Physics: free practice, theory and problems

Anything that moves back and forth around an equilibrium position – a mass on a spring, a pendulum, a bridge in the wind or a vibrating machine – can, to a first approximation, be described as a harmonic oscillator. A wave is an oscillation that travels through space: sound, water waves and waves on a string.

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Oscillations and waves
Electricity and magnetism
Energy and rotation

1. Oscillations and waves

What is it about?

Anything that moves back and forth around an equilibrium position – a mass on a spring, a pendulum, a bridge in the wind or a vibrating machine – can, to a first approximation, be described as a harmonic oscillator. A wave is an oscillation that travels through space: sound, water waves and waves on a string.

This matters to engineers because every structure has natural frequencies. If it is driven at a frequency close to one of them, the amplitude can become dangerously large (resonance).

Key quantities and formulas

Displacement: $x(t) = A\cos(\omega t + \varphi)$ , where $A$ is the amplitude (largest displacement).
Angular frequency $\omega$ (rad/s), frequency $f$ (Hz) and period $T$ (s): $\omega = 2\pi f$ and $T = 1/f$ .
Mass–spring system: $\omega = \sqrt{k/m}$ , so $T = 2\pi\sqrt{m/k}$ .
Simple pendulum (small swings): $T = 2\pi\sqrt{L/g}$ . The period does not depend on the mass.
Maximum speed and acceleration: $v_{max} = A\omega$ (at the equilibrium position) and $a_{max} = A\omega^2$ (at the turning points).
Energy: $E = \tfrac12 kA^2 = \tfrac12 m v_{max}^2$ . The energy alternates between kinetic and potential form.
Waves: $v = f\lambda$ . On a string with tension $F$ and mass per unit length $\mu$ , $v = \sqrt{F/\mu}$ .
Standing waves on a string fixed at both ends: $L = n\lambda/2$ , so $f_n = \dfrac{nv}{2L}$ with $n = 1, 2, 3, \dots$

How to solve the problems

Identify what oscillates and list the known quantities in SI units (cm → m, g → kg).
Find $\omega$ from the system ( $\sqrt{k/m}$ or $\sqrt{g/L}$ ).
Continue to $f = \omega/(2\pi)$ and $T = 1/f$ , or to $v_{max}$ , $a_{max}$ and the energy.
For waves: find the wave speed first, then $\lambda$ or $f$ from $v = f\lambda$ .
Check the order of magnitude: a small block on a spring typically oscillates a few times per second.

Example

A 2 kg block hangs from a spring with $k = 800$ N/m and oscillates with an amplitude of 10 cm. Find the period, the maximum speed and the energy.

$\omega = \sqrt{k/m} = \sqrt{800/2} = 20$ rad/s.
$T = 2\pi/\omega = 2\pi/20 \approx 0.314$ s, and $f = 1/T \approx 3.18$ Hz.
$v_{max} = A\omega = 0.10\cdot 20 = 2.0$ m/s.
$E = \tfrac12 kA^2 = \tfrac12\cdot 800\cdot 0.10^2 = 4.0$ J. Check: $\tfrac12 mv_{max}^2 = \tfrac12\cdot 2\cdot 2^2 = 4$ J.

Answer: $T \approx 0.314$ s, $v_{max} = 2.0$ m/s and $E = 4.0$ J.

Common mistakes

Mixing up $\omega$ (rad/s) and $f$ (Hz). Remember the factor $2\pi$ .
Forgetting to convert cm to m before computing the energy.
Believing that the period of a pendulum depends on its mass or amplitude (for small swings it does not).
Using $L = n\lambda$ instead of $L = n\lambda/2$ for standing waves.

Mass–spring:

\omega = \sqrt{k/m}

. Pendulum:

T = 2\pi\sqrt{L/g}

. Waves:

v = f\lambda

. The speed is greatest at the equilibrium position.

Concepts in this part

Harmonic oscillationA harmonic oscillator moves back and forth around an equilibrium position with a displacement that follows a c…Mass–spring and pendulumTwo common systems oscillate harmonically: a mass attached to a spring, and a pendulum swinging with small amp…Energy in oscillationsIn a harmonic oscillation the energy alternates between kinetic and potential form, while the total energy sta…Waves on a stringA wave is an oscillation that travels through space. On a string, the wave speed is set by the tension and the…

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2. Electricity and magnetism

What is it about?

Electric charges exert forces on each other, and moving charges (currents) create magnetic fields. A changing magnetic field in turn creates an electric voltage. These three ideas are the basis of electric motors, generators, transformers, sensors and the entire power grid.

The key is the field concept: a charge (or a current) creates a field in the space around it, and the field exerts a force on other charges (or currents) located there.

Key quantities and formulas

Charge $q$ is measured in coulombs (C). The elementary charge is $e = 1.602\cdot10^{-19}$ C.
Coulomb's law: $F = k\dfrac{q_1q_2}{r^2}$ with $k = 8.99\cdot10^9$ N·m²/C². Like charges repel, unlike charges attract.
Electric field: $E = F/q$ (N/C). From a point charge: $E = kq/r^2$ . Between two parallel plates: $E = U/d$ .
Potential (voltage): $U = W/q$ in volts. A charge that passes through the voltage $U$ gains the energy $qU$ .
Magnetic force on a charge: $F = qvB\sin\theta$ , perpendicular to both $\vec v$ and $\vec B$ .
Force on a current-carrying conductor: $F = BIL\sin\theta$ .
Field around a long, straight conductor: $B = \dfrac{\mu_0 I}{2\pi r}$ with $\mu_0 = 4\pi\cdot10^{-7}$ T·m/A.
Magnetic flux: $\Phi = BA\cos\theta$ (Wb).
Faraday's law: $\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}$ . The minus sign is Lenz's law: the induced current opposes the change.

How to solve the problems

Convert everything to SI: µC → $10^{-6}$ C, cm → m, cm² → $10^{-4}$ m².
Choose the right law: force between charges (Coulomb), force in a magnetic field ( $qvB$ or $BIL$ ) or induction (Faraday).
Compute the magnitude first, then determine the direction (sign, right-hand rule, Lenz).
For induction: find $\Delta\Phi$ , divide by $\Delta t$ and multiply by $N$ . The current is $I = \varepsilon/R$ .

Example

A coil has 200 turns and a cross-sectional area of 0.01 m². It is perpendicular to a magnetic field that increases uniformly from 0 to 0.5 T in 0.1 s. The coil has a resistance of 4 Ω. Find the induced voltage and current.

Change in flux per turn: $\Delta\Phi = A\,\Delta B = 0.01\cdot 0.5 = 0.005$ Wb.
Induced voltage: $\varepsilon = N\Delta\Phi/\Delta t = 200\cdot 0.005/0.1 = 10$ V.
Current: $I = \varepsilon/R = 10/4 = 2.5$ A.

Answer: 10 V and 2.5 A. The current flows so that its own magnetic field opposes the increase of the external field (Lenz).

Common mistakes

Forgetting to square the distance in Coulomb's law.
Putting µC or cm straight into the formulas without converting to SI.
Believing that a strong but constant magnetic field induces a voltage. It is the change in flux that counts.
Believing that the magnetic force does work on a charge. It is perpendicular to the velocity and only changes the direction.

Electric:

F = kq_1q_2/r^2

and

E = F/q

. Magnetic:

F = qvB

and

F = BIL

. Induction:

\varepsilon = -N\,\Delta\Phi/\Delta t

– only a change gives a voltage.

Concepts in this part

Coulomb's lawCharged particles exert an electric force on each other that falls off with the square of the distance. Like c…Electric field and potentialA charge creates an electric field in the space around it, and the field exerts a force on other charges locat…Magnetic forceA charge moving in a magnetic field feels a force perpendicular to both its velocity and the field, given by t…Induction (Faraday's law)Magnetic flux is a measure of how much magnetic field passes through a surface. Only a change in the flux indu…

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3. Energy and rotation

What is it about?

Energy is the most useful "currency" in physics: it cannot be created or destroyed, only converted from one form to another. With conservation of energy you can often find a speed without knowing anything about the forces and times along the way.

For rotating machines – wheels, flywheels, turbines and motors – there is a parallel world of quantities: angular velocity, moment of inertia, torque and angular momentum. Once you see the parallels to straight-line motion ( $m \leftrightarrow I$ , $v \leftrightarrow \omega$ , $F \leftrightarrow \tau$ ), rotation becomes easy.

Key quantities and formulas

Work: $W = Fs\cos\theta$ (J). Only the force component along the motion does work.
Kinetic energy $E_k = \tfrac12mv^2$ and potential energy $E_p = mgh$ .
Conservation of energy without friction: $E_k + E_p$ is constant. With friction, some of it turns into heat.
Power: $P = W/t = Fv$ (W).
Momentum $p = mv$ . It is conserved in collisions when external forces can be neglected. In a perfectly inelastic collision the bodies stick together, and kinetic energy is lost.
Rotation: angular velocity $\omega$ (rad/s), angular acceleration $\alpha$ (rad/s²), $v = \omega r$ and $\omega = 2\pi n/60$ when $n$ is in revolutions per minute.
Moment of inertia $I$ : point mass $mr^2$ , thin hoop $MR^2$ , solid cylinder $\tfrac12MR^2$ , solid sphere $\tfrac25MR^2$ .
Newton's second law for rotation: $\tau = I\alpha$ . Rotational kinetic energy: $\tfrac12I\omega^2$ . Angular momentum: $L = I\omega$ , conserved when there is no external torque.
Rolling without slipping: $E_k = \tfrac12mv^2 + \tfrac12I\omega^2$ with $\omega = v/R$ .

How to solve the problems

Decide whether the problem is about energy (speeds and heights), momentum (collisions) or rotation (torque and angular momentum).
Write down the energy or momentum before and after.
Set before equal to after (plus any losses) and solve for the unknown.
For rolling: remember that part of the energy goes into rotation.

Example

A solid cylinder is released from rest and rolls without slipping down an inclined plane. The height difference is 1.5 m. What is its speed at the bottom?

Conservation of energy: $mgh = \tfrac12mv^2 + \tfrac12I\omega^2$ .
With $I = \tfrac12mR^2$ and $\omega = v/R$ the rotational term becomes $\tfrac14mv^2$ , so $mgh = \tfrac34mv^2$ .
$v = \sqrt{4gh/3} = \sqrt{4\cdot 9.81\cdot 1.5/3} \approx 4.43$ m/s.

Answer: 4.43 m/s. A block sliding without friction would reach $\sqrt{2gh} \approx 5.42$ m/s. The difference is the energy stored in the rotation.

Common mistakes

Computing work for a force that is perpendicular to the motion (the work is zero).
Using conservation of kinetic energy in an inelastic collision. Use momentum there.
Forgetting the rotational energy when something rolls.
Using revolutions per minute directly instead of rad/s.

Energy before = energy after (+ losses). In collisions momentum is conserved. Rotation:

\tau = I\alpha

E = \tfrac12I\omega^2

and

L = I\omega

Concepts in this part

Work and energy conservationThe total work done on a body equals the change in its kinetic energy. Without friction, the sum of kinetic an…Momentum and collisionsThe momentum of a body is its mass times its velocity, and it is conserved in a collision when external forces…Rotation and moment of inertiaThe moment of inertia is rotation's counterpart to mass: it tells you how hard it is to change a body's angula…Torque and angular momentumNewton's second law has a rotational counterpart: torque gives angular acceleration, just as force gives accel…Rolling without slippingWhen something rolls without slipping, the kinetic energy is split between translation and rotation, and the a…

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Example problems with solutions

Here are some of the problems in physics. In the app, calculation problems get new numbers every time, so you can practise until it sticks – and take a graded practice exam before the real one.

Oscillations and waves: A spring with

k = 200

N/m carries a mass of 0.5 kg. What is the angular frequency?

Answer: 20 rad/s

$\omega = \sqrt{k/m} = \sqrt{400} = 20$ rad/s.

Electricity and magnetism: Two point charges of 1 µC are 1 m apart. What is the force between them, in mN? (

k = 8.99\cdot 10^9

)

Answer: 8.99 mN

$F = kq_1q_2/r^2 = 8.99\cdot10^9\cdot 10^{-12} = 8.99\cdot 10^{-3}$ N.

Energy and rotation: How much work is needed to lift 10 kg by 3 m? (

g = 9.81

)

Answer: 294.3 J

$W = mgh = 10\cdot 9.81\cdot 3 = 294.3$ J.

Oscillations and waves: What is the period of a simple pendulum (small swings)?

Answer: $T = 2\pi\sqrt{L/g}$

The period does not depend on the mass.

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Matches these university courses

The content covers the syllabus found in engineering degrees, for example:

MEK1400 (OsloMet)
TFY4104 (NTNU)
FYS101 (NMBU)

Physics: free practice, theory and problems

Contents

1. Oscillations and waves

What is it about?

Key quantities and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

2. Electricity and magnetism

What is it about?

Key quantities and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

3. Energy and rotation

What is it about?

Key quantities and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

Example problems with solutions

Matches these university courses