Waves, Sound & Optics — Chapter Practice, Grade 12

30 m/s

v = f × λ = 10 × 3 = 30 m/s.

250 Hz

f = 1 ÷ T = 1 ÷ 0.004 = 250 Hz.

3 m

λ = v ÷ f = 12 ÷ 4 = 3 m.

perpendicular to the wave's motion

Transverse means the vibration is perpendicular (at right angles) to the direction the wave travels.

they cancel to zero amplitude

Equal waves exactly out of phase interfere destructively and cancel — this is destructive interference.

5 m

λ = v ÷ f = 1500 ÷ 300 = 5 m.

Check v=fλ independently.

v=30 m/s.

State what the supplied observations or conditions establish, and separate that from any assumption.

Particles oscillate locally while interactions pass the disturbance and energy onward.

λ=v/f=0.50 m under the stated medium conditions.

Wave speed satisfies v=fλ for a steady periodic traveling wave. Frequency is cycles per second and wavelength is meters per cycle, so their product gives meters per second. In a nondispersive medium, changing source frequency changes wavelength while speed stays approximately fixed. In a dispersive medium, speed can depend on frequency, so “speed is always fixed” is too broad. Identify the medium, wave type and relevant approximation.

λ=v/f=0.50 m under the stated medium conditions.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

f=50 Hz and v=40 m/s. At100 Hz with the same 40 m/s speed, λ=0.40 m.

f=50 Hz and v=40 m/s.

At100 Hz with the same 40 m/s speed, λ=0.40 m.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

The equilibrium is about 20 mm and amplitude about 0.8 mm, measured from that mean rather than from zero. Roughly three cycles occupy 0.01 s, givingT≈0.0033 s andf≈300 Hz. These are approximate visual readings. Time at one source does not provide spatial wavelength or wave speed without additional information.

Read source labels, scope, axes and intentional omissions.

Use the complete local explanation to draw a justified mechanism or calculate a supported quantity.

State what is observed, what is assumed and what evidence remains needed.

Response criteria

• Interpret the source labels, units and scope accurately.
• Explain the relevant mechanism with a labeled model or supported calculation.
• Identify a meaningful assumption, uncertainty or missing comparison.

1020 m

distance = speed × time = 340 × 3 = 1,020 m.

frequency

Pitch depends on frequency — higher frequency means higher pitch.

the waves stretch out behind it, lowering the frequency you hear

For a receding source, the wavefronts stretch out, so the observed wavelength grows and the frequency (pitch) drops — the Doppler effect.

500 Hz

f = v ÷ λ = 340 ÷ 0.68 = 500 Hz.

Resonance is the large build-up of vibration that happens when a system is driven at its natural frequency. Pushing a swing in time with its natural back-and-forth rhythm makes it go higher and higher; other examples include a singer shattering a wine glass by matching its natural frequency, or a guitar string producing a loud clear note.

680 Hz

The source recedes, so f′ = f × v ÷ (v + vₛ) = 800 × 340 ÷ (340 + 60) = 800 × 340 ÷ 400 = 800 × 0.85 = 680 Hz.

Beat frequency is distinct from the approximately 442 Hz pitch region.

340×0.20/2=34 m, including the round trip.

State what the supplied observations or conditions establish, and separate that from any assumption.

Elastic and inertial properties of the medium affect propagation; speed is not fixed by loudness alone.

10^(30/10)=1000, provided the levels use the same intensity reference.

A sound-intensity level is β=10log₁₀(I/I₀) for a stated reference intensity I₀. Differences in decibels compare ratios: a 20 dB increase means 100 times the intensity, not 20 times. Decibels are not a direct linear loudness scale. The plotted points show selected calculated ratios on a linear horizontal axis; between-point connections are a guide rather than an exact smooth logarithmic curve.

10^(30/10)=1000, provided the levels use the same intensity reference.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

The intensity ratio is 10^1.5≈31.6. Echo distance=340×0.30/2=51 m; the two calculations use different models and quantities.

The intensity ratio is 10^1.5≈31.6.

Echo distance=340×0.30/2=51 m; the two calculations use different models and quantities.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

5 m

λ = c ÷ f = (3 × 10⁸) ÷ (6 × 10⁷) = 5 m.

gamma rays

Gamma rays sit at the short-wavelength end of the spectrum, giving them the highest frequency and energy.

a stream of particles called photons

The photoelectric effect requires the particle (photon) model: energy arrives in discrete packets whose energy depends on frequency.

6.63e-19 J

E = h f = (6.63 × 10⁻³⁴) × (1 × 10¹⁵) = 6.63 × 10⁻¹⁹ J.

Wave-particle duality means light behaves as both a wave and a stream of particles, depending on how you observe it. In experiments like interference and diffraction it acts like a wave, but in the photoelectric effect it acts like discrete particles (photons). Neither model alone is complete — you choose whichever one best describes the situation.

500000000000000 Hz

f = c ÷ λ = (3 × 10⁸) ÷ (6 × 10⁻⁷) = 0.5 × 10¹⁵ = 5 × 10¹⁴ Hz.

Keep powers of ten and length conversion consistent.

f=c/λ=1.0×10¹⁰ Hz.

State what the supplied observations or conditions establish, and separate that from any assumption.

Frequency stays constant across the boundary; speed and wavelength can change.

Photon energy, material interaction, absorbed amount and exposure conditions differ across situations.

The electromagnetic spectrum orders radiation by frequency or vacuum wavelength. Radio, microwave, infrared, visible, ultraviolet, X-ray and gamma labels cover ranges, not separate fundamental wave species. A material may transmit, reflect or absorb different ranges depending on its structure. Ionization removes an electron and requires enough interaction energy under the relevant conditions; heating and photochemical changes can occur through other mechanisms. Explain an application using energy, material response and dose rather than a color label alone.

Photon energy, material interaction, absorbed amount and exposure conditions differ across situations.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

λ=4.00×10⁻⁷ m; f=7.50×10¹⁴ Hz and E=4.97×10⁻¹⁹ J per photon. This is not total beam energy without photon number.

λ=4.00×10⁻⁷ m; f=7.50×10¹⁴ Hz and E=4.97×10⁻¹⁹ J per photon.

This is not total beam energy without photon number.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

the normal (the perpendicular to the surface)

All angles in optics are measured from the normal, the line perpendicular to the surface at the point of contact.

1.5

n = c ÷ v = (3 × 10⁸) ÷ (2 × 10⁸) = 1.5.

bends toward the normal and slows down

Entering a denser, higher-index medium, light slows down and bends toward the normal.

30 cm

1 ÷ dᵢ = 1 ÷ f − 1 ÷ dₒ = 1 ÷ 15 − 1 ÷ 30 = 2 ÷ 30 − 1 ÷ 30 = 1 ÷ 30, so dᵢ = 30 cm.

A real image forms where light rays actually converge, so it can be projected onto a screen (like the image on a camera sensor or a movie screen); it is usually inverted. A virtual image forms where rays only appear to come from — they never actually meet there — so it cannot be projected; it is upright, like the enlarged image seen through a magnifying glass or your reflection in a flat mirror.

6 cm

Magnification magnitude = dᵢ ÷ dₒ = 30 ÷ 10 = 3, so the image height = 3 × 2 cm = 6 cm.

State the sign convention; a negative image distance has a different interpretation.

1/d_i=1/10−1/20=1/20; d_i=20 cm.

State what the supplied observations or conditions establish, and separate that from any assumption.

Refraction changes propagation speed and wavelength while frequency is continuous across a stationary boundary.

di=−10 cm and m=+2: an upright enlarged virtual image on the object side.

In the stated thin-lens convention, 1/f=1/do+1/di, with f positive for a converging lens, do positive for a real object and di positive for a real image on the far side. Magnification m=−di/do gives image height divided by object height. For f=10 cm and do=20 cm, di=20 cm and m=−1: the real image is inverted and equal in height. State the convention rather than interpreting every negative sign as an error.

di=−10 cm and m=+2: an upright enlarged virtual image on the object side.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

1/di=1/12−1/36=1/18, so di=18 cm and m=−18/36=−0.50. It is real, inverted and half the object height in this paraxial model.

1/di=1/12−1/36=1/18, so di=18 cm and m=−18/36=−0.50.

It is real, inverted and half the object height in this paraxial model.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

Draw source-to-surface rays, then reflected or transmitted paths that actually reach the observer. Light enters the eye from the path; the eye does not emit rays that create sight. Angles are measured from the boundary normal, reflection uses equal angles and transmitted direction requires the indices and geometry.

Read source labels, scope, axes and intentional omissions.

Use the complete local explanation to draw a justified mechanism or calculate a supported quantity.

State what is observed, what is assumed and what evidence remains needed.

Response criteria

• Interpret the source labels, units and scope accurately.
• Explain the relevant mechanism with a labeled model or supported calculation.
• Identify a meaningful assumption, uncertainty or missing comparison.

Write the relationship first; substitute with consistent units.

0.6 m

Node spacing=λ/2.

No; it describes local combined displacement. The waves can continue and energy distribution must be considered.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

List the given quantities, select a relationship, and check units before substituting.

60 Hz

4×15=60.

Fixed, free, open, and closed boundaries allow different mode patterns.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

Boundary conditions determine allowed wavelengths.

f₂=2f₁=20 Hz.

State what the supplied observations or conditions establish, and separate that from any assumption.

Their displacements cancel locally by superposition; this does not imply energy is destroyed everywhere.

L=2λ/2, so λ=1.2 m.

A standing wave has nodes of zero displacement amplitude and antinodes of maximum amplitude, produced by counterpropagating coherent waves under appropriate conditions. A string fixed at both ends has nodes there and supports L=nλ/2. The displayed second harmonic has an additional middle node. Air columns with an open end have different displacement/pressure boundary conditions; do not transfer the fixed-string formula without identifying what is constrained.

L=2λ/2, so λ=1.2 m.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

f1=v/(2L)=10 Hz; f3=30 Hz. λ3=2L/3=1.0 m, with fixed-end nodes in this ideal string model.

f1=v/(2L)=10 Hz; f3=30 Hz. λ3=2L/3=1.0 m, with fixed-end nodes in this ideal string model.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.

Write the relationship first; substitute with consistent units.

514.71 Hz

500×350/340≈514.71.

No; it changes front spacing, while medium properties largely determine speed.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

List the given quantities, select a relationship, and check units before substituting.

320 Hz

300×320/(320−20)=320.

No. Light requires electromagnetic relativistic Doppler relationships.

Response criteria

• State a scientifically accurate conclusion.
• Show the calculation or mechanism that supports it.
• Include units, conditions, or limits where relevant.

The formula changes for moving observers or other geometries.

f'=500×340/(340+34)≈455 Hz.

State what the supplied observations or conditions establish, and separate that from any assumption.

Sound has a material medium and different source/observer roles; light requires the appropriate relativistic model.

√(0.40/1.60)=0.50; the observed frequency is half under the stated inertial model.

Light in vacuum does not use a stationary air-like medium. Its relativistic Doppler shift depends on the observer–source geometry and consistent inertial-frame definitions. For longitudinal recession, fobs=femit√[(1−β)/(1+β)] with β=v/c. This differs from simply replacing sound speed by c in the moving-source formula. The low-speed approximations can be similar, but their underlying assumptions and behavior at high speed differ.

√(0.40/1.60)=0.50; the observed frequency is half under the stated inertial model.

Response criteria

• Interpret labels, structures, axes, units and directions correctly.
• Explain the physical or environmental relationship or mechanism.
• Respect stated assumptions and distinguish observation from inference.

Receding:600×330/360=550 Hz. Approaching:600×330/300=660 Hz. The shifts are not symmetric about 600 in this finite-speed source model.

Receding:600×330/360=550 Hz.

Approaching:600×330/300=660 Hz.

The shifts are not symmetric about 600 in this finite-speed source model.

Response criteria

• Use the stated measurements or conditions, with correct units.
• Connect the result to the physical or environmental mechanism or evidence relationship.
• Identify the assumption, model limit or further evidence requested.