4.6 Waves — coverage pack

Answer

• transverse; the particle vibrations are perpendicular to the direction in which the wave travels

Method: Compare the two directions. The wave travels horizontally but the particles vibrate vertically, so the oscillations are perpendicular to energy transfer. The wave is transverse.

Answer

• The cork oscillates mainly up and down and returns close to its starting position; the ripple and energy travel across the tank, but the water does not travel across with them.

Method: Track the marker rather than the crest. The cork follows the local water motion, so it oscillates as the disturbance passes. Because it has no sustained motion to the far side, the observation shows that the wave transfers energy without a net transfer of the water.

Answer

• A coil oscillates backwards and forwards about a fixed position while the pulse and energy move along the spring; the oscillations are parallel to the direction of travel and the pulse contains compressed and less-compressed regions.

Method: Separate local motion from wave motion. Each coil moves to and fro along the spring rather than being carried with the pulse. This particle motion is parallel to the travelling disturbance, and the changing coil spacing forms compression and rarefaction regions. Those are the defining features of a longitudinal wave.

Answer

• $T=0.125\,\text{s}$

Method: The frequency is $f=8.0\,\text{Hz}$. Use $T=1/f$, giving $T=1/8.0=0.125\,\text{s}$.

Answer

• $\lambda=0.040\,\text{m}$
• $f=2.5\,\text{Hz}$
• $v=0.10\,\text{m s}^{-1}$

Method: Nine complete intervals give $\lambda=0.36/9=0.040\,\text{m}$. The frequency is $f=5/2.0=2.5\,\text{Hz}$. Therefore $v=f\lambda=(2.5)(0.040)=0.10\,\text{m s}^{-1}$.

Answer

• wavelength in air $=0.140\,\text{m}$
• wavelength in water $=0.600\,\text{m}$
• The wavelength increases because the speed increases while the frequency remains constant.

Method: Convert $2.40\,\text{kHz}$ to $2400\,\text{Hz}$. Rearrange $v=f\lambda$ to $\lambda=v/f$. In air, $\lambda=336/2400=0.140\,\text{m}$. In water, $\lambda=1440/2400=0.600\,\text{m}$. The source fixes the frequency, so the higher speed in water requires a proportionally longer wavelength.

Answer

• angle of reflection $=35^\circ$; both angles are measured from the normal

Method: For reflection, the angle of reflection equals the angle of incidence. The incident angle is already measured from the normal, so the reflected ray is also at $35^\circ$ to the normal.

Answer

• $15\%$ is absorbed; energy travels into the second material, returns into the first material and is transferred to the materials at the boundary.

Method: The incident energy represents $100\%$. The absorbed fraction is $100-68-17=15\%$. Transmission carries energy onward through the boundary, reflection carries energy back, and absorption transfers wave energy to the material.

Answer

• Direct the same light beam at each surface at the same angle; measure reflected light intensity with a light sensor at equal angles and distances; control source brightness, ambient light and illuminated area; repeat and compare mean reflected intensities or angular distributions.

Method: Fix a lamp or ray box and mark one incidence angle. Put the sensor the same distance from the point of incidence and take readings at matching reflection angles for each surface, shielding the setup from ambient light. Keep the source, brightness, colour, illuminated area and geometry unchanged. Repeat each reading, calculate means and compare either the peak reflected intensity or readings across several detector angles. The smooth tile should give a more concentrated reflected beam, whereas the rough card scatters light over more directions.

Answer

• $20\,\text{Hz}$ to $20\,\text{kHz}$

Method: Recall the standard range: the lower limit is about $20\,\text{Hz}$ and the upper limit is about $20\,000\,\text{Hz}$, which is $20\,\text{kHz}$.

Answer

• The electrical signal makes the loudspeaker cone vibrate; the cone produces longitudinal sound waves in the air; the sound waves make the eardrum vibrate.

Method: Follow each conversion in order. The alternating signal drives the solid cone backwards and forwards. The cone creates compressions and rarefactions in the air, transferring sound energy across the room. At the listener, the pressure changes exert forces on the eardrum and make that solid membrane vibrate.

Answer

• Ear structures respond effectively only over a limited frequency range; this person's detectable interval is narrower than the normal range of about $20\,\text{Hz}$ to $20\,\text{kHz}$, with both low-frequency and high-frequency loss.

Method: The eardrum and other ear structures must be driven into vibration for a tone to be detected. Their mechanical response is limited, so frequencies outside the effective range do not produce a sufficient vibration or sensation. Compared with $20\,\text{Hz}$ to $20\,\text{kHz}$, the measured lower limit is higher and the upper limit is lower, showing reduced hearing at both ends.

Answer

• Ultrasound has a frequency above $20\,\text{kHz}$; the pulse is partially reflected at the boundary.

Method: The upper limit of normal human hearing is about $20\,\text{kHz}$, so ultrasound lies above it. A change of medium forms an interface, where some of the wave is reflected and can return to a detector.

Answer

• depth $=63\,\text{m}$

Method: The pulse travels down and back, so its total distance is twice the depth. Calculate $vt=(1500)(0.084)=126\,\text{m}$, then divide by $2$: depth $=126/2=63\,\text{m}$.

Answer

• P-waves are longitudinal and can pass through solids and liquids, while transverse S-waves cannot pass through liquids; missing S-waves therefore indicate a liquid layer, and changes in P-wave speed, direction and arrival time reveal the positions and sizes of boundaries.

Method: Use the different transmission properties as the test. Both wave types can cross solid rock, but only P-waves continue through a liquid. If a detector geometry should receive both in an all-solid Earth but receives only P-waves, the S-wave path has been blocked by liquid. P-wave speed and direction change at material boundaries, so arrival-time patterns from many earthquakes and detectors can be used to map those interfaces and estimate the size of the liquid region.

Answer

• infrared is below; ultraviolet is above

Method: Use the frequency order radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. The neighbours of visible light are therefore infrared on the lower-frequency side and ultraviolet on the higher-frequency side.

Answer

• $\lambda=6.0\times10^{-3}\,\text{m}$; microwave

Method: Rearrange $c=f\lambda$ to $\lambda=c/f$. Then $\lambda=(3.00\times10^8)/(5.0\times10^{10})=6.0\times10^{-3}\,\text{m}$. A wavelength of $6.0\,\text{mm}$ is in the microwave region.

Answer

• radio wavelength $=4.0\,\text{m}$
• microwave wavelength $=0.10\,\text{m}$
• They travel at the same speed in air, but the higher-frequency microwave has the shorter wavelength.

Method: Convert the frequencies: $75\,\text{MHz}=75\times10^6\,\text{Hz}$ and $3.0\,\text{GHz}=3.0\times10^9\,\text{Hz}$. Using $\lambda=v/f$, the radio wavelength is $(3.00\times10^8)/(75\times10^6)=4.0\,\text{m}$ and the microwave wavelength is $(3.00\times10^8)/(3.0\times10^9)=0.10\,\text{m}$. Both are electromagnetic waves, so they have the same speed in air; the frequency-wavelength product stays constant.

Answer

• The ray bends towards the normal; angles are measured from the normal.

Method: Apply the refraction ray rule for light entering glass from air: draw the refracted ray closer to the normal. The normal is perpendicular to the boundary at the point where the ray enters, and both angles are referenced to it.

Answer

• $v_B/v_A=0.625$
• The closer wavefronts show that the wavelength and speed are smaller in medium B.

Method: Since $v=f\lambda$ and $f$ is unchanged, the speed ratio equals the wavelength ratio: $v_B/v_A=\lambda_B/\lambda_A=1.5/2.4=0.625$. The wavefronts are closer together in B because each cycle travels a shorter distance there, so B is the slower medium.

Answer

• Use identical cans with equal water volumes at the same starting temperature; measure temperature at regular times or infrared intensity at a fixed distance; control surroundings, exposed area and detector position; repeat and compare mean cooling rates or intensities.

Method: Fill identical cans to the same level with equal volumes of water at the same starting temperature. Put them side by side away from draughts. Either log each temperature at equal time intervals and find the gradient of temperature-time data, or place the same infrared detector at a fixed distance and angle from each surface. Keep can dimensions, exposed area, water volume, starting temperature, room conditions and detector geometry constant. Repeat the experiment, swap positions to reduce location bias and compare mean cooling rates or mean detector readings. The matt-black surface should be the better emitter.

Answer

• $0.180\,\text{Sv}$

Method: Use $1000\,\text{mSv}=1\,\text{Sv}$. Divide by $1000$: $180\,\text{mSv}=180/1000=0.180\,\text{Sv}$.

Answer

• Procedure A gives $27\,\text{mSv}$ in total; on dose alone it presents the greater risk because $27\,\text{mSv}>18\,\text{mSv}$.

Method: Add repeated exposures by multiplication: total for A is $6(4.5)=27\,\text{mSv}$. The dose for B is $18\,\text{mSv}$. Since radiation dose measures risk of harm, A has the greater indicated risk if the radiation type and other conditions are comparable.

Answer

• Electrical oscillations produce the radio wave; absorption by the receiver induces an alternating current at the same frequency; gamma rays are produced by changes in an atomic nucleus.

Method: The alternating charges and currents in the transmitter form an oscillating electrical circuit, which emits radio-frequency electromagnetic waves. When the $92\,\text{MHz}$ wave is absorbed by a suitable receiving circuit, it induces electrical oscillations and an alternating current at the same $92\,\text{MHz}$ frequency. Gamma radiation is also electromagnetic, but it is generated by a change in the nucleus rather than by a macroscopic electrical circuit.

Answer

• microwaves; infrared; X-rays

Method: Recall the standard application pairs. Satellite links use microwaves, thermal cameras detect infrared radiation, and bone imaging uses X-rays.

Answer

• All objects emit infrared and hotter regions emit more infrared in a given time; the camera detects differences in infrared intensity, whereas visible light from the building depends mainly on illumination and may be absent in darkness.

Method: A warm surface emits infrared radiation, and increasing temperature increases the amount emitted per second. An infrared detector converts the differing intensities into a thermal image, so hotter areas can be distinguished. In darkness there may be no visible light incident on the building to be reflected, so an ordinary visible-light image does not directly reveal the emitted thermal pattern.

Answer

• Microwaves penetrate into the food and are absorbed within it, while infrared is absorbed mainly near the surface; microwaves heat deeper regions and infrared can brown or heat the outside, so combining them gives more even internal heating with a hot surface.

Method: Link each wave to where its energy is absorbed. Microwaves can enter the food and transfer energy within a greater depth, so they heat material below the surface. Infrared is absorbed strongly at the exposed surface, transferring energy there. Microwave heating alone may leave less surface browning, while infrared alone heats inward more slowly. Using both provides internal heating and a hotter outer layer.

Answer

• magnification $=3.0$

Method: Use magnification $=\text{image height}/\text{object height}$. Therefore magnification $=3.6/1.2=3.0$. It is a ratio, so it has no unit.

Answer

• Draw a ray through the optical centre undeviated and a parallel ray refracted as if it came from the near focus; extend the diverging rays backwards; the image is virtual, upright and diminished.

Method: Start both rays at the top of the object. Send one through the optical centre without deviation. Draw the other parallel to the principal axis, then refract it outwards so its backward extension passes through the principal focus on the object side. Extend the rays backwards until they meet. Their apparent intersection gives an upright, smaller virtual image on the same side of the lens as the object.

Answer

• object height $=2.4\,\text{cm}$; convex lens; the screen shows the image is real, and a concave lens produces only virtual images

Method: Rearrange magnification $=\text{image height}/\text{object height}$ to object height $=7.2/3.0=2.4\,\text{cm}$. A screen can intercept only a real image. A concave lens always produces a virtual image, so the lens must be convex.

Answer

• The reflecting card appears white; the absorbing card appears black.

Method: White light contains the visible wavelength range. Equal reflection of all those wavelengths makes the first card look white. With no visible wavelengths reflected to the eye, the second card looks black.

Answer

• The blue filter transmits blue light and absorbs most other colours; the red book reflects red but absorbs most blue, so little light from the book passes through the filter to the eye.

Method: Work from the object to the observer. Under white light the book reflects mainly red wavelengths. A blue filter does not transmit those red wavelengths. Any blue light reaching the book is mostly absorbed by the red surface, so very little visible light completes the path to the eye and the book appears dark.

Answer

• It appears mainly green in white light, bright green through the green filter, and dim red or very dark through the red filter; the filters transmit only their colour while the surface reflects green much more strongly than red and absorbs blue.

Method: In white light, all three colour ranges arrive, but strong green reflection dominates, so the surface appears green. A green filter passes the strongly reflected green light and absorbs most other wavelengths, so the surface remains clearly green. A red filter blocks green and passes only the weak red reflection, so much less light reaches the eye and the surface looks dim red or nearly dark. Blue contributes nothing because the surface absorbs it.

Answer

• The object at $75\,{}^\circ\text{C}$ emits more infrared radiation each second.

Method: The surfaces are identical, so temperature is the relevant difference. A hotter object radiates more infrared energy in a given time, so the $75\,{}^\circ\text{C}$ object emits more.

Answer

• The matt-black can cools faster because a matt-black surface is a better emitter of infrared radiation than a polished-silver surface, so it transfers energy to the surroundings at a greater rate.

Method: Because the cans and starting conditions are identical, compare only their emitting surfaces. Matt black is a good infrared emitter, while polished silver is a poor emitter. The black can therefore loses internal energy by radiation more rapidly and its temperature falls faster.

Answer

• Expose equal plates at equal distance and angle for equal times from the same lamp; measure equal-mass plates' temperature rises with identical contact thermometers; control starting temperature and surroundings; repeat; the surface with the greatest mean temperature rise is the best absorber.

Method: Use plates of the same material, mass and area, differing only in surface finish. Let them reach the same starting temperature, place each at the same distance and orientation to the lamp, and expose each for the same time at constant lamp power. Measure the temperature before and after with the same sensor. Shield the setup from draughts and other heat sources, repeat each finish and calculate its mean temperature rise. With the energy input controlled, the largest mean rise indicates the greatest infrared absorption.

Answer

• the intensity; the wavelength distribution

Method: All objects emit radiation. The specification identifies two temperature-dependent properties of that emission: how intense it is and how the emitted energy is distributed across wavelengths.

Answer

• net energy change $=-50\,\text{W}$; the body cools

Method: Take absorbed power minus emitted power: $480-530=-50\,\text{W}$. The negative result means the body loses $50\,\text{J}$ of energy each second. Its internal energy and temperature therefore decrease.

Answer

• initial net power $=0\,\text{W m}^{-2}$
• new net power $=+12\,\text{W m}^{-2}$
• Earth warms until increased emission restores the balance.

Method: Initially the absorbed solar power is $340-102=238\,\text{W m}^{-2}$, equal to the $238\,\text{W m}^{-2}$ emitted, so the net power is zero. After the reflection change, absorption is $340-90=250\,\text{W m}^{-2}$. The initial new imbalance is $250-238=+12\,\text{W m}^{-2}$, so Earth gains energy and warms. As temperature rises, emitted radiation increases until outgoing power again matches absorbed incoming power.