Oscillations and Radiation Problem Solving

This final lesson brings together all the material from the course into exam-style problem solving. We will work through the types of calculations and analysis questions that regularly appear in Edexcel A-Level Physics papers on oscillations and nuclear radiation. The focus is on selecting the right equation, interpreting graphs, and combining concepts.

SHM Calculation Strategy

When faced with an SHM problem, follow this approach:

Identify what you know: amplitude A, frequency f or period T, mass m, spring constant k, displacement x, velocity v, or acceleration a
Calculate ω if not given: ω = 2πf = 2π/T, or ω = √(k/m), or ω = √(g/L)
Choose the right equation based on what you need to find
Check units — especially converting cm to m, minutes to seconds, and ensuring calculator is in radians

flowchart TD
    A["SHM Problem"] --> B["What do you know?"]
    B --> C["Given f or T?\nCalculate ω = 2πf"]
    B --> D["Given k and m?\nCalculate ω = √(k/m)"]
    B --> E["Given L and g?\nCalculate ω = √(g/L)"]
    C --> F["What do you need?"]
    D --> F
    E --> F
    F --> G["Displacement at time t?\nx = A cos(ωt)"]
    F --> H["Speed at displacement x?\nv = ω√(A² − x²)"]
    F --> I["Maximum speed?\nv_max = Aω"]
    F --> J["Maximum acceleration?\na_max = Aω²"]
    F --> K["Energy?\nE = ½mω²A²"]
    F --> L["Period?\nT = 2π/ω"]

SHM Formulae Summary

Equation	What It Describes	When to Use
a = −ω²x	Defining equation of SHM	Relating acceleration to displacement
x = A cos(ωt)	Displacement vs time (starts at +A)	Finding x at a given time
x = A sin(ωt)	Displacement vs time (starts at 0)	Finding x at a given time
v = ±ω√(A² − x²)	Speed at displacement x	Finding speed at a given position
v_max = Aω	Maximum speed (at x = 0)	Finding the fastest speed
a_max = Aω²	Maximum acceleration (at x = ±A)	Finding the largest acceleration
E_total = ½mω²A² = ½kA²	Total energy in SHM	Energy calculations
KE = ½mω²(A² − x²)	Kinetic energy at displacement x	Energy partition problems
PE = ½mω²x²	Potential energy at displacement x	Energy partition problems
T = 2π√(m/k)	Period of mass-spring system	Spring problems
T = 2π√(L/g)	Period of simple pendulum	Pendulum problems

Worked Example 1: Complete SHM Analysis

A 0.25 kg mass on a spring (k = 100 N m⁻¹) is pulled 0.08 m from equilibrium and released.

(a) Find the period.

ω = √(k/m) = √(100/0.25) = √400 = 20 rad s⁻¹

T = 2π/ω = 2π/20 = π/10 ≈ 0.314 s

(b) Find the maximum speed.

v_max = Aω = 0.08 × 20 = 1.6 m s⁻¹

(c) Find the speed when x = 0.05 m.

v = ω√(A² − x²) = 20 × √(0.08² − 0.05²) = 20 × √(0.0064 − 0.0025) = 20 × √0.0039 = 20 × 0.0625 = 1.25 m s⁻¹

(d) Find the total energy.

E = ½kA² = ½ × 100 × 0.08² = 0.32 J

(e) Find the displacement after 0.20 s.

x = A cos(ωt) = 0.08 × cos(20 × 0.20) = 0.08 × cos(4.0) = 0.08 × (−0.654) = −0.052 m

(f) Find the time to first reach x = 0.04 m.

0.04 = 0.08 cos(20t) cos(20t) = 0.5 20t = π/3 t = π/60 ≈ 0.052 s

Worked Example 2: Pendulum and g

A student measures the period of a pendulum for various lengths:

Length L (m)	Period T (s)	T² (s²)
0.20	0.90	0.81
0.40	1.27	1.61
0.60	1.56	2.43
0.80	1.80	3.24
1.00	2.01	4.04

Plotting T² against L gives a straight line with gradient ≈ 4.04 s² m⁻¹.

Since T² = (4π²/g)L, the gradient = 4π²/g.

So g = 4π²/gradient = 4 × 9.87/4.04 = 39.48/4.04 ≈ 9.77 m s⁻².

This is close to the accepted value of 9.81 m s⁻², within experimental uncertainty.

Percentage difference = |9.81 − 9.77|/9.81 × 100% = 0.4%

Worked Example 3: Energy Partition in SHM

A mass-spring system oscillates with amplitude 0.12 m and total energy 0.36 J.

(a) Find the displacement where KE = 3 × PE.

KE + PE = E_total, so if KE = 3PE, then 3PE + PE = 0.36, giving PE = 0.09 J.

PE = ½mω²x² and E_total = ½mω²A², so PE/E_total = x²/A².

x²/A² = 0.09/0.36 = 0.25, so x = ±0.5A = ±0.06 m.

(b) Find the speed at this point if the mass is 0.50 kg.

KE = 3 × 0.09 = 0.27 J

½mv² = 0.27, so v = √(2 × 0.27/0.50) = √1.08 = 1.04 m s⁻¹

Worked Example 4: Resonance Curve Analysis

A resonance curve shows amplitude vs driving frequency for a mass-spring system with natural frequency 5 Hz. The peak amplitude is 12 cm at f_d = 5 Hz.

At f_d = 3 Hz, the amplitude is approximately 2 cm
At f_d = 5 Hz, the amplitude is 12 cm (resonance)
At f_d = 7 Hz, the amplitude is approximately 1.5 cm

If damping were increased:

The peak would be lower (e.g., 6 cm instead of 12 cm)
The peak would be broader (significant response over a wider range of frequencies)
The peak would shift very slightly below 5 Hz (usually negligible)

If the mass were increased (keeping k the same):

The natural frequency would decrease: f₀ = (1/2π)√(k/m)
The resonance peak would shift to a lower frequency
The peak amplitude would depend on the new damping ratio

Worked Example 5: Inverse Square Law with Background

A gamma source gives a corrected count rate of 900 counts per minute at a distance of 0.15 m. The background count rate is 36 counts per minute.

(a) Find the corrected count rate at 0.45 m.

C₂ = C₁ × (r₁/r₂)² = 900 × (0.15/0.45)² = 900 × (1/3)² = 900 × 1/9 = 100 cpm

(b) What would the GM tube actually read (including background)?

Actual reading = corrected count rate + background = 100 + 36 = 136 cpm

(c) At what distance would the corrected count rate equal the background count rate?

36 = 900 × (0.15/r)²

(0.15/r)² = 36/900 = 0.04

0.15/r = 0.2

r = 0.15/0.2 = 0.75 m

Beyond 0.75 m, the source count rate is less than the background, making detection unreliable.

Worked Example 6: Combined Half-Life and Inverse Square Law

A gamma source has an activity of 8000 Bq and a half-life of 6 hours. A GM tube at 2 m gives a corrected count rate of 200 counts per minute.

(a) What will the corrected count rate be at 2 m after 18 hours?

18 hours = 3 half-lives. Activity drops by factor 2³ = 8.

New corrected count rate = 200/8 = 25 cpm (at same distance).

(b) At what distance should the GM tube be placed (after 18 hours) to obtain the same count rate of 200 cpm as originally?

The source is now 8× weaker. To compensate, the detector must be √8 ≈ 2.83 times closer.

New distance = 2/√8 = 2/(2√2) = 1/√2 ≈ 0.71 m

Check: count rate ∝ Activity/r². Originally: 200 ∝ A/4. After 18 h at r: 200 ∝ (A/8)/r². So r² = 4/8 = 0.5, r = √0.5 ≈ 0.71 m. ✓

Worked Example 7: Dose Equivalent Comparison

A worker is exposed to 0.5 mGy of gamma radiation and 0.025 mGy of alpha radiation in one week. Calculate the total dose equivalent and compare each contribution.

Solution:

H_gamma = D × Q = 0.5 × 1 = 0.5 mSv

H_alpha = D × Q = 0.025 × 20 = 0.5 mSv

Total H = 0.5 + 0.5 = 1.0 mSv

Despite receiving 20× less absorbed dose from alpha, the biological effect is the same as from gamma. This demonstrates why the quality factor is essential — absorbed dose alone does not reflect biological risk.

Radiation Properties Quick-Reference

Property	Alpha (α)	Beta (β⁻)	Gamma (γ)
Stopped by	Paper	3 mm Al	cm of Pb
Ionising power	Very high	Moderate	Low
Range in air	~5 cm	~1 m	Follows 1/r²
External hazard	Low	Moderate	High
Internal hazard	Extreme	Moderate	Lower
Quality factor Q	20	1	1
Deflected by fields?	Yes (small)	Yes (large)	No

Common Exam Mistakes to Avoid

flowchart TD
    A["Common Mistakes\nin Oscillations &\nRadiation"] --> B["SHM Mistakes"]
    A --> C["Radiation Mistakes"]
    B --> B1["Calculator in degrees\nnot radians"]
    B --> B2["Confusing f and ω\n(ω = 2πf, NOT ω = f)"]
    B --> B3["Writing v_max = Aω²\n(should be Aω)"]
    B --> B4["Forgetting E ∝ A²\n(2A gives 4E, not 2E)"]
    B --> B5["Not checking initial\nconditions for sin/cos"]
    C --> C1["Not subtracting\nbackground before\napplying 1/r²"]
    C --> C2["Applying 1/r² to\nalpha or beta"]
    C --> C3["Confusing Gy and Sv\n(Sv = Gy × Q)"]
    C --> C4["Confusing irradiation\nand contamination"]
    C --> C5["Forgetting to convert\nunits (cm→m, min→s)"]

Key Equations Summary

Equation	What It Describes
a = −ω²x	Defining equation of SHM
x = A cos(ωt)	Displacement vs time (starts at +A)
v = ±ω√(A² − x²)	Speed at displacement x
v_max = Aω	Maximum speed (at x = 0)
a_max = Aω²	Maximum acceleration (at x = ±A)
E_total = ½mω²A²	Total energy in SHM
T = 2π√(m/k)	Period of mass-spring system
T = 2π√(L/g)	Period of simple pendulum
I ∝ 1/r²	Inverse square law for gamma
H = D × Q	Dose equivalent (Sv)

Exam Strategy Tips

Read the question carefully — identify all given quantities before reaching for an equation
Write down what you know — list all given values with correct units
Show your working — even if you make an arithmetic error, method marks are available
Check your answer — does it have sensible units? Is the magnitude reasonable?
For graph questions — always label axes with quantities and units; draw smooth curves, not dot-to-dot
For "explain" questions — use correct physics terminology and refer to specific equations where relevant
For practical questions — mention repeated measurements, calculating averages, and sources of uncertainty

A-Level Deep Dive: Oscillations and Radiation Problem Solving

Spec mapping

This synoptic capstone draws on Edexcel 9PH0 specification Topic 11 — Astrophysics-style applications of SHM (oscillations, damping, resonance and forced motion) together with Topic 13 — Nuclear radiation (radioactive decay, activity, half-life, the inverse-square law for gamma intensity, and biological dose) (refer to the official specification document for exact wording). Examiners build synoptic problem sets that deliberately cross these topics because both phenomena are governed by first-order linear differential equations with exponential solutions: damped SHM amplitude decays as $A(t) = A_0 e^{-\gamma t}$ , while nuclear activity decays as $N(t) = N_0 e^{-\lambda t}$ . Recognising the shared mathematical scaffold is an explicit AO2 expectation in Paper 3 and is also tested in Paper 1 multi-step calculations. The Edexcel formula booklet supplies the SHM definitions, $T = 2\pi\sqrt{m/k}$ , $T = 2\pi\sqrt{L/g}$ , and the radioactive decay relations; it does not supply the inverse-square-law form, the dose-equivalent definition, or the damped-amplitude expression — these must be reconstructed from physical reasoning.

Worked example with full mark scheme

Question (12 marks): A research apparatus uses a vertical mass-spring oscillator (mass $m = 0.250$ kg, spring constant $k = 40.0$ N m $^{-1}$ ) carrying a small sealed source of $^{60}$ Co that emits gamma photons. The oscillator is released from rest with amplitude $A_0 = 8.0$ cm and the amplitude decays exponentially with time constant $\tau_{\text{SHM}} = 45$ s. A gamma detector sits a fixed perpendicular distance $d = 0.50$ m from the equilibrium position. The half-life of $^{60}$ Co is $T_{1/2} = 5.27$ years.

(a) Calculate the natural angular frequency $\omega_0$ and period $T_0$ of the oscillator. (2)

(b) The amplitude has decayed to $1.0$ cm. Show that the time elapsed is approximately $94$ s. (3)

(c) During this $94$ s, calculate the fractional decrease in the activity of the $^{60}$ Co source. Comment on whether the decrease in detected gamma count rate is dominated by the source decay or by the change in oscillator amplitude. (4)

(d) The $94$ s exposure delivers an absorbed dose of $D = 1.2 \times 10^{-5}$ Gy to a small tissue sample. Using a quality factor $Q = 1$ for gamma radiation, calculate the dose equivalent. (3)

Solution with mark scheme:

(a) $\omega_0 = \sqrt{k/m} = \sqrt{40.0/0.250} = \sqrt{160} = 12.65$ rad s $^{-1}$ .

M1 — correct application of $\omega = \sqrt{k/m}$ for a mass-spring system.

$T_0 = 2\pi/\omega_0 = 2\pi/12.65 = 0.497$ s.

A1 — both numerical answers correct to 3 s.f. with units.

(b) Damped-amplitude model: $A(t) = A_0 e^{-t/\tau_{\text{SHM}}}$ .

Set $A(t) = 1.0$ cm with $A_0 = 8.0$ cm:

$\frac{1.0}{8.0} = e^{-t/45} \implies \ln(0.125) = -t/45 \implies t = 45 \ln 8 = 45 \times 2.079 = 93.6 \text{ s}.$

M1 — setting up the exponential-decay equation for the amplitude. M1 — taking natural logarithms correctly. A1 — answer rounded appropriately to $94$ s. (Use of $\log_{10}$ without conversion to $\ln$ loses both M marks.)

(c) Source decay constant: $\lambda = \ln 2 / T_{1/2}$ . Convert $T_{1/2}$ to seconds: $5.27 \times 365.25 \times 24 \times 3600 = 1.66 \times 10^{8}$ s.

$\lambda = \frac{0.693}{1.66 \times 10^{8}} = 4.17 \times 10^{-9} \text{ s}^{-1}.$

M1 — correct conversion of half-life and use of $\lambda = \ln 2 / T_{1/2}$ .

Fractional decrease over $94$ s: $1 - e^{-\lambda t} \approx \lambda t = 4.17 \times 10^{-9} \times 94 = 3.9 \times 10^{-7}$ , i.e. about $4 \times 10^{-5}$ % — negligible.

M1 — using the small- $\lambda t$ approximation $1 - e^{-\lambda t} \approx \lambda t$ (valid when $\lambda t \ll 1$ ). A1 — numerical answer with order of magnitude correct.

B1 — comment: over $94$ s the source decay is utterly negligible (5.27 years vs 94 seconds), so any change in detector count rate must be attributed to the geometry — chiefly the changing perpendicular distance from source to detector as the oscillator moves through its cycle, not radioactive decay.

(d) $H = D \times Q = 1.2 \times 10^{-5} \times 1 = 1.2 \times 10^{-5}$ Sv $= 12$ μSv.

M1 — quoting $H = DQ$ . A1 — numerical answer. B1 — correct unit (Sv or μSv) explicitly stated, with conversion shown.

Total: 12 marks (M5 A4 B3).

Specimen question modelled on the Edexcel 9PH0 Paper 3 format

Question (10 marks): A simple pendulum of length $L = 1.20$ m carries a bob of mass $m = 0.150$ kg into which a tiny $^{14}$ C tracer ( $T_{1/2} = 5730$ years) has been embedded for a tracking experiment. The bob is displaced through a small angle and released; over $30$ minutes the amplitude decays from $\theta_0 = 6.0°$ to $\theta = 1.5°$ .

(a) State two assumptions required for the small-angle SHM model to apply. (2)

(b) Calculate the period of the pendulum and hence the number of complete oscillations in $30$ minutes. (3)

(d) Justify quantitatively why the activity of the $^{14}$ C tracer can be treated as constant throughout the experiment. (2)

Mark scheme decomposition by AO:

(a) B1 (AO1.1) — small-angle approximation $\sin\theta \approx \theta$ (in radians). B1 (AO1.1) — air resistance / pivot friction either neglected or modelled as a small linear-drag perturbation; pendulum mass concentrated at the bob.

(b) M1 (AO1.1b) — $T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1.20/9.81} = 2.20$ s. A1 (AO1.1b) — period stated. A1 (AO2.1) — number of oscillations $= 1800/2.20 \approx 818$ .

(c) M1 (AO2.1) — recognise exponential decay model $\theta(t) = \theta_0 e^{-t/\tau}$ . M1 (AO1.1b) — solve $1.5/6.0 = e^{-1800/\tau}$ , giving $\tau = 1800/\ln 4 = 1800/1.386 = 1299$ s. A1 (AO2.5) — answer to appropriate s.f., $\tau \approx 1.3 \times 10^{3}$ s.

(d) M1 (AO3.1a) — calculate $\lambda t$ where $t = 1800$ s and $T_{1/2} \approx 1.81 \times 10^{11}$ s, giving $\lambda t \approx 7 \times 10^{-9}$ . A1 (AO3.2a) — conclude the activity changes by a fraction of order $10^{-9}$ , far smaller than any measurement uncertainty, so treating activity as constant is justified.

Total: 10 marks split AO1 = 4, AO2 = 4, AO3 = 2. Paper 3 favours synoptic AO3 reasoning in the closing parts; the candidate is expected to quantify the assumption rather than merely assert it.

Synoptic links

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