Required practicals: A-level Physics AQA revision notes

RP1

Stationary waves on a string

Key results

First harmonic: $f = \frac{1}{2 L} \frac{T}{μ}$ .
Tension from the hanging masses: $T = m g$ .
Mass per unit length: $μ = \frac{mass of a measured length of string}{that length}$ .
Linear graphs: $f$ against $1/ L$ (gradient $\frac{1}{2} T / μ$ ); $f^{2}$ against $T$ (gradient $\frac{1}{4 L ^{2} μ}$ ).

Method

Set up the string between the vibration generator and the pulley, with masses on the hanger to set the tension.
Measure the vibrating length $L$ (vibrator to pulley) with a metre rule.
Increase the frequency from low until the first harmonic (one large loop) forms with maximum amplitude; read $f$ from the signal generator.
Change $L$ (move the vibrator), or the tension (add masses), or use strings of different $μ$ , and repeat; plot the appropriate straight-line graph.

In practiceA string 0.60 m long has tension 4.9 N and mass per unit length $5.0 \times 1 0^{- 4}$ kg m⁻¹. Find its first-harmonic frequency.

$v = \frac{T}{μ} = \frac{4.9}{5.0 \times 1 0 ^{- 4}} = 99.0 m s^{- 1}$
$f = \frac{v}{2 L} = \frac{99.0}{1.20} = 82.5 Hz$ first harmonic: λ = 2L

Notes

Judging maximum amplitude is subjective; approach the resonance from both sides and take the mean, or use a strobe to see the nodes clearly.
Safety: masses can fall if the string snaps; keep feet clear and use a sand tray.
Exam trap: measuring $L$ to the end of the hanging string rather than to where the string first touches the top of the pulley (the node).

A string fixed at both ends and driven at one end forms stationary waves when a whole number of half-wavelengths fits its length. The first harmonic has one loop, so $λ = 2 L$ . Since $v = T / μ$ for a string, the first-harmonic frequency depends on the length, the tension and the mass per unit length, and each can be varied in turn.

RP2

Young's slits and the diffraction grating

Interference: Young's double slits and the diffraction grating

Key results

Double slit: $w = \frac{λ D}{s}$ , so $λ = \frac{w s}{D}$ (for $D ≫ s$ ).
Grating: $d sin θ = nλ$ , with $d = 1/ N$ for $N$ lines per metre.
Angle of the $n$ th order: $tan θ = x / D$ , where $x$ is its distance from the central maximum on a screen a distance $D$ away.

Method

Double slit: shine the laser through the slits onto a screen 1–2 m away; measure $D$ with a tape measure.
Measure across as many fringe spacings as possible (e.g. ten) with a metre rule, and divide by the number of spacings to find $w$ .
Read the slit separation $s$ from the slide or measure it with a travelling microscope.
Grating: measure the distance between matching maxima on either side of the centre for each order, halve it to get $x$ , and find $θ$ ; or use a spectrometer to measure $θ$ directly.

In practiceA grating has 600 lines per mm; the first-order maximum is at 22.3°. Find the wavelength.

$d = \frac{1}{600 \times 1 0 ^{3}} = 1.67 \times 1 0^{- 6} m$
$λ = d sin θ = 1.67 \times 1 0^{- 6} \times sin 22. 3^{\circ} = 6.33 \times 1 0^{- 7} m$ n = 1

Notes

Measuring across many fringes reduces the percentage uncertainty in $w$ .
Measuring between matching maxima on either side avoids locating the centre, and measuring $2 x$ halves the percentage uncertainty in $x$ .
Safety: never look into the laser beam or let it reflect into eyes; remove reflective jewellery; use a low-power (class 2) laser and warning signs.
Exam trap: using $sin θ \approx tan θ$ for a grating, where angles are large.

Coherent light through two slits produces equally spaced bright and dark fringes; through a grating it produces sharp bright maxima at angles that depend on the wavelength. Both give a value for the wavelength of laser light.

RP3

g by free fall

Measuring g with a falling ball

Key results

$h = \frac{1}{2} g t^{2}$ , so a graph of $h$ against $t^{2}$ has gradient $g /2$ ; $g = 2 \times gradient$ .
Equivalently, $2 h$ against $t^{2}$ has gradient $g$ .

Method

Hold a steel ball on an electromagnet above a trapdoor; switching off the magnet starts the timer, and the ball opening the trapdoor stops it.
Measure $h$ from the bottom of the ball to the trapdoor with a metre rule, using a set square to avoid parallax.
Record $t$ for each height several times and take the mean.
Repeat for at least six heights and plot $h$ against $t^{2}$ .

In practiceA ball falls 0.800 m from rest in 0.404 s. Find g.

$h = \frac{1}{2} g t^{2} \Rightarrow g = \frac{2 h}{t ^{2}}$
$g = \frac{2 \times 0.800}{0.40 4 ^{2}} = 9.80 m s^{- 2}$ a graph of many heights is more reliable

Notes

Residual magnetism delays the release, adding a constant time $δ$ to every reading. Then $t = 2/ g h + δ$ : a graph of $t$ against $h$ is straight, the delay is its intercept, and its gradient $2/ g$ still gives $g$ . (On an $h$ – $t^{2}$ graph the same delay bends the line and lowers its gradient.)
Air resistance is negligible for a small dense ball over short drops; a ping-pong ball would give a low value.
Exam trap: measuring $h$ from the top of the ball, which makes every height too large by one diameter.

An object dropped from rest falls a height $h$ in time $t$ , with $h = \frac{1}{2} g t^{2}$ . Timing the fall for several heights and plotting $h$ against $t^{2}$ gives a straight line of gradient $g /2$ . Electronic timing (an electromagnet and a trapdoor, or light gates) removes reaction time.

RP4

The Young modulus

Stretching a wire: the Young modulus

Key results

$E = \frac{stress}{strain} = \frac{F L}{A Δ L}$ , with $A = \frac{π d ^{2}}{4}$ .
Plot $F$ against $Δ L$ : gradient $= \frac{E A}{L}$ , so $E = gradient \times \frac{L}{A}$ .

Method

Clamp a long wire (2–3 m) at one end; run it over a pulley with a hanger; fix a marker on the wire next to a metre rule (or use a vernier on a Searle's apparatus).
Apply a small initial load to straighten the wire and take this as zero.
Measure the original length $L$ from the clamp to the marker with a tape measure.
Measure the diameter with a micrometer at several points along the wire, in two perpendicular directions, and take the mean.
Add loads in equal steps, recording the extension each time; then remove them, checking that the wire returns to its original length (elastic).

In practiceA 40 N load stretches a 2.00 m wire of diameter 0.50 mm by 1.6 mm. Find the Young modulus.

$A = π (0.25 \times 1 0^{- 3})^{2} = 1.96 \times 1 0^{- 7} m^{2}$ radius, in metres
$σ = \frac{40}{1.96 \times 1 0 ^{- 7}} = 2.04 \times 1 0^{8} Pa, ε = \frac{1.6 \times 1 0 ^{- 3}}{2.00} = 8.0 \times 1 0^{- 4}$
$E = \frac{σ}{ε} = 2.5 \times 1 0^{11} Pa$

Notes

The diameter is small, so its percentage uncertainty is large, and it is squared in $A$ : it usually dominates the uncertainty in $E$ .
Safety: wear eye protection in case the wire snaps; keep feet away from the masses.
Exam trap: using the diameter in place of the radius in $π r^{2}$ .

A long, thin wire is stretched by known loads and its extension measured. Within the limit of proportionality, stress is proportional to strain and the ratio is the Young modulus. A long wire gives a larger extension, reducing its percentage uncertainty; a thin wire has a small area, so a modest load gives a large stress and a measurable extension (at the cost of a larger percentage uncertainty in its diameter).

RP5

Resistivity of a wire

Key results

$R = \frac{ρ L}{A}$ , so the gradient of $R$ against $L$ is $ρ / A$ and $ρ = gradient \times A$ .
$A = \frac{π d ^{2}}{4}$ ; $R = V / I$ from the meters.

Method

Measure the diameter with a micrometer at several points and orientations; calculate the mean and $A$ .
Tape the wire to a metre rule; connect a crocodile clip at zero and a flying lead at the chosen length.
Measure the current and the potential difference across that length; calculate $R$ .
Repeat for several lengths and plot $R$ against $L$ .

In practiceA 0.500 m length of wire of diameter 0.27 mm has resistance 4.5 Ω. Find the resistivity.

$A = π (0.135 \times 1 0^{- 3})^{2} = 5.73 \times 1 0^{- 8} m^{2}$
$ρ = \frac{R A}{L} = \frac{4.5 \times 5.73 \times 1 0 ^{- 8}}{0.500} = 5.2 \times 1 0^{- 7} Ω m$ one length only: a graph of R against L is better

Notes

Keep the current small and switch off between readings: a hot wire has a higher resistance.
A zero error from contact resistance shifts the line up but not its gradient, so the gradient method gives a better $ρ$ than a single reading.
Exam trap: forgetting to convert the diameter from mm to m before squaring.

The resistance of a uniform wire is proportional to its length, $R = ρ L / A$ . Measuring $R$ for several lengths and plotting $R$ against $L$ gives a straight line of gradient $ρ / A$ ; the cross-sectional area comes from the diameter.

RP6

EMF and internal resistance

EMF and internal resistance of a cell

Key results

$V = ε - I r$ .
Graph of $V$ against $I$ : $y$ -intercept $= ε$ ; gradient $= - r$ .

Method

Connect the cell in series with an ammeter, a variable resistor and a switch; connect a voltmeter across the cell's terminals.
Set the variable resistor, close the switch, read $V$ and $I$ , and open the switch again.
Repeat over a range of resistances; plot $V$ against $I$ .

In practiceA cell gives 1.40 V at 0.20 A and 1.10 V at 0.80 A. Find its internal resistance and EMF.

$r = - \frac{Δ V}{Δ I} = \frac{1.40 - 1.10}{0.80 - 0.20} = 0.50 Ω$ minus the gradient
$ε = V + I r = 1.40 + 0.20 \times 0.50 = 1.50 V$

Notes

Open the switch between readings and keep currents modest: a cell warms up when it delivers current and its internal resistance changes.
A protective resistor in series stops the current becoming too large if the variable resistor is set near zero.
Exam trap: reading the EMF as the terminal p.d. at a non-zero current.

Some of a cell's EMF is used driving current through its own internal resistance, so the terminal potential difference falls as the current rises. Varying the load and plotting terminal p.d. against current gives a straight line whose intercept is the EMF and whose gradient is minus the internal resistance.

RP7

Simple harmonic motion

Simple harmonic motion: mass–spring and pendulum

Key results

Mass–spring: $T = 2 π \frac{m}{k}$ ; $T^{2}$ against $m$ has gradient $4 π^{2} / k$ .
Pendulum: $T = 2 π \frac{l}{g}$ ; $T^{2}$ against $l$ has gradient $4 π^{2} / g$ .

Method

Displace the mass (or the pendulum bob, by less than about 10°) and release it.
Use a fiducial marker at the equilibrium position, where the object moves fastest, to start and stop the timer.
Time 10 (or 20) complete oscillations and divide to find $T$ ; repeat and take the mean.
Vary $m$ (or $l$ , measured to the centre of mass of the bob) and plot $T^{2}$ against it.

In practiceFor a mass on a spring, the graph of $T^{2}$ against $m$ has gradient 0.160 s² kg⁻¹. Find the spring constant.

$T^{2} = \frac{4 π ^{2}}{k} m \Rightarrow k = \frac{4 π ^{2}}{gradient}$
$k = \frac{39.5}{0.160} = 247 N m^{- 1}$

Notes

Timing many oscillations divides the reaction-time uncertainty by the number of oscillations.
Large pendulum amplitudes break the small-angle approximation and lengthen the period.
Exam trap: counting a 'swing' (half an oscillation) as one oscillation.

Both a mass on a spring and a simple pendulum oscillate with simple harmonic motion (the pendulum only for small angles). Their periods depend on different things, and squaring the period gives straight-line graphs from which $k$ or $g$ is found.

RP8

Boyle's law and Charles's law

Key results

Boyle's law: $p \propto 1/ V$ , so $p$ against $1/ V$ is a straight line through the origin.
Charles's law: $V$ against temperature in °C is a straight line meeting $V = 0$ at about −273 °C.
In a uniform tube, $V \propto$ the length of the gas column, so length can stand in for volume.

Method

Boyle: trap air in a sealed glass tube connected to a pressure gauge and pump; read the volume from the scale.
Change the pressure in steps; wait a minute after each change for the air to return to room temperature before reading.
Charles: trap a column of air in a capillary tube with a bead of liquid; immerse the tube in a water bath beside a thermometer.
Heat in steps, stirring, and wait for the temperature to settle; record the column length at each temperature.

In practiceAir at 120 kPa occupies 30.0 cm³; it expands slowly at constant temperature to 45.0 cm³. Find the new pressure.

$p_{1} V_{1} = p_{2} V_{2}$ Boyle's law
$p_{2} = \frac{120 \times 30.0}{45.0} = 80 kPa$

Notes

Compressing a gas warms it; reading too soon gives pressures that are too high.
Safety: the pressurised tube should be behind a screen and the pressure kept below the maximum on the equipment; hot water can scald.
Exam trap: plotting $V$ against temperature in °C and expecting a line through the origin.

A fixed mass of gas obeys $p V$ = constant at constant temperature (Boyle) and $V \propto T$ at constant pressure (Charles), with $T$ in kelvin. Extrapolating Charles's-law data to zero volume gives an estimate of absolute zero.

RP9

Capacitor charge and discharge

Charging and discharging a capacitor

Key results

$ln V = ln V_{0} - \frac{t}{RC}$ : a graph of $ln V$ against $t$ has gradient $- 1/ RC$ .
Time constant $RC$ : the time to fall to $1/ e$ (about 37%) of the initial value. Half-life $t_{1/2} = RC ln 2 \approx 0.69 RC$ .
Charging: $V = V_{0} (1 - e^{- t / RC})$ .

Method

Connect the capacitor to a d.c. supply through a two-way switch to charge it, with a voltmeter across it.
Switch to discharge through the resistor and start the stopwatch at once.
Record $V$ at regular intervals (e.g. every 10 s), or use a data logger for fast discharges.
Plot $ln V$ against $t$ and find the gradient.

In practiceA capacitor's p.d. falls from 9.0 V to 3.3 V in 20 s while discharging through a resistor. Find the time constant.

$ln \frac{3.3}{9.0} = - \frac{t}{RC}$ from V = V₀e^(−t/RC)
$- 1.003 = - \frac{20}{RC} \Rightarrow RC = 19.9 s$ about 20 s: a fall to about 37% in one time constant

Notes

Choose $R$ and $C$ so that $RC$ is tens of seconds when timing by hand; a data logger is needed for short time constants.
Electrolytic capacitors must be connected the right way round, and their capacitance tolerance is often ±20%: the measured value may differ from the label.
Exam trap: finding $RC$ from the gradient without the minus sign, or using log₁₀ instead of ln.

When a capacitor discharges through a resistor, the potential difference falls exponentially, $V = V_{0} e^{- t / RC}$ . Taking natural logs gives a straight line, and its gradient gives the time constant $RC$ ; with $R$ known, the capacitance follows.

RP10

Force on a current-carrying wire

Force on a current-carrying wire in a magnetic field

Key results

$F = B I L$ , with $F = (Δ m) g$ from the change in the balance reading $Δ m$ (in kg).
Graph of $F$ against $I$ : gradient $= B L$ , so $B = \frac{gradient}{L}$ .
$L$ is the length of wire within the field, approximately the length of the pole faces.

Method

Stand the magnets on the balance with the wire clamped horizontally through the gap, not touching the magnets; zero the balance.
Pass a current through the wire, measured by an ammeter, and record the balance reading.
Increase the current in steps and record each reading; reverse the current to check that the change reverses.
Vary $L$ by using magnets of different lengths, or keep $I$ fixed and change the number of magnets.

In practiceWith 3.0 A in a 0.040 m length of wire in the field, the balance reading changes by 2.4 g. Find B.

$F = Δ m g = 2.4 \times 1 0^{- 3} \times 9.81 = 0.0235 N$ grams to kilograms
$B = \frac{F}{I L} = \frac{0.0235}{3.0 \times 0.040} = 0.20 T$

Notes

The wire warms if the current is large: keep the current on only while reading.
The field is not uniform near the edges of the poles, so the effective length is uncertain.
Exam trap: forgetting to convert the balance reading from grams to kilograms before multiplying by $g$ .

A wire carrying a current across a magnetic field experiences a force $F = B I L$ . Putting the magnets on a top-pan balance turns this force into a change in the balance reading: by Newton's third law, the magnets feel an equal and opposite force to the wire.

RP11

Search coil and flux linkage

Key results

Flux linkage $N Φ = B A N cos θ$ , with $θ$ between the field and the normal to the coil.
For a sinusoidal field, the peak emf is proportional to $cos θ$ : a graph of peak emf against $cos θ$ is a straight line through the origin.
On the oscilloscope, peak emf = (peak height in divisions) × (volts per division).

Method

Connect the search coil to an oscilloscope; place it at the centre of the field coil, which is driven by a signal generator.
Align the coil so its plane is perpendicular to the field ( $θ = 0$ ); measure the peak-to-peak height and halve it.
Rotate the coil in steps (e.g. 10°) using a protractor, recording the peak emf each time.
Plot peak emf against $cos θ$ .

In practiceThe peak emf is 6.0 mV with the field along the coil's normal. Predict it at 45° and at 90°.

$ε_{45} = 6.0 cos 4 5^{\circ} = 4.2 mV$ peak emf ∝ cos θ
$ε_{90} = 6.0 cos 9 0^{\circ} = 0$ plane of coil parallel to the field

Notes

Keep the search coil at the same position and the supply frequency and current constant: the emf also depends on both.
Measuring peak-to-peak and halving reduces the uncertainty in locating the zero line.
Exam trap: measuring $θ$ from the plane of the coil and getting $sin θ$ instead of $cos θ$ .

An alternating current in a large coil produces an alternating magnetic field. A small search coil placed in it has an alternating emf induced, whose peak value is proportional to the peak flux linkage. Rotating the search coil changes the angle between its normal and the field, and the flux linkage varies as $cos θ$ .

RP12

Inverse-square law for gamma radiation

The inverse-square law for gamma radiation

Key results

$I \propto \frac{1}{d ^{2}}$ , so corrected count rate $C = \frac{k}{d ^{2}}$ .
The true distance includes an unknown offset $e$ inside the source holder and tube: $C = \frac{k}{( d + e ) ^{2}}$ , so $\frac{1}{C}$ against $d$ is a straight line whose intercept on the $d$ -axis is $- e$ .

Method

Measure the background count rate with the source in its lead box, over a long time.
Place the source (handled with tongs) a measured distance from the GM tube, along a metre rule.
Record the count over a fixed time (e.g. 1–5 minutes) at each distance; subtract background to get the corrected count rate.
Plot $1/ C$ against $d$ .

In practiceThe corrected count rate is 360 per minute at 0.15 m. Predict it at 0.30 m and 0.45 m if the inverse-square law holds.

$C \propto \frac{1}{d ^{2}} : C_{0.30} = \frac{360}{2 ^{2}} = 90 min^{- 1}$ distance doubled
$C_{0.45} = \frac{360}{3 ^{2}} = 40 min^{- 1}$ distance tripled

Notes

Radioactive decay is random, so longer counts reduce the percentage uncertainty (about $1/ N$ for $N$ counts).
Safety: use tongs, keep the source at arm's length and pointed away from people, minimise exposure time, and return it to the lead box as soon as possible.
Exam trap: forgetting to subtract background, which matters most at large distances where the count rate is small.

Gamma radiation from a point source spreads out in all directions and is hardly absorbed by air, so its intensity falls as the inverse square of the distance. Measuring the count rate at several distances with a Geiger–Müller tube, after subtracting background, tests this.