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overall transfer."},{"text":"A wave transfers matter overall without transferring energy.","correct":false,"explanation":"This reverses the key idea: waves transfer energy, but particles do not move with the wave overall. Matter may oscillate locally, but it is not transported overall."},{"text":"A wave transfers energy and matter overall without transferring nothing.","correct":false,"explanation":"The first part incorrectly says matter is transferred overall, a common confusion between particle motion and wave travel. The correct distinction is energy transfer without overall matter transfer."},{"text":"A wave transfers matter overall without transferring anything.","correct":false,"explanation":"A wave must transfer energy, so claiming that nothing is transferred misses the purpose of wave motion. The medium’s particles vibrate locally rather than carrying matter overall."}]},{"id":"93676096-5d4a-4132-9e50-a7e725a4cde1","cardType":"basic","points":1,"front":"In a wave through a medium, what do the particles do overall?","back":"They oscillate about rest positions; they do not travel with the wave overall.","mcqOptions":[{"text":"They oscillate about rest positions; they do not travel with the wave overall.","correct":true,"explanation":"Particles in the medium vibrate around their equilibrium or rest positions. The disturbance and its energy move through the medium, but the particles do not travel with it overall."},{"text":"They travel with the wave continuously; they do not return to rest positions.","correct":false,"explanation":"This confuses particle motion with wave motion. Particles oscillate around rest positions, repeatedly returning towards them, while the wave disturbance travels through the medium."},{"text":"They remain permanently at rest; only the energy moves.","correct":false,"explanation":"Although energy moves through the medium, its particles must oscillate to pass the disturbance on. They are not permanently stationary, even though they do not travel overall."},{"text":"They move from the source to the receiver; they do not oscillate locally.","correct":false,"explanation":"Particles generally move back and forth or up and down about fixed positions rather than travelling from source to receiver. The wave transfers energy across the medium."}]},{"id":"8d3da579-1ff5-406c-80f9-689f50f27cf5","cardType":"basic","points":1,"front":"What does sound need to travel through?","back":"A medium, such as air.","mcqOptions":[{"text":"A medium, such as air.","correct":true,"explanation":"Sound is a mechanical wave, so it requires particles to vibrate and pass the disturbance along. It can travel through gases, liquids and solids, but not through a vacuum."},{"text":"A vacuum, such as outer space.","correct":false,"explanation":"A vacuum contains no particles to vibrate and transfer the sound disturbance. The correct answer is a medium, such as air, although sound can also travel through liquids and solids."},{"text":"Only a solid, such as metal.","correct":false,"explanation":"Sound can travel through solids, but it is not restricted to them. It also travels through liquids and gases, so the broader requirement is a medium."},{"text":"An electromagnetic field, such as light.","correct":false,"explanation":"Sound is mechanical and needs vibrating particles, unlike electromagnetic waves which can travel through a vacuum. The required substance is a medium, including air."}]},{"id":"fc587b23-ae66-4cee-aa47-907f0dcddc4b","cardType":"basic","points":1,"front":"Which waves can travel through a vacuum?","back":"Electromagnetic waves.","mcqOptions":[{"text":"Electromagnetic waves.","correct":true,"explanation":"Electromagnetic waves consist of changing electric and magnetic fields, so they do not require particles in a medium. They can therefore travel through a vacuum."},{"text":"Mechanical waves.","correct":false,"explanation":"Mechanical waves require particles in a medium to oscillate, so they cannot travel through a vacuum. Electromagnetic waves are the type that can propagate without a medium."},{"text":"Sound waves.","correct":false,"explanation":"Sound is a mechanical wave and needs particles in a medium to pass on its vibrations. A vacuum has no particles, so electromagnetic waves are the correct choice."},{"text":"Water waves.","correct":false,"explanation":"Water waves depend on water particles and therefore need a material medium. The waves able to travel through a vacuum are electromagnetic waves."}]},{"id":"76261c5a-0723-44df-8f0d-1c79b6c45a3e","cardType":"cloze","points":null,"front":"In a transverse wave, vibrations are [ ] to energy transfer; in a longitudinal wave, they are [ ].","back":"In a transverse wave, vibrations are perpendicular to energy transfer; in a longitudinal wave, they are parallel.","mcqOptions":[{"text":"In a transverse wave, vibrations are perpendicular to energy transfer; in a longitudinal wave, they are parallel.","correct":true,"explanation":"Transverse vibrations are at right angles to the direction of energy transfer, whereas longitudinal vibrations are along the same direction as the energy transfer."},{"text":"In a transverse wave, vibrations are parallel to energy transfer; in a longitudinal wave, they are perpendicular.","correct":false,"explanation":"The two definitions have been reversed. Transverse vibrations are perpendicular to energy transfer, while longitudinal vibrations are parallel to it."},{"text":"In a transverse wave, vibrations are parallel to energy transfer; in a longitudinal wave, they are parallel.","correct":false,"explanation":"The transverse description is incorrect because its vibrations are at right angles to energy transfer. Only the longitudinal description uses parallel motion."},{"text":"In a transverse wave, vibrations are perpendicular to energy transfer; in a longitudinal wave, they are perpendicular.","correct":false,"explanation":"The transverse description is correct, but longitudinal vibrations are not at right angles to energy transfer. They occur parallel to the direction in which energy travels."}]},{"id":"4bcdb59e-6dba-40c3-b020-cd2bd6cd4580","cardType":"basic","points":1,"front":"A wave travels left to right and particles move up and down. What type is it?","back":"Transverse.","mcqOptions":[{"text":"Transverse.","correct":true,"explanation":"The particles move up and down while the wave travels horizontally, so the vibrations are perpendicular to the direction of energy transfer. This makes the wave transverse."},{"text":"Longitudinal.","correct":false,"explanation":"Longitudinal waves have particle vibrations parallel to the direction of travel. Here the particles move vertically while the wave travels horizontally, so the wave is transverse."},{"text":"Surface.","correct":false,"explanation":"Surface describes waves that travel along a boundary, not the vibration direction shown here. Perpendicular particle motion relative to travel identifies this as a transverse wave."},{"text":"Electromagnetic.","correct":false,"explanation":"Electromagnetic describes the nature of the wave and does not follow simply from the stated particle motion. The perpendicular motion makes the wave transverse."}]},{"id":"f2ff0d1a-c253-45a7-b02c-b7dbd9acf706","cardType":"basic","points":1,"front":"A wave travels left to right and particles move left and right. What type is it?","back":"Longitudinal.","mcqOptions":[{"text":"Longitudinal.","correct":true,"explanation":"The particles move back and forth in the same direction as the wave travels. Vibrations parallel to energy transfer are the defining feature of a longitudinal wave."},{"text":"Transverse.","correct":false,"explanation":"Transverse waves have vibrations perpendicular to the direction of travel. Since the particles here move in the same left-right direction as the wave, it is longitudinal."},{"text":"Surface.","correct":false,"explanation":"Surface refers to propagation along a boundary and does not identify the particle direction in this description. Parallel particle motion indicates a longitudinal wave."},{"text":"Electromagnetic.","correct":false,"explanation":"Electromagnetic is a classification based on electric and magnetic fields, not simply the stated particle motion. Motion parallel to travel identifies a longitudinal wave."}]},{"id":"ed0722be-3e6d-4c9d-ab56-3152f6de1eff","cardType":"cloze","points":null,"front":"In a longitudinal wave, particles close together form [ ]; particles spread apart form [ ].","back":"In a longitudinal wave, particles close together form compressions; particles spread apart form rarefactions.","mcqOptions":[{"text":"In a longitudinal wave, particles close together form compressions; particles spread apart form rarefactions.","correct":true,"explanation":"Compressions are regions where particles are crowded together, while rarefactions are regions where particles are spread apart. These alternating regions characterise longitudinal waves."},{"text":"In a longitudinal wave, particles close together form rarefactions; particles spread apart form compressions.","correct":false,"explanation":"The two terms have been swapped. Crowded particles produce compressions, while particles that are farther apart produce rarefactions."},{"text":"In a longitudinal wave, particles close together form compressions; particles spread apart form transverse vibrations.","correct":false,"explanation":"The first term correctly describes crowded particles, but the second should name a low-density region. Spread-out particles form rarefactions, not a type of vibration."},{"text":"In a longitudinal wave, particles close together form crest; particles spread apart form troughs.","correct":false,"explanation":"Crests and troughs describe features of transverse waves. Longitudinal waves instead contain crowded regions called compressions and spread-out regions called rarefactions."}]},{"id":"5833d7ef-8e15-4b36-8acd-15fb16d2e9c8","cardType":"basic","points":1,"front":"On a displacement diagram, what distance is the amplitude?","back":"Maximum displacement from the rest position.","mcqOptions":[{"text":"Maximum displacement from the rest position.","correct":true,"explanation":"Amplitude is the greatest displacement of a point on the wave from its equilibrium or rest position, measured vertically on a displacement diagram."},{"text":"Total displacement from crest to trough.","correct":false,"explanation":"The total displacement from crest to trough equals twice the amplitude. Amplitude is the maximum displacement from the rest position to either extreme."},{"text":"Horizontal distance from crest to crest.","correct":false,"explanation":"Horizontal distance from one crest to the next represents wavelength, not amplitude. Amplitude is the maximum displacement from the rest position."},{"text":"Maximum displacement from crest to trough.","correct":false,"explanation":"Displacement from crest to trough covers both extremes and is twice the amplitude. Amplitude is measured from the rest position to one extreme."}]},{"id":"72e6ea24-b48b-40f4-a41c-a76568a8f3fe","cardType":"basic","points":1,"front":"How is wavelength measured on a transverse wave diagram?","back":"From one crest to the next crest.","mcqOptions":[{"text":"From one crest to the next crest.","correct":true,"explanation":"A wavelength is the distance between successive points in phase. Two adjacent crests are in phase, so their separation is one complete wavelength."},{"text":"From one crest to the next trough.","correct":false,"explanation":"A crest and the next trough are half a cycle apart, so this distance is half a wavelength. A full wavelength runs between adjacent crests."},{"text":"From one rest position to the next crest.","correct":false,"explanation":"The rest position to the next crest spans only one quarter of a cycle. Wavelength is measured between corresponding points, such as adjacent crests."},{"text":"From one crest to the second-next crest.","correct":false,"explanation":"A crest to the second following crest covers two complete cycles, not one. Wavelength is the distance from one crest to the next crest."}]},{"id":"59494b84-9ea8-4d99-bd17-6f9d5d71e654","cardType":"basic","points":1,"front":"How is wavelength usually measured in a longitudinal wave?","back":"From compression to compression.","mcqOptions":[{"text":"From compression to compression.","correct":true,"explanation":"One wavelength is the distance between consecutive points in the same phase. The centres of two adjacent compressions are corresponding points in a longitudinal wave."},{"text":"From compression to rarefaction.","correct":false,"explanation":"A compression and the adjacent rarefaction are half a cycle apart, giving half a wavelength. A full wavelength runs from one compression to the next."},{"text":"From one edge of a compression to the other.","correct":false,"explanation":"The width of one compressed region is not a complete repeating cycle. Wavelength is measured between corresponding points in adjacent compressions."},{"text":"From particle to adjacent particle.","correct":false,"explanation":"Adjacent particles are not necessarily one complete cycle apart. Wavelength is normally found from the centre of one compression to the next compression."}]},{"id":"4ce5b445-d325-49da-8174-d84a8540e732","cardType":"cloze","points":null,"front":"Frequency is the number of complete waves passing a point each second, measured in [ ].","back":"Frequency is the number of complete waves passing a point each second, measured in Hz.","mcqOptions":[{"text":"Hz","correct":true,"explanation":"Frequency counts complete cycles per second, and the SI unit for cycles per second is the hertz, written Hz."},{"text":"s","correct":false,"explanation":"Seconds measure time and are used for the time period, not frequency. Frequency is measured in hertz, Hz, meaning complete cycles per second."},{"text":"m","correct":false,"explanation":"Metres measure distances such as wavelength, not the rate at which waves pass a point. Frequency is measured in hertz, Hz."},{"text":"m/s","correct":false,"explanation":"Metres per second measure wave speed rather than cycles passing each second. The correct unit of frequency is hertz, written Hz."}]},{"id":"ce0d5c4b-9104-4e0b-bd8e-087b24a79e93","cardType":"basic","points":1,"front":"What does the time period of a wave measure?","back":"Time for one complete cycle.","mcqOptions":[{"text":"Time for one complete cycle.","correct":true,"explanation":"The time period is the time taken for one full oscillation or wave cycle to occur, usually measured in seconds."},{"text":"Time for half a complete cycle.","correct":false,"explanation":"Half a cycle accounts for only one half-oscillation, not the period. The time period is the time taken for one complete cycle."},{"text":"Time for two complete cycles.","correct":false,"explanation":"Counting two cycles would give twice the time period. The period is defined as the time taken for exactly one complete cycle."},{"text":"Time for all cycles in one second.","correct":false,"explanation":"The number of cycles completed each second describes frequency, not time period. Time period measures the time taken for one complete cycle."}]},{"id":"63f78464-6d0d-4ab2-af7c-21feff5e9dab","cardType":"basic","points":1,"front":"What is the relationship between frequency and time period?","back":"$$f = \\frac{1}{T}$","mcqOptions":[{"text":"$$f = \\frac{1}{T}$","correct":true,"explanation":"Frequency and time period are reciprocals: if one cycle takes $T$ seconds, the number of cycles completed each second is $f = \\frac{1}{T}$."},{"text":"$$f = T$","correct":false,"explanation":"Equating frequency with period ignores their reciprocal relationship and gives incompatible units. The correct relationship is $f = \\frac{1}{T}$."},{"text":"$$f = \\frac{1}{T^2}$","correct":false,"explanation":"Squaring the period is an unnecessary operation and produces incorrect units. Frequency is simply the reciprocal of the time period: $f = \\frac{1}{T}$."},{"text":"$$f = 2T$","correct":false,"explanation":"Multiplying the period by two does not convert time per cycle into cycles per second. Frequency is found using $f = \\frac{1}{T}$."}]},{"id":"70454810-fbed-4faa-b2f7-06f670a08f12","cardType":"basic","points":1,"front":"If a wave’s time period gets shorter, what happens to its frequency?","back":"The frequency gets higher.","mcqOptions":[{"text":"The frequency gets higher.","correct":true,"explanation":"Frequency and time period are inversely proportional through $f = \\frac{1}{T}$. A shorter period therefore means that more cycles occur each second."},{"text":"The frequency gets lower.","correct":false,"explanation":"This reverses the inverse relationship between frequency and period. Since $f = \\frac{1}{T}$, reducing the period makes the frequency higher, not lower."},{"text":"The frequency stays unchanged.","correct":false,"explanation":"Frequency remains unchanged only when the time period remains unchanged. Because frequency is the reciprocal of period, a shorter period produces a higher frequency."},{"text":"The frequency gets doubled.","correct":false,"explanation":"A shorter period does increase frequency, but it doubles only if the period is exactly halved. The definite conclusion is that frequency gets higher."}]},{"id":"d4f44fa9-b04e-4a40-8aff-b8f7200bb500","cardType":"basic","points":2,"front":"Calculate frequency if $T = 0.0040\\ \\text{s}$.","back":"$$250\\ \\text{Hz}$","mcqOptions":[{"text":"$$250\\ \\text{Hz}$","correct":true,"explanation":"Using $f = \\frac{1}{T}$ gives $f = \\frac{1}{0.0040} = 250\\ \\text{Hz}$. The unit is hertz because frequency counts cycles per second."},{"text":"$$25\\ \\text{Hz}$","correct":false,"explanation":"This results from a decimal-place error when taking the reciprocal. Calculating $1 \\div 0.0040$ correctly gives $250\\ \\text{Hz}$."},{"text":"$$0.0040\\ \\text{Hz}$","correct":false,"explanation":"This treats the period’s numerical value as though it were already the frequency. Frequency is the reciprocal, so $1 \\div 0.0040 = 250\\ \\text{Hz}$."},{"text":"$$4000\\ \\text{Hz}$","correct":false,"explanation":"This incorrectly removes the decimal places without taking the reciprocal accurately. Applying $f = \\frac{1}{T}$ gives $1 \\div 0.0040 = 250\\ \\text{Hz}$."}]},{"id":"585cdf3c-7031-4572-8a3d-853cc2f5d9b6","cardType":"cloze","points":null,"front":"Wave speed is given by [ ].","back":"Wave speed is given by $v = f \\times \\lambda$.","mcqOptions":[{"text":"$$v = f \\times \\lambda$","correct":true,"explanation":"Wave speed equals frequency multiplied by wavelength. Frequency is measured in hertz and wavelength in metres, giving wave speed in metres per second."},{"text":"$$v = \\frac{f}{\\lambda}$","correct":false,"explanation":"Dividing frequency by wavelength confuses the wave equation’s relationship. The correct equation is $v = f \\times \\lambda$, because each cycle advances the wave by one wavelength."},{"text":"$$v = \\frac{\\lambda}{f}$","correct":false,"explanation":"Dividing wavelength by frequency introduces an extra factor of time and gives the wrong dimensions. Wave speed is calculated using $v = f \\times \\lambda$."},{"text":"$$v = f + \\lambda$","correct":false,"explanation":"Adding frequency and wavelength combines quantities with incompatible units. The correct relationship multiplies them: $v = f \\times \\lambda$."}]},{"id":"d4b2f7df-5834-40e9-89ea-09579f7ab5f1","cardType":"basic","points":2,"front":"Calculate wave speed for $f = 50\\ \\text{Hz}$ and $\\lambda = 0.80\\ \\text{m}$.","back":"$$40\\ \\text{m/s}$","mcqOptions":[{"text":"$$40\\ \\text{m/s}$","correct":true,"explanation":"Using $v = f\\lambda$ gives $v = 50 \\times 0.80 = 40\\ \\text{m/s}$. Hertz multiplied by metres gives metres per second."},{"text":"$$62.5\\ \\text{m/s}$","correct":false,"explanation":"This results from dividing $50$ by $0.80$ instead of multiplying. Wave speed is $v = f\\lambda = 50 \\times 0.80 = 40\\ \\text{m/s}$."},{"text":"$$400\\ \\text{m/s}$","correct":false,"explanation":"This treats $0.80$ as $8.0$, making the answer ten times too large. The correct calculation is $50 \\times 0.80 = 40\\ \\text{m/s}$."},{"text":"$$4.0\\ \\text{m/s}$","correct":false,"explanation":"This treats $0.80$ as $0.08$, making the result ten times too small. Multiplying the stated values gives $50 \\times 0.80 = 40\\ \\text{m/s}$."}]},{"id":"b3849b46-95ba-4e56-be6a-b7217ba2d17c","cardType":"basic","points":1,"front":"Before using $v = f \\times \\lambda$, what must kHz be converted to?","back":"Hz.","mcqOptions":[{"text":"Hz.","correct":true,"explanation":"The standard frequency unit used in the wave equation is hertz. Multiply a value in kilohertz by $1000$ to convert it to Hz."},{"text":"MHz.","correct":false,"explanation":"Megahertz is a larger unit, not the required standard unit. Before substitution, kHz should be converted to Hz by multiplying by $1000$."},{"text":"seconds.","correct":false,"explanation":"Seconds measure time rather than frequency. A frequency given in kHz must be converted to Hz before using $v = f \\times \\lambda$."},{"text":"metres.","correct":false,"explanation":"Metres measure wavelength, not frequency. Kilohertz must be converted to hertz, with $1\\ \\text{kHz} = 1000\\ \\text{Hz}$."}]},{"id":"746aba9b-d88c-4d8b-872d-9ce6a231e0b5","cardType":"cloze","points":null,"front":"A source moving towards an observer gives a [ ] and a [ ].","back":"A source moving towards an observer gives a shorter observed wavelength and a higher observed frequency.","mcqOptions":[{"text":"A source moving towards an observer gives a shorter observed wavelength and a higher observed frequency.","correct":true,"explanation":"As the source approaches, successive wavefronts are emitted closer together. The observer therefore detects a shorter wavelength and more wavefronts each second, producing a higher frequency."},{"text":"A source moving towards an observer gives a longer observed wavelength and a higher observed frequency.","correct":false,"explanation":"Approaching motion compresses the wavefronts rather than spreading them out. The observed wavelength becomes shorter, while the observed frequency becomes higher."},{"text":"A source moving towards an observer gives a shorter observed wavelength and a lower observed frequency.","correct":false,"explanation":"A shorter wavelength from an approaching source corresponds to a higher, not lower, observed frequency because wavefronts reach the observer more frequently."},{"text":"A source moving towards an observer gives a longer observed wavelength and a lower observed frequency.","correct":false,"explanation":"Longer wavelength and lower frequency describe a source moving away. A source moving towards the observer produces a shorter observed wavelength and a higher observed frequency."}]},{"id":"2e2fd0a6-926f-4ede-9753-64ab1a5257ba","cardType":"basic","points":1,"front":"For sound, what pitch is heard when the source moves away?","back":"A lower pitch.","mcqOptions":[{"text":"A lower pitch.","correct":true,"explanation":"A receding source produces wavefronts that are more widely spaced at the observer. The resulting lower observed frequency is heard as a lower pitch."},{"text":"A higher pitch.","correct":false,"explanation":"A higher pitch is heard when the source moves towards the observer. Moving away reduces the observed frequency, so the sound has a lower pitch."},{"text":"An unchanged pitch.","correct":false,"explanation":"The pitch would remain unchanged only with no relative radial motion. A source moving away causes a lower observed frequency and therefore a lower pitch."},{"text":"An alternating pitch.","correct":false,"explanation":"Steady motion away causes a consistently reduced observed frequency, not alternating changes. The observer therefore hears a lower pitch."}]},{"id":"dfff776d-23df-497d-98f7-46920418545d","cardType":"basic","points":1,"front":"In the Doppler effect, which frequency changes: source or observed?","back":"The observed frequency changes.","mcqOptions":[{"text":"The observed frequency changes.","correct":true,"explanation":"The source continues emitting at its own fixed frequency, but relative motion changes how frequently wavefronts reach the observer. Therefore, the observed frequency changes."},{"text":"The source frequency changes.","correct":false,"explanation":"The source’s emission frequency remains fixed unless the source itself changes its oscillation. In the Doppler effect, relative motion changes the observed frequency."},{"text":"Both source and observed frequencies change.","correct":false,"explanation":"Relative motion does not alter the source’s actual rate of oscillation. It changes only the frequency measured by the observer, not both frequencies."},{"text":"Neither source nor observed frequency changes.","correct":false,"explanation":"The Doppler effect specifically changes the frequency detected because wavefronts arrive at a different rate. The source frequency remains fixed while the observed frequency changes."}]},{"id":"032110e2-e289-4eed-a176-b6a306225ee0","cardType":"basic","points":1,"front":"What happens in reflection at a boundary?","back":"The wave bounces off the boundary.","mcqOptions":[{"text":"The wave bounces off the boundary.","correct":true,"explanation":"Reflection occurs when a wave reaches a boundary and returns into the original medium. In simple terms, the wave bounces off the boundary."},{"text":"The wave passes through the boundary.","correct":false,"explanation":"Passing through a boundary describes transmission, not reflection. During reflection, the wave bounces off the boundary and returns into its original medium."},{"text":"The wave is absorbed by the boundary.","correct":false,"explanation":"Being taken in by a material describes absorption. Reflection instead occurs when the wave bounces off the boundary."},{"text":"The wave refracts through the boundary.","correct":false,"explanation":"Changing direction while crossing into another medium describes refraction. In reflection, the wave does not cross the boundary but bounces off it."}]},{"id":"515da45d-88ce-4ac7-8754-a784f498ffee","cardType":"cloze","points":null,"front":"For reflection, the angle of incidence equals the [ ], measured from the [ ].","back":"For reflection, the angle of incidence equals the angle of reflection, measured from the normal.","mcqOptions":[{"text":"For reflection, the angle of incidence equals the angle of reflection, measured from the normal.","correct":true,"explanation":"The law of reflection states that the angle of incidence equals the angle of reflection. Both angles are measured from the normal, which is perpendicular to the surface."},{"text":"For reflection, the angle of incidence equals the angle of refraction, measured from the normal.","correct":false,"explanation":"The angle of refraction applies when a wave enters another medium. For reflection, incidence equals the angle of reflection, with both measured from the normal."},{"text":"For reflection, the angle of incidence equals the angle of reflection, measured from the surface.","correct":false,"explanation":"Although incidence equals reflection, these angles are measured from the normal rather than the surface. Measuring from the surface gives complementary angles instead."},{"text":"For reflection, the angle of incidence equals the angle of refraction, measured from the surface.","correct":false,"explanation":"This confuses reflection with refraction and uses the wrong reference line. The angle of incidence equals the angle of reflection, measured from the normal."}]},{"id":"07ccaa32-22d2-47b1-9ab2-94f8827a66ad","cardType":"basic","points":1,"front":"What causes refraction when a wave enters a different medium?","back":"A change in wave speed.","mcqOptions":[{"text":"A change in wave speed.","correct":true,"explanation":"Refraction occurs because a wave travels at a different speed in the new medium. This speed change can alter its wavelength and, unless it enters normally, its direction."},{"text":"A change in wave frequency.","correct":false,"explanation":"The source determines the wave’s frequency, so crossing a boundary does not change it. Refraction is caused by a change in wave speed in the new medium."},{"text":"A change in wave amplitude.","correct":false,"explanation":"Amplitude may change because some energy is reflected or absorbed, but that does not cause refraction. Refraction results from a change in wave speed between media."},{"text":"A change in wave energy.","correct":false,"explanation":"The wave’s energy may be partly transferred or reflected at a boundary, but this is not the cause of refraction. The cause is a change in wave speed."}]},{"id":"78d22ed7-ffd5-4a66-9c7e-bea20eb6d231","cardType":"basic","points":1,"front":"During refraction, what stays the same because it is set by the source?","back":"Frequency.","mcqOptions":[{"text":"Frequency.","correct":true,"explanation":"The source fixes how many wave cycles are produced each second. Therefore, the frequency remains unchanged when the wave crosses into a different medium."},{"text":"Wavelength.","correct":false,"explanation":"Wavelength is not fixed by the source alone because it depends on wave speed as well as frequency. Frequency remains constant when the wave enters another medium."},{"text":"Wave speed.","correct":false,"explanation":"Wave speed is determined by the properties of the medium, so it commonly changes during refraction. Frequency stays the same because the source sets it."},{"text":"Amplitude.","correct":false,"explanation":"Amplitude can change at a boundary if some energy is reflected, transmitted or absorbed. Frequency, rather than amplitude, is fixed by the wave source."}]},{"id":"5e9153bb-97b3-4d27-a5d2-dbe33ea9f33f","cardType":"basic","points":1,"front":"If a wave enters a new medium along the normal, what happens to its direction?","back":"It does not change direction.","mcqOptions":[{"text":"It does not change direction.","correct":true,"explanation":"A wave travelling along the normal strikes the boundary at an angle of incidence of zero. Its speed and wavelength may change, but it does not bend."},{"text":"It bends towards the normal.","correct":false,"explanation":"Bending towards the normal occurs when a wave enters more slowly at a non-zero angle. Along the normal, the wave continues without changing direction."},{"text":"It bends away from the normal.","correct":false,"explanation":"Bending away from the normal requires the wave to speed up while entering at a non-zero angle. A wave entering along the normal does not change direction."},{"text":"It reverses its direction.","correct":false,"explanation":"Reversal describes reflection, not the path of the transmitted wave during refraction. When the wave enters along the normal, it continues in the same direction."}]},{"id":"3fe12d80-9b7e-4a1d-ab80-af3943f6a2e8","cardType":"basic","points":2,"front":"If a wave slows down during refraction and frequency is unchanged, what happens to wavelength?","back":"The wavelength decreases.","mcqOptions":[{"text":"The wavelength decreases.","correct":true,"explanation":"The wave equation is $v=f\\lambda$. With frequency unchanged, a decrease in wave speed requires a proportional decrease in wavelength."},{"text":"The wavelength increases.","correct":false,"explanation":"An increase would be expected if wave speed increased while frequency stayed constant. Since $v=f\\lambda$ and the wave slows down, its wavelength decreases."},{"text":"The wavelength stays the same.","correct":false,"explanation":"Keeping wavelength constant while frequency is unchanged would also keep wave speed constant. Because the speed decreases, the wavelength must decrease according to $v=f\\lambda$."},{"text":"The wavelength becomes zero.","correct":false,"explanation":"Slowing down does not mean the wave speed becomes zero, so its wavelength does not become zero. With constant frequency, a lower speed produces a shorter wavelength."}]}]}],"examCountdown":{"examDate":"2027-05-25","paperLabel":"Paper 1"},"heading":"Properties of waves","title":"Properties of waves Flashcards","topbar":{"backHref":"/igcse/physics/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"00af75fc-d090-5386-ae0a-97e2717e3a63","walled":"$L38"}],"$L39","$L3a"]