Foundations of Physics: free practice, theory and problems

1. Quantities, units and measurement

What is it about?

All of physics is about measuring and calculating with quantities such as length, time, mass, speed and force. A physical quantity is always a number times a unit. "The length is 3" tells you nothing – 3 mm and 3 km are very different things.

For an engineer, a unit error can be expensive: a beam calculated in N where it should have said kN is a thousand times too weak. This unit therefore covers the SI system, prefixes such as kilo and milli, how to convert between units, and how many digits it is reasonable to keep in an answer.

Concepts and formulas

• Quantity = numerical value × unit, for example $m = 2.5\ \mathrm{kg}$. The symbol ($m$) is written in italics, the unit (kg) upright.
• SI base units: meter (m) for length, kilogram (kg) for mass, second (s) for time, ampere (A) for electric current, kelvin (K) for temperature, mole (mol) for amount of substance and candela (cd) for luminous intensity.
• Derived units are built from the base units: velocity in m/s, acceleration in m/s², force in newtons ($1\ \mathrm{N} = 1\ \mathrm{kg\cdot m/s^2}$), energy in joules ($1\ \mathrm{J} = 1\ \mathrm{N\cdot m}$), power in watts ($1\ \mathrm{W} = 1\ \mathrm{J/s}$) and pressure in pascals ($1\ \mathrm{Pa} = 1\ \mathrm{N/m^2}$).
• Prefixes are powers of ten placed in front of the unit: giga (G) $= 10^9$, mega (M) $= 10^6$, kilo (k) $= 10^3$, centi (c) $= 10^{-2}$, milli (m) $= 10^{-3}$, micro (µ) $= 10^{-6}$ and nano (n) $= 10^{-9}$.
• Area and volume: the conversion factor must be squared or cubed as well. Since $1\ \mathrm{m} = 100\ \mathrm{cm}$, we get $1\ \mathrm{m^2} = 100^2\ \mathrm{cm^2} = 10^4\ \mathrm{cm^2}$ and $1\ \mathrm{m^3} = 10^6\ \mathrm{cm^3}$. Liquids are measured in liters: $1\ \mathrm{L} = 1\ \mathrm{dm^3}$ and $1\ \mathrm{m^3} = 1000\ \mathrm{L}$.
• Speed: $1\ \mathrm{m/s} = 3.6\ \mathrm{km/h}$. To go from km/h to m/s, divide by 3.6.
• Significant figures are the digits that carry information. Leading zeros do not count ($0.0052$ has 2), while zeros between other digits and trailing zeros after the decimal point do ($4.05$ and $3.40$ have 3). When multiplying or dividing, the result gets as many significant figures as the factor with the fewest.
• Measurement uncertainty: a measurement is written $x \pm \Delta x$. The relative uncertainty is $\Delta x / x$, often given in percent. When multiplying or dividing, we add the relative uncertainties.

How to solve the problems

1. Write down every given quantity with its number and its unit.
2. Convert to SI base units (m, kg, s) before substituting into a formula. Replace prefixes with powers of ten.
3. Convert by multiplying by a fraction equal to 1, for example $\frac{1000\ \mathrm{m}}{1\ \mathrm{km}}$ or $\frac{1\ \mathrm{h}}{3600\ \mathrm{s}}$. Units cancel just like numbers.
4. Check the unit of the answer. If you get m/s when you wanted a length, the formula is wrong.
5. Round to a sensible number of significant figures at the end – not along the way.

Example

A car drives at a constant 72 km/h. How far does it travel in 2.5 minutes?

1. Speed in SI: $72\ \mathrm{km/h} = \frac{72\cdot 1000\ \mathrm{m}}{3600\ \mathrm{s}} = 20\ \mathrm{m/s}$.
2. Time in SI: $2.5\ \mathrm{min} = 2.5\cdot 60\ \mathrm{s} = 150\ \mathrm{s}$.
3. Distance: $s = vt = 20\ \mathrm{m/s}\cdot 150\ \mathrm{s} = 3000\ \mathrm{m}$.
4. Unit check: $\mathrm{m/s}\cdot\mathrm{s} = \mathrm{m}$, which is a length.

The answer is 3.0 km. We give two significant figures because 2.5 has two.

Example: measurement uncertainty

A plate is measured as $a = 2.50 \pm 0.02$ m and $b = 1.20 \pm 0.01$ m. The area is $A = ab = 3.00\ \mathrm{m^2}$. The relative uncertainties are $0.02/2.50 = 0.8\ \%$ and $0.01/1.20 \approx 0.8\ \%$, about $1.6\ \%$ in total. So $\Delta A \approx 0.016\cdot 3.00 \approx 0.05\ \mathrm{m^2}$, and $A = 3.00 \pm 0.05\ \mathrm{m^2}$.

Common mistakes

• Plugging km/h straight into formulas. Convert to m/s first (divide by 3.6).
• Thinking that $1\ \mathrm{m^2} = 100\ \mathrm{cm^2}$. It is $10\,000\ \mathrm{cm^2}$, because both the length and the width must be converted.
• Mixing up mass (kg) and weight (N). A kilogram is not a force.
• Copying every digit from the calculator. The answer cannot be more precise than the measurements it is based on.
• Counting leading zeros as significant: $0.004$ has only one.
• Confusing m for milli with m for meter: mN is a millinewton, while N·m (newton meter) is force times lever arm.

Concepts in this part

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2. Motion

What is it about?

Kinematics describes how things move – without asking why. We use four quantities: position (where), velocity (how fast), acceleration (how quickly the velocity changes) and time. With them you can calculate the braking distance of a car, how long a falling load takes to reach the ground, or how hard a robot arm has to accelerate.

Here we study motion along a straight line. A sign is then enough to describe direction: you choose a positive direction, and quantities pointing the other way become negative.

Concepts and formulas

• Displacement $s$ (m): how far the body has moved from its starting point, with sign.
• Average velocity: $\bar v = \dfrac{\Delta s}{\Delta t}$, in m/s. The velocity has a sign that shows the direction; its size is called the speed.
• Acceleration: $a = \dfrac{\Delta v}{\Delta t}$, in m/s². If $a$ has the same sign as the velocity, the body speeds up. The opposite sign means it slows down.
• The equations of motion hold only when the acceleration is constant. With initial velocity $v_0$ at time $t = 0$:

• Free fall: without air resistance, every body has the acceleration $g = 9.81\ \mathrm{m/s^2}$ straight down, regardless of its mass.
• Graphs: in an $s$-$t$ graph the slope is the velocity. In a $v$-$t$ graph the slope is the acceleration, and the area under the graph is the displacement.
• Projectiles: the horizontal and vertical motions are independent. An object thrown horizontally falls just as fast as one that is simply dropped.

How to solve the problems

1. Make a quick sketch and choose a positive direction (for example the direction of motion, or upward).
2. Write down which of $v_0$, $v$, $a$, $s$ and $t$ you know, with signs, and what you are looking for. Convert km/h to m/s.
3. Choose the equation that contains the known quantities and the unknown – and leaves out the one quantity you neither know nor need.
4. Solve for the unknown and substitute the numbers.
5. Check the sign, the unit and whether the answer is reasonable.

Example

A car traveling at 72 km/h brakes with constant acceleration and stops in 4.0 s. How long is the braking distance?

1. The positive direction is the direction of motion. $v_0 = 72/3.6 = 20\ \mathrm{m/s}$, $v = 0$ and $t = 4.0\ \mathrm{s}$.
2. Acceleration: $a = \dfrac{v - v_0}{t} = \dfrac{0 - 20}{4.0} = -5.0\ \mathrm{m/s^2}$. The minus sign means braking.
3. Distance: $s = \tfrac12 (v_0 + v)\,t = \tfrac12 (20 + 0)\cdot 4.0 = 40\ \mathrm{m}$.
4. Check with another equation: $s = \dfrac{v^2 - v_0^2}{2a} = \dfrac{0 - 400}{2\cdot(-5.0)} = 40\ \mathrm{m}$.

The braking distance is 40 m. In a $v$-$t$ graph this is the area of the triangle with base 4.0 s and height 20 m/s.

Common mistakes

• Using km/h directly in the formulas. Always convert to m/s first.
• Forgetting the minus sign on the acceleration when braking, or on $g$ when upward is chosen as positive.
• Using the equations of motion when the acceleration is not constant.
• Taking the average velocity as the mean of two speeds. The correct value is total distance divided by total time.
• Believing that both velocity and acceleration are zero at the top of a vertical throw. The velocity is zero, but the acceleration is still $g$ downward.
• Believing that heavy objects fall faster. Without air resistance everything falls with the same acceleration.

Concepts in this part

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3. Forces and Newton's laws

What is it about?

A force is a push or a pull. A force has both a magnitude and a direction (it is a vector) and is measured in newtons (N). Newton's three laws tell us how forces change motion. They are the foundation of all the mechanics you will meet later, from equilibrium of beams (statics) to accelerating machines (dynamics).

The key is to look at one body at a time and ask: which forces act on this body? Their sum determines the acceleration.

Concepts and formulas

• Newton's first law (law of inertia): if the net force is zero, the body stays at rest or keeps moving at constant velocity in a straight line.
• Newton's second law: $\sum F = ma$. The net force gives the acceleration, and they point in the same direction. $1\ \mathrm{N} = 1\ \mathrm{kg\cdot m/s^2}$.
• Newton's third law: when A exerts a force on B, B exerts an equally large, opposite force on A. The two forces act on different bodies and therefore never cancel each other.
• Weight (gravitational force): $F_g = mg$, pointing straight down, with $g = 9.81\ \mathrm{m/s^2}$. The mass (kg) is the same everywhere, while the weight (N) depends on where you are.
• **Normal force $N$:** the force from a surface, perpendicular to the surface. On a horizontal surface with no other vertical forces, $N = mg$. On an incline with angle $\theta$, $N = mg\cos\theta$, and the component of the weight along the incline is $mg\sin\theta$.
• **Friction $f$:** acts along the surface, opposing the motion (or the direction in which the body "wants" to slide). When the body slides, $f = \mu_k N$. When it is at rest, $f \le \mu_s N$, where $\mu_s$ is the coefficient of static friction.
• **Tension $T$:** a taut rope pulls on the body along the rope. Over a light, frictionless pulley the tension is the same at both ends.
• Free-body diagram: a sketch of the body on its own, with every force acting on it drawn as an arrow.

How to solve the problems

1. Choose the body you are studying and draw it on its own.
2. Draw every force acting on it: weight, normal force, friction, tensions and any other external forces. Leave out the forces the body itself exerts on other things.
3. Choose axes: one along the acceleration (the direction of motion) and one perpendicular to it. Resolve forces that act at an angle into components.
4. Write Newton's second law for each axis: $\sum F_x = ma$ and $\sum F_y = 0$ (no acceleration perpendicular to the surface).
5. Solve the equations. It often pays to find $N$ first, then the friction $f = \mu N$, and finally the acceleration. With several bodies (for example connected by a rope), write one equation per body.

Example

A 20 kg crate is pulled across a horizontal floor by a horizontal force of 100 N. The coefficient of kinetic friction is $\mu_k = 0.30$. Find the acceleration.

1. Forces on the crate: the weight $F_g$ downward, the normal force $N$ upward, the pulling force $F = 100$ N forward and the friction $f$ backward.
2. Vertical: $\sum F_y = 0$ gives $N = mg = 20\cdot 9.81 = 196.2\ \mathrm{N}$.
3. Friction: $f = \mu_k N = 0.30\cdot 196.2 \approx 58.9\ \mathrm{N}$.
4. Horizontal: $\sum F_x = F - f = ma$ gives $a = \dfrac{100 - 58.9}{20} \approx 2.1\ \mathrm{m/s^2}$.

The crate accelerates at about 2.1 m/s². Without friction the acceleration would have been $100/20 = 5.0\ \mathrm{m/s^2}$.

Common mistakes

• Mixing up mass and weight: 50 kg is the mass; the weight is $50\cdot 9.81 \approx 491$ N.
• Assuming the normal force is always $mg$. It is smaller on an incline, and it changes when someone pulls at an angle or an elevator accelerates.
• Believing a net force is needed to keep the velocity constant. Constant velocity means the net force is zero – the pulling force only balances friction.
• Letting action and reaction (third law) cancel. They act on different bodies.
• Drawing $ma$ as a separate force in the free-body diagram. $ma$ is the result of the forces, not a force itself.
• Using $\mu_s$ when the body is sliding (then $\mu_k$ applies), or computing friction from $mg$ when $N \ne mg$.

Concepts in this part

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4. Work, energy and power

What is it about?

Energy is the ability to do work. It comes in many forms: energy of motion (kinetic energy), energy of position (potential energy), heat, chemical energy in fuel and electrical energy. The most important principle in all of physics is that energy cannot be created or destroyed – it can only change from one form into another.

For an engineer this is a very powerful tool. With energy methods you can find the speed of an object without knowing every detail of its motion, work out how strong a motor an elevator needs, and find out how much electricity a machine uses. You will use these ideas again in physics, mechanics and thermodynamics.

Key quantities and formulas

• Work $W$ is measured in joules (J), and $1\text{ J} = 1\text{ N}\cdot\text{m}$. When a force $F$ acts while something moves a distance $s$, $W = Fs\cos\theta$, where $\theta$ is the angle between the force and the direction of motion.
• If the force acts straight along the motion, $\cos 0^\circ = 1$ and $W = Fs$. If the force is perpendicular to the motion, the work is zero. If it acts against the motion (like friction), the work is negative.
• Kinetic energy (energy of motion): $E_k = \tfrac12 mv^2$. Twice the speed gives four times the energy.
• Potential energy (energy of position) in the gravitational field: $E_p = mgh$, where $h$ is the height above a reference level you choose yourself, and $g = 9.81\ \mathrm{m/s^2}$.
• Work and kinetic energy: the total work done on an object equals its change in kinetic energy, $W_{\text{net}} = \Delta E_k$.
• Conservation of energy without friction: $E_{k1} + E_{p1} = E_{k2} + E_{p2}$. With friction, part of the energy turns into heat: $E_1 = E_2 + W_f$, where $W_f = F_f\cdot s$ is the work done by friction.
• Power $P$ is work or energy per unit time, measured in watts: $P = W/t$, and $1\text{ W} = 1\text{ J/s}$. At constant speed $v$, $P = Fv$.
• Kilowatt-hour: $1\text{ kWh} = 1000\text{ W}\cdot 3600\text{ s} = 3.6\text{ MJ}$. Electricity bills are measured in kWh.
• Efficiency: $\eta = P_{\text{out}}/P_{\text{in}}$, that is, useful power divided by input power. It is always less than 1 (100 %), because some energy always ends up as heat.

How to solve the problems

1. Write down what is given and convert to SI units: km/h to m/s (divide by 3.6), minutes to seconds, grams to kilograms.
2. Decide what is asked for: work, energy, speed, power or efficiency.
3. If you need a speed or a height, choose a reference level for height and write down the energy at the start and at the end.
4. Set the energy before equal to the energy after, plus any work done by friction, and solve for the unknown.
5. For power: first find the work or energy, then divide by the time. If an efficiency is given, divide the useful power by $\eta$ to get the input power.
6. Check the unit and the order of magnitude.

Example

A construction crane lifts a 500 kg concrete block straight up 12 m in 20 s at constant speed. The motor has an efficiency of 75 %. How much electrical power does the motor draw?

1. The work done on the block is the increase in potential energy: $W = mgh = 500\cdot 9.81\cdot 12 = 58\,860$ J.
2. Useful power: $P_{\text{out}} = W/t = 58\,860/20 = 2943$ W.
3. Input power: $P_{\text{in}} = P_{\text{out}}/\eta = 2943/0.75 \approx 3924$ W.

Answer: The motor draws about 3.9 kW. The rest, almost 1 kW, becomes heat in the motor and gearbox.

Now imagine the block hanging at rest 12 m up when the cable snaps. How fast does it hit the ground? Conservation of energy gives $mgh = \tfrac12 mv^2$, so $v = \sqrt{2gh} = \sqrt{2\cdot 9.81\cdot 12} \approx 15.3$ m/s. Notice that the mass cancels out.

Common mistakes

• Using the length along a slope instead of the vertical height in $mgh$.
• Forgetting the square or the factor $\tfrac12$ in $E_k = \tfrac12 mv^2$.
• Calculating with km/h or minutes instead of m/s and seconds.
• Mixing up work and power: the joule is energy, the watt is energy per second.
• Dividing the wrong way with the efficiency. The input power is always larger than the useful power.
• Believing that heavy objects fall faster. Without air resistance, all objects reach the same speed from the same height.

Concepts in this part

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5. Pressure, density and heat

What is it about?

This unit is about how materials behave: how heavy they are compared with their size (density), how a force is spread over a surface (pressure), why things float (buoyancy), and how much energy it takes to heat, melt or boil something (heat).

Engineers use this all the time: pressure in hydraulics, pipes and tanks, buoyancy in ships and buoys, and heat calculations in everything from engines and heat pumps to electronics that must be cooled. It is the foundation for thermodynamics and fluid mechanics.

Key quantities and formulas

• Density: $\rho = m/V$, measured in kg/m³. Water has $\rho \approx 1000$ kg/m³, which is 1 kg per liter. Remember that $1\text{ L} = 1\text{ dm}^3 = 0.001\text{ m}^3$.
• Pressure is force per unit area: $p = F/A$, measured in pascals, $1\text{ Pa} = 1\text{ N/m}^2$. We often use kPa, MPa or bar, where $1\text{ bar} = 10^5$ Pa. Air pressure at sea level is about 101.3 kPa.
• Pressure in a liquid increases with depth: $p = p_0 + \rho g h$, where $p_0$ is the pressure at the surface. The term $\rho g h$ is called the gauge pressure. In water the pressure rises by about 1 bar per 10 m.
• Buoyancy (Archimedes' principle): an object in a fluid feels an upward force equal to the weight of the fluid it displaces: $F_B = \rho_{\text{fluid}} V g$, where $V$ is the volume below the surface. An object floats if its density is less than that of the fluid.
• Temperature is measured in degrees Celsius (°C) or kelvin (K): $T = t + 273.15$. A change of 1 °C is the same as 1 K, so $\Delta T$ is equal on both scales.
• Heat $Q$ is energy transferred because of a temperature difference, measured in joules. Changing the temperature takes $Q = mc\Delta T$, where $c$ is the specific heat capacity. Water: 4180 J/(kg·K), aluminum: 900, iron: 450, copper: 385.
• Phase changes (melting, boiling): the temperature stays constant while the substance changes phase, and the heat is $Q = mL$. For water the latent heat of fusion is 334 kJ/kg and the latent heat of vaporization is 2257 kJ/kg.
• Heat balance: when something hot and something cold are mixed without heat loss, the heat released by one part equals the heat absorbed by the other.

How to solve the problems

1. Convert to SI units: liters to m³, cm² to m² ($1\text{ cm}^2 = 10^{-4}\text{ m}^2$), kJ to J and grams to kg.
2. Pressure: find the force (often the weight $mg$) and the area, and divide. In liquids, use $\rho g h$, and add $p_0$ if you need the absolute pressure.
3. Buoyancy: find the volume below the surface, and use $F_B = \rho V g$ with the density of the fluid, not of the object.
4. Heat: split the process into steps. A temperature change gives $mc\Delta T$, a phase change gives $mL$. Add up the steps.
5. Heating time: $t = Q/P$, where $P$ is the power delivered to the substance.

Example 1: pressure in a tank

An open water tank is 4.0 m deep. What are the gauge pressure and the absolute pressure at the bottom?

1. Gauge pressure: $\rho g h = 1000\cdot 9.81\cdot 4.0 = 39\,240$ Pa $\approx 39.2$ kPa.
2. Absolute pressure: $p = p_0 + \rho g h = 101.3 + 39.2 = 140.5$ kPa.

Example 2: electric kettle

A 2000 W kettle heats 1.5 kg of water from 15 °C to 100 °C. How long does it take if all the energy goes into the water?

1. $\Delta T = 100 - 15 = 85$ K.
2. $Q = mc\Delta T = 1.5\cdot 4180\cdot 85 = 532\,950$ J $\approx 533$ kJ.
3. $t = Q/P = 532\,950/2000 \approx 266$ s, which is about 4.4 minutes.

To boil the water away as well, you would need another $Q = mL = 1.5\cdot 2257 \approx 3386$ kJ – more than six times the energy used for the heating itself.

Common mistakes

• Forgetting to convert liters and cm² to m³ and m².
• Using the density of the object instead of the fluid in the buoyancy formula.
• Mixing up gauge pressure and absolute pressure.
• Adding 273 to a temperature change. A rise of 20 °C is a rise of 20 K.
• Forgetting the phase change: ice at 0 °C must melt (334 kJ/kg) before its temperature can rise.
• Mixing up kJ and J when $c$ is given in kJ/(kg·K).

Concepts in this part

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6. Electricity

What is it about?

All electricity is based on charge. Matter consists of atoms with positive protons in the nucleus and negative electrons around it. In metals, some of the electrons can move freely. When they are driven through a conductor we get an electric current that can heat a radiator, drive a motor or charge a phone.

Three quantities appear all the time: voltage (how hard the charges are "pushed"), current (how much charge passes per second) and resistance (how much a component opposes the current). The relation between them is Ohm's law, and it is the foundation for everything you will do in electric circuits.

Key quantities and formulas

• Charge $Q$ is measured in coulombs (C). An electron has the charge $-e$, where $e = 1.602\cdot 10^{-19}$ C. Like charges repel each other, opposite charges attract each other.
• Current $I$ is charge per unit time: $I = Q/t$, measured in amperes (A), where 1 A = 1 C/s. The current direction is taken from plus to minus through the circuit outside the source; the electrons actually move the opposite way.
• Voltage $U$ is energy per charge: $U = W/Q$, measured in volts (V), where 1 V = 1 J/C. A 9 V battery gives 9 J of energy to every coulomb that passes. (Many English texts write $V$ for voltage; this course uses $U$.)
• Resistance $R$ is measured in ohms (Ω).
• Ohm's law: $U = RI$. Then also $I = U/R$ and $R = U/I$.
• Power: $P = UI$. Using Ohm's law this also becomes $P = RI^2 = U^2/R$. The energy is $E = Pt$, often measured in kWh.
• Series connection (one after another): the same current flows through all of them, the voltages add up, and $R = R_1 + R_2 + \dots$
• Parallel connection (side by side): the same voltage across all of them, the currents add up, and $1/R = 1/R_1 + 1/R_2 + \dots$ For two resistors, $R = R_1R_2/(R_1+R_2)$. The total resistance is always smaller than the smallest resistor.
• Kirchhoff's laws: the current into a node equals the current out, and around a closed loop the sum of the voltage rises equals the sum of the voltage drops.

How to solve the problems

1. Draw the circuit and mark which resistors are in series and which are in parallel.
2. Combine parallel groups and series groups step by step into a single total resistance.
3. Find the current from the source: $I = U/R_{\text{tot}}$.
4. Work back through the circuit: the voltage across a series resistor is $RI$, and the voltage across a parallel group is the same for all its branches.
5. Find the current in each branch with Ohm's law, and the power with $P = UI$. Check that the currents into and out of every node agree.

Example

A 12 V battery is connected to $R_1 = 4$ Ω in series with a parallel combination of $R_2 = 6$ Ω and $R_3 = 12$ Ω.

1. The parallel group: $R_{23} = \dfrac{6\cdot 12}{6+12} = 4$ Ω.
2. In total: $R_{\text{tot}} = 4 + 4 = 8$ Ω.
3. Current from the battery: $I = 12/8 = 1.5$ A.
4. Voltage across $R_1$: $U_1 = 4\cdot 1.5 = 6$ V. That leaves $12 - 6 = 6$ V across the parallel group.
5. The branch currents: $I_2 = 6/6 = 1$ A and $I_3 = 6/12 = 0.5$ A. The sum is 1.5 A, exactly as it should be.
6. Power delivered by the battery: $P = UI = 12\cdot 1.5 = 18$ W. Check: $R_1$ takes 9 W, $R_2$ takes 6 W and $R_3$ takes 3 W, 18 W in total.

Common mistakes

• Adding parallel resistors as if they were in series.
• Forgetting to take the inverse at the end of $1/R = 1/R_1 + 1/R_2$.
• Putting the full source voltage across a single resistor in a series connection.
• Calculating with mA and kΩ without converting: 1 mA = 0.001 A and 1 kΩ = 1000 Ω. (Kilohms times milliamperes actually gives volts directly.)
• Believing that the current is "used up" in a lamp. It is the energy that is converted; the current is the same going in and coming out.

Concepts in this part

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7. Waves, sound and light

What is it about?

Sound, light, radio signals and vibrations in machines are all waves. The same few relationships describe them all: how fast the wave travels, how long it is and how often it oscillates. Engineers use this in everything from ultrasound measurement and noise calculations to fibre optics.

Concepts and formulas

• The frequency $f$ (Hz) is the number of oscillations per second. The period is the time for one oscillation: $T = 1/f$.
• The wavelength $\lambda$ is the distance between two crests. The wave speed is

• The speed of sound in air is about 343 m/s (at 20 °C), in water about 1480 m/s. The speed of light in vacuum is $c = 3.00\cdot 10^8$ m/s.
• Echo: the sound travels there and back, so the distance is $s = vt/2$.
• Refractive index $n = c/v$. At a boundary, light bends (Snell's law):

• Sound level in decibels: $+3$ dB is double intensity, $+10$ dB is ten times the intensity.

How to solve the problems

1. Write down what you know: $v$, $f$, $\lambda$ or $T$.
2. Use $v = f\lambda$ and $T = 1/f$ and rearrange as needed.
3. Echo: remember to divide by 2. Refraction: the angles are measured from the normal (perpendicular to the surface).

Example

The note A has a frequency of 440 Hz. How long is the sound wave in air?

1. $v = 343$ m/s and $f = 440$ Hz.
2. $\lambda = v/f = 343/440 \approx 0.78$ m.
3. The period is $T = 1/440 \approx 2.3$ ms.

Common mistakes

• Forgetting to divide by 2 for an echo.
• Believing the frequency changes when a wave enters a new material. It is the speed and the wavelength that change.
• Measuring the angle from the surface instead of from the normal in Snell's law.

Concepts in this part

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