Senior 5 practice book · Unit 1 of 11

Simple Harmonic Motion

50 questions that complete the Senior 5 quiz for this unit: 20 core and 30 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

A pendulum swinging through a bigger arc takes longer for each swing.For small angles the period does not depend on the amplitude: a bigger arc is covered at a higher speed, so each swing takes the same time.
Where the oscillator moves fastest, the force on it must be largest.In SHM the force is largest at the ends (F = −kx), where the body is momentarily at rest; at the centre the force is zero and the speed is greatest.
Any motion that repeats itself at regular intervals is simple harmonic.SHM needs a restoring acceleration proportional to the displacement (a = −ω²x). Many periodic motions, such as a ball bouncing on a floor, do not obey this rule.
A stiffer spring makes a mass bounce more slowly.A stiffer spring gives a larger restoring force for the same stretch, so the mass oscillates faster: T = 2π√(m/k) gets smaller as k grows.
Doubling the amplitude doubles the energy of the oscillator.The energy is proportional to the square of the amplitude, E = ½kA², so doubling A gives four times the energy.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Definition and conditions of SHM: restoring force and a = −ω²x
Displacement equation x = A cos(ωt + φ): amplitude, phase and phase constant
Velocity and acceleration in SHM: v = ω√(A² − x²), v_max = Aω, a_max = ω²A
Graphs of displacement, velocity and acceleration against time
Energy in SHM: exchange of kinetic and potential energy, E = ½kA²
Simple pendulum: T = 2π√(l/g) and its uses
Mass–spring system: T = 2π√(m/k), horizontal and vertical
Springs in series and in parallel: effective spring constant
Other SHM systems: liquid in a U-tube and a floating body
Experiments: finding g and k from T² graphs, accuracy and errors
Small-angle approximation (sin θ ≈ θ) and derivation of the pendulum period
SHM as the projection of uniform circular motion (reference circle)

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 30 harder ones are marked ★ Challenge.

1True or false
Two identical springs joined end to end form a combination that is stiffer than one of the springs alone.
Show answer
Answer: False
End to end (series), each spring carries the full load and stretches, so the total extension doubles and the constant halves: k/2.
Common mistakeLearners often think 'two springs are always stronger than one'; that is true only when they share the load side by side.
2True or false · ★ Challenge
Averaged over a whole cycle, the kinetic energy of an undamped SHM oscillator equals half of its total energy.
Show answer
Answer: True
KE = E sin²(ωt), and the average of sin² over a cycle is ½, so the average KE (and the average PE) is E/2.
Common mistakeLearners often expect the average KE to equal the total energy or zero; KE and PE share the energy equally on average.
3True or false · ★ Challenge
Two oscillators with the same amplitude but different frequencies have the same maximum speed.
Show answer
Answer: False
v_max = Aω = 2πfA, so the oscillator with the higher frequency has the larger maximum speed.
Common mistakeLearners often think the amplitude alone fixes the top speed; the faster oscillator must cover the same distance in less time.
4True or false
In SHM, the speed is greatest at the point where the acceleration is zero.
Show answer
Answer: True
At x = 0 the acceleration −ω²x is zero and the speed ω√(A² − x²) is at its maximum Aω.
Common mistakeLearners often link large speed with large acceleration; in SHM they are greatest at different places.
5True or false · ★ Challenge
A simple pendulum released from 60° has the same period as when it is released from 5°.
Show answer
Answer: False
At 60° sin θ is much smaller than θ, the restoring force is weaker than SHM predicts and the period is about 7% longer; isochronism only holds for small angles.
Common mistakeLearners often apply 'the period does not depend on amplitude' to any angle; it is only true when sin θ ≈ θ.
6Fill in the blank
In a pendulum experiment, a pin or mark placed at the lowest point of the swing, used to count oscillations accurately, is called a ______ marker.
Show answer
Answer: fiducial
A fiducial marker is a fixed reference point; timing as the bob passes it reduces errors in counting and timing.
Common mistakeLearners sometimes place the mark at the end of the swing; it should be at the centre, where the bob moves fastest past it.
7Multiple choice · ★ Challenge
Two oscillators of the same frequency move as x₁ = A sin ωt and x₂ = A cos ωt. Which statement is correct?
1. Ax₁ leads x₂ by π/2
2. BThey are exactly in phase
3. CThey are π out of phase
4. Dx₂ leads x₁ by π/2
Show answer
Answer: D. x₂ leads x₁ by π/2
cos ωt = sin(ωt + π/2), so x₂ reaches each stage of the cycle a quarter-cycle earlier: x₂ leads by π/2.
Common mistakeChoosing 'x₁ leads' comes from seeing that sine starts first at t = 0 from zero; it is the cosine that is already at its peak at t = 0, a quarter cycle ahead.
8Multiple choice · ★ Challenge
A pendulum bob of mass m hangs on a string and is displaced so the string makes an angle θ with the vertical. Which force pulls it back along its arc towards the lowest point?
1. Amg cos θ
2. Bmg tan θ
3. Cmg sin θ
4. Dmg
Show answer
Answer: C. mg sin θ
The weight mg is split into mg cos θ along the string (balanced by tension) and mg sin θ along the arc; mg sin θ is the restoring force.
Common mistakeChoosing mg cos θ takes the component along the string; that part is balanced by the tension and does not restore the bob.
9True or false
In the reference-circle model of SHM, the phase of the oscillation at any instant equals the angle turned through by the radius of the circle.
Show answer
Answer: True
x = A cos θ where θ = ωt + φ is the angle of the radius, so the phase and the angle are the same quantity.
Common mistakeLearners sometimes think phase is measured in seconds; it is an angle, measured in radians.
10Fill in the blank · ★ Challenge
If the amplitude of an oscillator is reduced from 6 cm to 2 cm, its total energy becomes ______ of its original value.
Show answer
Answer: one ninth (1/9)
E ∝ A²; the amplitude is divided by 3, so the energy is divided by 3² = 9.
Common mistakeWriting one third forgets that energy depends on the SQUARE of the amplitude.
11Multiple choice · ★ Challenge
A marble rolls to and fro in a smooth V-shaped groove. Why is its motion periodic but NOT simple harmonic?
1. AThe force down each slope has the same size wherever the marble is
2. BThe marble has no equilibrium position in the groove
3. CThe marble stops for an instant at each end of its path
4. DThe period of the motion is too short to measure
Show answer
Answer: A. The force down each slope has the same size wherever the marble is
On a straight slope the component of weight along it is constant (mg sin α), so the restoring force does not grow with displacement as SHM requires.
Common mistakeChoosing 'it stops at each end' describes every oscillator, including SHM ones, so it cannot be the reason.
12Fill in the blank
In the equation a = −ω²x, the SI unit of ω² is ______.
Show answer
Answer: s⁻² (rad²/s²)
a is in m/s² and x in m, so ω² = −a/x has unit (m/s²) ÷ m = s⁻².
Common mistakeWriting m/s² for ω² forgets that it has been divided by a displacement in metres.
13Fill in the blank · ★ Challenge
At θ = 10°, θ = 0.1745 rad and sin θ = 0.1736. Using θ instead of sin θ therefore makes an error of about ______ %.
Show answer
Answer: 0.5
Error = (0.1745 − 0.1736) ÷ 0.1736 × 100 = 0.0009 ÷ 0.1736 × 100 ≈ 0.5%.
Common mistakeWriting 0.09% divides by 1 instead of by sin θ; a percentage error must be found relative to the true value.
14Multiple choice
The derivation of T = 2π√(l/g) uses the approximation sin θ ≈ θ. When is this approximation good?
1. AWhen θ is in degrees and less than 45°
2. BFor any angle, as long as the string is light
3. CWhen θ is close to 90°, so the bob moves horizontally
4. DWhen θ is in radians and small, below about 10°
Show answer
Answer: D. When θ is in radians and small, below about 10°
sin θ ≈ θ only works with θ in radians: at 10° (0.175 rad) sin θ = 0.174, an error of about 0.5%, but at larger angles the error grows quickly.
Common mistakeChoosing 'degrees' is a common slip: sin 10° = 0.17, not 10, so the approximation only makes sense in radians.
15Fill in the blank · ★ Challenge
Springs of constant 30 N/m and 60 N/m are joined end to end. The spring constant of the combination is ______ N/m.
Show answer
Answer: 20
1/k = 1/30 + 1/60 = 3/60, so k = 60 ÷ 3 = 20 N/m.
Common mistakeWriting 90 N/m adds the constants as for parallel springs; in series the combination is softer than the softer spring.
16Multiple choice · ★ Challenge
A point moves round a circle of radius 0.10 m at a steady angular speed of 5.0 rad/s. Its projection on a diameter moves in SHM. What is the maximum acceleration of the projection?
1. A0.50 m/s²
2. B250 m/s²
3. C2.5 m/s²
4. D0.020 m/s²
Show answer
Answer: C. 2.5 m/s²
The projection's maximum acceleration equals the centripetal acceleration of the point: ω²r = 5.0² × 0.10 = 2.5 m/s².
Common mistakeChoosing 0.50 m/s² calculates ωr, which is the maximum speed, not the maximum acceleration.
17Fill in the blank
A pendulum has a period of 2.0 s on Earth. On a planet where g is four times as large as on Earth, the same pendulum has a period of ______ s.
Show answer
Answer: 1.0
T ∝ 1/√g; g is 4 times larger, so T is divided by √4 = 2: T = 2.0 ÷ 2 = 1.0 s.
Common mistakeWriting 0.5 s divides by 4 instead of by √4; the period depends on the square root of g.
18Multiple choice · ★ Challenge
An oscillator of period 1.2 s passes through its centre. What is the shortest time it takes to reach half of its amplitude?
1. A0.30 s
2. B0.10 s
3. C0.15 s
4. D0.20 s
Show answer
Answer: B. 0.10 s
x = A sin ωt = A/2 gives ωt = π/6, which is 1/12 of a cycle: t = T/12 = 1.2 ÷ 12 = 0.10 s.
Common mistakeChoosing 0.15 s (T/8) assumes the speed is constant; the oscillator is fastest near the centre, so the first half of the amplitude takes less than half of T/4.
19Multiple choice
A shopkeeper hangs a basket from two identical springs placed side by side, each of spring constant 40 N/m. What is the spring constant of the pair?
1. A20 N/m
2. B40 N/m
3. C1600 N/m
4. D80 N/m
Show answer
Answer: D. 80 N/m
Springs side by side (in parallel) share the load and their constants add: k = 40 + 40 = 80 N/m.
Common mistakeChoosing 20 N/m uses the rule for springs joined end to end (series); side-by-side springs are stiffer, not softer.
20Multiple choice · ★ Challenge
Pendulums of length 1.00 m and 0.64 m are released together, in step. What is the smallest number of complete oscillations of the SHORTER pendulum after which they are in step again?
1. A4
2. B8
3. C5
4. D25
Show answer
Answer: C. 5
T ∝ √l, so T_long/T_short = √(1.00/0.64) = 1.25 = 5/4. In the time the long one makes 4 oscillations the short one makes 5, and they are in step again.
Common mistakeChoosing 4 gives the number for the LONGER pendulum, which swings more slowly and completes fewer oscillations in the same time.
21Multiple choice · ★ Challenge
A coin rests on a platform that vibrates up and down in SHM at 5.0 Hz (g = 9.8 m/s²). What is the largest amplitude for which the coin stays in contact with the platform?
1. A99 mm
2. B0.31 m
3. C0.39 m
4. D9.9 mm
Show answer
Answer: D. 9.9 mm
The coin leaves when the downward acceleration at the top exceeds g, so ω²A = g: A = 9.8 ÷ (2π × 5.0)² = 9.8 ÷ 987 = 9.9 × 10⁻³ m.
Common mistakeChoosing 0.39 m uses f² instead of ω² = (2πf)²; ω is in rad/s and is 2π times bigger than f.
22True or false
A fishing boat on Lake Kivu that bobs up and down on calm water after a wave has passed can be modelled as SHM, because the extra upthrust is proportional to how far the boat is pushed down.
Show answer
Answer: True
For a boat with nearly vertical sides, the extra volume under water is proportional to the extra depth, so the restoring force is proportional to displacement.
Common mistakeSome learners think floating objects cannot do SHM because there is no spring; any restoring force proportional to displacement gives SHM.
23Multiple choice · ★ Challenge
A 0.50 kg mass hangs from two identical springs, each of spring constant 50 N/m, joined end to end in a vertical line. What is the period of its vertical oscillations?
1. A0.89 s
2. B0.44 s
3. C0.63 s
4. D0.14 s
Show answer
Answer: A. 0.89 s
In series 1/k = 1/50 + 1/50, so k = 25 N/m; T = 2π√(m/k) = 2π√(0.50 ÷ 25) = 0.89 s.
Common mistakeChoosing 0.44 s adds the constants (k = 100 N/m) as for parallel springs; springs end to end each stretch, so the pair is softer.
24Multiple choice
A peg is fixed to the edge of a turntable that turns steadily at 2.0 revolutions per second. A lamp throws the shadow of the peg onto a wall, and the shadow moves to and fro in SHM. What is the frequency of the shadow's motion?
1. A4.0 Hz
2. B2.0 Hz
3. C1.0 Hz
4. D12.6 Hz
Show answer
Answer: B. 2.0 Hz
Each revolution of the peg gives one complete oscillation of the shadow, so f = 2.0 Hz (and ω = 2πf = 12.6 rad/s).
Common mistakeChoosing 12.6 Hz confuses the angular speed ω (in rad/s) with the frequency f (in Hz).
25Short answer · ★ Challenge
A spring of spring constant k is cut into two equal halves. (a) What is the spring constant of each half? (b) A mass m has period T on the whole spring. What is its period on one half, and on the two halves used side by side?
Show answer
Model answer: (a) Each half stretches half as much for the same force, so its constant is 2k. (b) On one half T′ = 2π√(m/2k) = T/√2 ≈ 0.71T. The two halves side by side give 2k + 2k = 4k, so the period is T/2.
Common mistakeMany learners think a shorter piece of the same spring is weaker; in fact each coil stretches the same amount, so fewer coils give less total extension and a stiffer spring.
26Multiple choice · ★ Challenge
A 0.80 kg trolley on a smooth bench is tied between two fixed posts by a spring of 30 N/m on its left and a spring of 50 N/m on its right. What is its period of oscillation?
1. A1.30 s
2. B0.79 s
3. C0.63 s
4. D0.10 s
Show answer
Answer: C. 0.63 s
Moving the trolley stretches one spring and compresses the other, and both push it back, so k = 30 + 50 = 80 N/m; T = 2π√(0.80 ÷ 80) = 0.63 s.
Common mistakeChoosing 1.30 s treats the springs as being in series; springs on opposite sides of the trolley both pull it back, so they act in parallel.
27Multiple choice
A mass on a spring is pulled to x = +A and released from rest at t = 0. Which equation describes its displacement?
1. Ax = A sin ωt
2. Bx = A cos ωt
3. Cx = −A cos ωt
4. Dx = A sin(ωt + π)
Show answer
Answer: B. x = A cos ωt
At t = 0 the equation must give x = +A; only A cos ωt does this (A cos 0 = A).
Common mistakeChoosing A sin ωt gives x = 0 at t = 0, which describes a mass starting at the centre, not at the end.
28Multiple choice · ★ Challenge
A 2.0 kg mass oscillates on a spring at 3.0 Hz. What extra mass must be added so that it oscillates at 2.0 Hz on the same spring?
1. A2.5 kg
2. B4.5 kg
3. C1.0 kg
4. D0.89 kg
Show answer
Answer: A. 2.5 kg
f ∝ 1/√m, so m₂ = m₁(f₁/f₂)² = 2.0 × (3.0/2.0)² = 4.5 kg. Extra mass = 4.5 − 2.0 = 2.5 kg.
Common mistakeChoosing 4.5 kg gives the total mass needed, not the extra mass that must be added.
29Short answer
Water in a U-tube is blown down in one arm and released. What provides the restoring force, and why is it proportional to the displacement x of the water surface?
Show answer
Model answer: When one surface goes down by x the other rises by x, so there is a level difference of 2x. The weight of this extra column (2xAρg) is unbalanced and pushes the water back. It is proportional to x, so the motion is SHM.
Common mistakeLearners often take the level difference as x instead of 2x; one side falls by x while the other rises by x.
30Multiple choice · ★ Challenge
A learner measures each pendulum length only to the top of the bob, so every length is 1.0 cm too short. She plots T² against her lengths. What does she find?
1. AThe gradient is unchanged, so g is still correct, but the line cuts the T² axis above the origin
2. BThe line passes through the origin and the value of g is too small
3. CThe gradient is bigger, so the value of g comes out too large
4. DThe line passes through the origin and the value of g is too large
Show answer
Answer: A. The gradient is unchanged, so g is still correct, but the line cuts the T² axis above the origin
T² = (4π²/g)(l + 0.010): the constant 0.010 m only shifts the line, giving a positive intercept on the T² axis; the gradient 4π²/g, and so g, is not affected.
Common mistakeChoosing 'g too small' is true for a single reading worked out with g = 4π²l/T², but the graph method separates the constant error into the intercept.
31True or false
The same mass on the same spring has the same period whether it oscillates horizontally on a smooth table or vertically.
Show answer
Answer: True
T = 2π√(m/k) depends only on m and k; in the vertical case gravity only shifts the equilibrium position, it does not change the restoring force per unit displacement.
Common mistakeLearners often think gravity must make the vertical oscillation faster or slower; it only stretches the spring to a new centre.
32Multiple choice · ★ Challenge
Water in a glass U-tube is made to slosh up and down. The total length of the water column, measured along the tube, is 0.40 m. Using T = 2π√(L/2g) with g = 9.8 m/s², what is the period?
1. A1.27 s
2. B0.63 s
3. C0.14 s
4. D0.90 s
Show answer
Answer: D. 0.90 s
T = 2π√(L/2g) = 2π√(0.40 ÷ 19.6) = 2π × 0.143 = 0.90 s.
Common mistakeChoosing 1.27 s uses T = 2π√(L/g) as for a pendulum; in the U-tube the restoring weight comes from a level difference of 2x, which brings in the factor 2.
33Multiple choice · ★ Challenge
A loaded test tube of mass 0.020 kg floats upright in water (density 1000 kg/m³). Its cross-section area is 2.0 × 10⁻⁴ m² and g = 9.8 m/s². What is the period of its small up-and-down oscillations?
1. A1.99 s
2. B0.10 s
3. C0.63 s
4. D9.9 s
Show answer
Answer: C. 0.63 s
Effective k = ρAg = 1000 × 2.0 × 10⁻⁴ × 9.8 = 1.96 N/m; T = 2π√(m/k) = 2π√(0.020 ÷ 1.96) = 0.63 s.
Common mistakeChoosing 1.99 s leaves g out of k = ρAg; the restoring force is the weight of the extra water displaced, so g must be included.
34Short answer
Explain how SHM is related to uniform circular motion. Say what the radius, the angular speed and the angle turned by the radius represent in the SHM.
Show answer
Model answer: If a point moves round a circle of radius A at constant angular speed ω, its projection (shadow) on a diameter moves as x = A cos(ωt + φ), which is SHM. The radius of the circle is the amplitude, the angular speed of the point is the angular frequency, and the angle turned by the radius at any instant is the phase (ωt + φ).
Common mistakeLearners often say the shadow moves at constant speed like the point; in fact the shadow is fastest at the centre and stops at the ends.
35Short answer · ★ Challenge
A test tube loaded with sand floats upright in water. It is pushed down a small distance x and released. Explain why it moves in SHM and write an expression for its period (mass m, cross-section area A, water density ρ).
Show answer
Model answer: When pushed down by x, the tube displaces an extra volume Ax, so the upthrust increases by ρAxg. This extra force acts upwards, opposite to x, and is proportional to x: F = −(ρAg)x. So a = −(ρAg/m)x, which is SHM with ω² = ρAg/m and T = 2π√(m/ρAg).
Common mistakeLearners often say the restoring force is the weight of the tube; the weight is constant and is balanced at equilibrium, it is only the EXTRA upthrust that restores.
36Short answer · ★ Challenge
Sketch, or describe, how the kinetic energy of an SHM oscillator varies with time over one period, and compare its frequency with that of the displacement.
Show answer
Model answer: KE is never negative: it rises from zero at an end to a maximum at the centre and falls back to zero at the other end. This happens twice in every period, so KE oscillates at twice the frequency of the displacement (a sin² shape between 0 and E).
Common mistakeLearners often draw KE as a sine curve going negative with the same period as x; kinetic energy cannot be negative and peaks twice each cycle.
37Short answer
A child on a swing is pulled back to the highest point on the left and let go. Ignoring friction, describe the energy changes during one complete oscillation.
Show answer
Model answer: At the highest point on the left all the energy is potential (KE = 0). Going down, PE changes to KE until at the lowest point KE is maximum and PE minimum. Rising to the right, KE changes back to PE until she stops at the highest point on the right. The same happens on the way back. The total energy stays constant all the time.
Common mistakeLearners often say energy is 'used up' during each swing; without friction it only changes form, and the total stays the same.
38Multiple choice · ★ Challenge
A class plots T² (in s²) against pendulum length l (in m) and gets a straight line through the origin with gradient 4.0 s²/m. What value of g does this give?
1. A0.10 m/s²
2. B9.9 m/s²
3. C39 m/s²
4. D0.25 m/s²
Show answer
Answer: B. 9.9 m/s²
Gradient = 4π²/g, so g = 4π² ÷ gradient = 39.5 ÷ 4.0 = 9.9 m/s².
Common mistakeChoosing 0.10 m/s² divides the gradient by 4π² instead of dividing 4π² by the gradient; check that the answer is close to 10 m/s².
39Short answer
In a pendulum experiment, explain why a learner (a) times 20 complete oscillations instead of one, and (b) starts and stops the stopwatch as the bob passes the lowest point.
Show answer
Model answer: (a) The reaction-time error (about 0.2 s) is spread over 20 oscillations, so the error in one period becomes about 20 times smaller. (b) The bob moves fastest at the lowest point, so the moment it passes is sharp and easy to judge; at the ends it moves slowly and the turning instant is hard to see.
Common mistakeLearners often start timing at the end of the swing because the bob is still there; that is exactly where the instant is hardest to judge.
40Short answer · ★ Challenge
A 4.0 kg block on a smooth floor is acted on by a net force F = −200x (F in N, x in m). Show that it moves in SHM and find its period.
Show answer
Model answer: a = F/m = −200x ÷ 4.0 = −50x. This has the form a = −ω²x, so the motion is SHM with ω² = 50 s⁻², ω = 7.07 rad/s. T = 2π/ω = 2π ÷ 7.07 = 0.89 s.
Common mistakeLearners often take ω² = 200 and forget to divide by the mass; ω² comes from a = −ω²x, not from F = −kx.
41Short answer
A learner wants to build a mass–spring timer that oscillates with a period of exactly 1.0 s, using a spring of constant 20 N/m. What mass must she hang on it?
Show answer
Model answer: T = 2π√(m/k), so m = kT²/4π² = 20 × 1.0² ÷ 39.5 = 0.51 kg.
Common mistakeLearners often rearrange to m = kT/2π and forget to square T; square both sides of T = 2π√(m/k) first.
42Multiple choice · ★ Challenge
The acceleration–time graph of a vibrating machine part has peaks of 8.0 m/s² and repeats every 0.50 s. What is the amplitude of its displacement?
1. A0.64 m
2. B2.0 m
3. C5.1 cm
4. D2.5 cm
Show answer
Answer: C. 5.1 cm
ω = 2π/T = 12.6 rad/s; A = a_max/ω² = 8.0 ÷ 158 = 0.051 m = 5.1 cm.
Common mistakeChoosing 0.64 m divides by ω instead of ω²; a_max = ω²A, so the angular frequency must be squared.
43Short answer · ★ Challenge
Show that, for small angles, the bob of a simple pendulum of length l moves with SHM, and deduce its period.
Show answer
Model answer: Restoring force F = −mg sin θ. For small θ (in radians), sin θ ≈ θ = x/l, where x is the arc displacement. So F = −(mg/l)x and a = −(g/l)x. This is SHM with ω² = g/l, so T = 2π/ω = 2π√(l/g).
Common mistakeLearners often forget the minus sign or write x = lθ with θ in degrees; the arc length formula only works with θ in radians.
44Multiple choice
On the displacement–time graph of an oscillator, what does the gradient at any point give?
1. AThe velocity at that instant
2. BThe acceleration at that instant
3. CThe amplitude of the motion
4. DThe restoring force on the body
Show answer
Answer: A. The velocity at that instant
Gradient of x–t = Δx/Δt = velocity; it is zero at the peaks and greatest where the curve crosses the time axis.
Common mistakeChoosing 'acceleration' mixes up graphs: acceleration is the gradient of the velocity–time graph, not of the displacement–time graph.
45Multiple choice · ★ Challenge
A graph of potential energy against displacement for a mass–spring oscillator is a parabola. The total energy line meets the parabola at x = ±0.040 m, where the potential energy is 0.080 J. What is the spring constant?
1. A50 N/m
2. B4.0 N/m
3. C2.0 N/m
4. D100 N/m
Show answer
Answer: D. 100 N/m
At the amplitude all the energy is potential: ½kA² = E, so k = 2E/A² = 2 × 0.080 ÷ 0.040² = 100 N/m.
Common mistakeChoosing 50 N/m leaves out the ½ in E = ½kA², so the factor 2 is lost.
46Multiple choice
To find g with a simple pendulum, a learner varies the length l and measures the period T. Which graph should give a straight line through the origin?
1. AT² against l
2. BT against l
3. CT against 1/l
4. DT² against 1/l
Show answer
Answer: A. T² against l
T = 2π√(l/g) gives T² = (4π²/g)l, so T² is proportional to l and the graph of T² against l is a straight line through the origin with gradient 4π²/g.
Common mistakeChoosing T against l forgets that T depends on the square root of l, so that graph is a curve, not a straight line.
47Short answer · ★ Challenge
A float on a fish pond starts at its highest point, 6.0 cm above its rest level, and then bobs as y = 0.06 cos(πt) (y in metres, t in seconds). After how long is it first only 3.0 cm above the rest level?
Show answer
Model answer: 0.03 = 0.06 cos(πt), so cos(πt) = 0.5 and πt = π/3 rad. t = 1/3 s ≈ 0.33 s.
Common mistakeLearners often use degrees (πt = 60) on a calculator set to degrees; the phase here is in radians.
48Short answer · ★ Challenge
For a load on a spring, a graph of T² (y-axis) against the added mass m (x-axis) is a straight line of gradient 2.0 s²/kg. It does not pass through the origin but cuts the m-axis at m = −0.015 kg. Find the spring constant and explain the intercept.
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Model answer: T² = (4π²/k)(m + m₀), so the gradient is 4π²/k and k = 4π² ÷ 2.0 = 19.7 N/m ≈ 20 N/m. The intercept shows that an extra 0.015 kg is also oscillating even when no load is added: this is the effect of the spring's own mass (and the hanger), which also moves.
Common mistakeLearners often say the intercept means the experiment failed; a graph method is chosen precisely because the gradient is still correct when there is an extra constant mass.
49Short answer
A pendulum clock keeps correct time in the cool season. On hot days its metal rod expands slightly. Explain whether the clock gains or loses time, and suggest one way to prevent the error.
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Model answer: Expansion makes the length l larger, so T = 2π√(l/g) increases. Each swing takes longer, the clock counts fewer swings per day and it loses time (runs slow). The rod can be made of a material that hardly expands (such as invar), or the bob can be moved up slightly to keep l constant.
Common mistakeLearners often say a longer pendulum swings 'faster' because it covers a bigger arc; a longer pendulum has a longer period.
50Short answer · ★ Challenge
Suggest how you could find the mass of a stone using a spring, a known 0.20 kg mass and a stopwatch, but no balance. Give the calculation you would do.
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Model answer: Hang the 0.20 kg mass, time 20 oscillations and find T₁; then k = 4π² × 0.20 ÷ T₁². Replace it by the stone, time 20 oscillations and find T₂. Then m = kT₂²/4π², or simply m = 0.20 × (T₂/T₁)².
Common mistakeLearners often write m = 0.20 × T₂/T₁, using the ratio of the periods directly; since T² ∝ m, the periods must be squared: m = 0.20 × (T₂/T₁)².