 7 practice test: Oscillations
12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value; data sets share one table, graph or scene diagram.
Covers: Defining Simple Harmonic Motion; Frequency and Period of SHM; Representing and Analyzing SHM; Energy of Simple Harmonic Oscillators.
Data table
Energy of an oscillating block
A block oscillates on a horizontal spring with negligible friction. Sensors record its kinetic energy and the spring's potential energy at four moments.
| Moment | Kinetic energy K (J) | Spring potential energy U_s (J) |
|---|---|---|
| 1 | 0.00 | 0.36 |
| 2 | 0.27 | 0.09 |
| 3 | 0.36 | 0.00 |
| 4 | ? | 0.16 |
1. At which moment is the block at its equilibrium position?
- Moment 3
- Moment 1
- Moment 2
- Moment 4
Show the answer
At equilibrium the spring is at its relaxed length, so U_s = 0 and all the energy is kinetic. That is moment 3.
- Correct: Moment 3: Right: U_s = 0 means x = 0.
- Moment 1: At moment 1 K = 0: the block is at a turning point.
- Moment 2: At moment 2 the spring still holds 0.09 J, so the block is displaced.
- Moment 4: At moment 4 the spring holds 0.16 J.
2. At moment 2, what is the block's displacement as a fraction of the amplitude A?
- A/2
- A/4
- 3A/4
- A/√2
Show the answer
U_s = ½kx² and the total is ½kA². At moment 2, U_s/E = 0.09/0.36 = 1/4, so x²/A² = 1/4 and x = A/2.
- Correct: A/2: Right: a quarter of the energy in the spring means half the amplitude.
- A/4: This treats U_s as proportional to x; it goes as x².
- 3A/4: This uses the kinetic share (3/4) as if it gave the position.
- A/√2: A/√2 is where the energy splits half and half.
3. The block is stopped and set oscillating again with twice the amplitude. What is the total energy now?
- 1.44 J
- 0.72 J
- 0.36 J
- 0.18 J
Show the answer
E = ½kA². Doubling A multiplies E by 2² = 4: 4 × 0.36 J = 1.44 J.
- Correct: 1.44 J: Right: total energy goes as the amplitude squared.
- 0.72 J: This doubles E; it should be multiplied by 4 because E depends on A².
- 0.36 J: The total energy depends on the amplitude; it is not fixed by the spring.
- 0.18 J: A larger amplitude stores more energy, not less.
Diagram
A pendulum at three points
The bob is released from rest at X and swings through Y to Z and back. Air resistance is negligible.
4. Which point is the equilibrium position of the pendulum, and why?
- Y, where nothing pushes the bob along its path
- X, where the bob is momentarily at rest
- Z, where the bob stops before turning back
- Y, because the bob is moving fastest at that point
Show the answer
At Y the string pulls straight up and Earth pulls straight down, so no force acts along the arc. That is the equilibrium position: the bob would stay there if placed at rest.
- Correct: Y, where nothing pushes the bob along its path: Right: at the equilibrium position the restoring force along the path is zero.
- X, where the bob is momentarily at rest: Being at rest for an instant does not make a point the equilibrium position. At X a restoring force pulls the bob back toward Y.
- Z, where the bob stops before turning back: At Z the restoring force is largest, which is the opposite of equilibrium.
- Y, because the bob is moving fastest at that point: Y is the right point, but speed does not define equilibrium; the zero restoring force does.
5. What is the amplitude of the pendulum's motion, measured as an angle?
- 8°
- 16°
- 4°
- 0°
Show the answer
Amplitude is the largest displacement from equilibrium, measured from Y. The bob reaches 8° on each side.
- Correct: 8°: Right: from Y to X (or to Z) is 8°.
- 16°: This is the angle from one extreme to the other, which is twice the amplitude.
- 4°: This halves the amplitude; the bob reaches the full 8° on each side.
- 0°: The amplitude is not the displacement at equilibrium; it is the largest displacement.
6. Why is this pendulum a good example of simple harmonic motion?
- At small angles the restoring force along the arc is close to proportional to the bob's displacement
- The string pulls the bob toward the pivot with a force that stays the same at each point
- The bob's weight pulls it down, and gravity is a force that is the same everywhere
- The bob moves along a curved path, so it returns to its starting point each cycle
Show the answer
Along the arc, gravity has a part of size mg sin θ pointing back toward Y. For small angles sin θ ≈ θ (in radians), so the restoring force is close to mgθ = (mg/ℓ)s, proportional to the displacement s along the arc.
- Correct: At small angles the restoring force along the arc is close to proportional to the bob's displacement: Right: a restoring force proportional to the displacement is what SHM needs.
- The string pulls the bob toward the pivot with a force that stays the same at each point: The string force is not constant (it is largest at Y) and it points along the string, not along the path, so it is not the restoring force.
- The bob's weight pulls it down, and gravity is a force that is the same everywhere: A constant downward force alone does not give a force proportional to the displacement along the arc.
- The bob moves along a curved path, so it returns to its starting point each cycle: Returning to the start shows periodic motion, not simple harmonic motion.
7. The same pendulum is released from 60° instead. Which statement is best?
- The motion is periodic but noticeably less like SHM, because sin 60° is well below 60° in radians
- The motion is still SHM, because a pendulum oscillates about Y at any angle
- The motion is no longer periodic, because the bob speeds up too much at Y
- The motion is closer to SHM, because a larger angle gives a larger restoring force
Show the answer
60° is 1.05 rad but sin 60° = 0.87, so the restoring force (mg sin θ) is 17% smaller than a force proportional to θ would be. The bob still swings back and forth, but the motion is no longer a close match to SHM.
- Correct: The motion is periodic but noticeably less like SHM, because sin 60° is well below 60° in radians: Right: the small-angle approximation sin θ ≈ θ breaks down at large angles.
- The motion is still SHM, because a pendulum oscillates about Y at any angle: Oscillating about Y makes the motion periodic; SHM also needs the restoring force to be proportional to the displacement.
- The motion is no longer periodic, because the bob speeds up too much at Y: Without friction the bob repeats the same swing, so the motion is still periodic.
- The motion is closer to SHM, because a larger angle gives a larger restoring force: A larger force is not the test; the force must stay proportional to the displacement, and at 60° it falls short.
8. A child on a swing completes 12 full swings (out and back) in 30.0 s. What is the frequency of the swinging? Give the value and its unit.
Type a number and its unit.
Show the answer
f = number of cycles / time = 12 / 30.0 s = 0.40 Hz. The period is T = 1/f = 2.5 s.
- Answer: 0.40 Hz
9. An object in simple harmonic motion is at x = +A/2 and moving toward equilibrium. Which describes its velocity and acceleration?
- Velocity −, acceleration −, so it is speeding up
- Velocity −, acceleration +, so it is slowing down
- Velocity +, acceleration −, so it is slowing down
- Velocity −, acceleration zero, so its speed is constant
Show the answer
Moving from +A/2 toward x = 0 means moving in −x, so v is −. The acceleration points toward equilibrium, opposite to x; x is +, so a is −. Velocity and acceleration in the same direction means the object is speeding up, as it should on its way to equilibrium.
- Correct: Velocity −, acceleration −, so it is speeding up: Right: v and a both point toward x = 0, so the object speeds up.
- Velocity −, acceleration +, so it is slowing down: The acceleration is opposite to the displacement, not to the velocity. With x positive, a is negative.
- Velocity +, acceleration −, so it is slowing down: Moving from +A/2 toward x = 0 is moving in the − direction, so v is negative.
- Velocity −, acceleration zero, so its speed is constant: The acceleration is zero only at x = 0. At x = +A/2 the restoring force is −kA/2, half its largest value.
10. A ball bounces straight up and down on a hard floor, rising to the same height each time. Why is this periodic motion but not simple harmonic motion?
- In the air the force on it is its constant weight, not a force proportional to its distance from a center
- The ball is not attached to a spring, so there is no restoring force acting on it at any point of its motion
- The ball does not return to the same height, so the motion does not repeat
- The ball moves fastest at the top, so its speed pattern is reversed
Show the answer
Between bounces the only force is the ball's weight, mg downward, which does not change with height. SHM needs a restoring force that grows in proportion to the displacement from equilibrium.
- Correct: In the air the force on it is its constant weight, not a force proportional to its distance from a center: Right: a constant force is not a restoring force proportional to displacement.
- The ball is not attached to a spring, so there is no restoring force acting on it at any point of its motion: A pendulum is SHM with no spring; what matters is the form of the restoring force.
- The ball does not return to the same height, so the motion does not repeat: The stem says the ball rises to the same height each time, so the motion does repeat.
- The ball moves fastest at the top, so its speed pattern is reversed: The ball is slowest at the top, not fastest.
11. What length of simple pendulum has a period of 2.0 s? Use g = 9.8 m/s². Give the value and its unit.
Type a number and its unit.
Show the answer
Square T = 2π√(ℓ/g): T² = 4π²ℓ/g, so ℓ = gT²/(4π²) = (9.8 m/s²)(2.0 s)²/(4π²) = 0.99 m.
- Answer: 0.99 m
12. A cart on a spring moves according to x = (0.12 m) cos(2π · 0.50 t). What is the first time after t = 0 at which the cart is at x = +0.060 m? Give the value and its unit.
Type a number and its unit.
Show the answer
Set (0.12 m) cos(πt) = 0.060 m: cos(πt) = 0.5, so πt = π/3 and t = 1/3 s ≈ 0.33 s. The period is 2.0 s, so the cart covers the first half of its distance to equilibrium in one sixth of a period: it starts slowly.
- Answer: 0.33 s
Keep going
Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8.