Waves | HSC - Wyatt's Notes

HSC physics study notes - Waves

flowchart TD
    A[Waves] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Wave speed equation: $v = f\lambda$

Frequency: $f = \frac{1}{T}$ (where $T$ is the period)

Wave types:

• Transverse: oscillations perpendicular to wave direction (e.g., light, water waves)
• Longitudinal: oscillations parallel to wave direction (e.g., sound, compressions)

Speed of sound in air: $v \approx 340\,\text{m/s}$ at $20°C$

Sound intensity: $I = \frac{P}{4\pi r^2}$ (inverse square law)

Sound level (decibels): $\beta = 10\log_{10}\left(\frac{I}{I_0}\right)$ where $I_0 = 10^{-12}\,\text{W/m}^2$

Speed of EM waves in vacuum: $c = 3 \times 10^8\,\text{m/s}$

Energy of a photon: $E = hf = \frac{hc}{\lambda}$

Relationship: $c = f\lambda$

Superposition principle: When two waves meet, the resultant displacement is the sum of individual displacements.

Constructive interference: Path difference $= n\lambda$ ($n = 0, 1, 2, \ldots$)

Destructive interference: Path difference $= (n + \frac{1}{2})\lambda$ ($n = 0, 1, 2, \ldots$)

Single slit: Central maximum is twice the width of other maxima.

Diffraction grating: $d\sin\theta = n\lambda$

Problem: A wave has frequency $50\,\text{Hz}$ and wavelength $0.8\,\text{m}$. Find the wave speed.

Solution:

Step 1: Apply the wave speed equation: $v = f\lambda = 50 \times 0.8 = 40\,\text{m/s}$

Answer: The wave speed is $40\,\text{m/s}$

Problem: A sound has intensity $10^{-4}\,\text{W/m}^2$. Find the sound level in decibels.

Solution:

Step 1: Apply the decibel formula: $\beta = 10\log_{10}\left(\frac{I}{I_0}\right) = 10\log_{10}\left(\frac{10^{-4}}{10^{-12}}\right)$

Step 2: Simplify: $\beta = 10\log_{10}(10^8) = 10 \times 8 = 80\,\text{dB}$

Answer: The sound level is $80\,\text{dB}$

Problem: Light of wavelength $600\,\text{nm}$ passes through a diffraction grating with $5000$ lines per cm. Find the angle of the first-order maximum.

Solution:

Step 1: Find the grating spacing: $d = \frac{1}{5000} = 2 \times 10^{-4}\,\text{cm} = 2 \times 10^{-6}\,\text{m}$

Step 2: Convert wavelength: $\lambda = 600\,\text{nm} = 6 \times 10^{-7}\,\text{m}$

Step 3: Apply the grating equation for $n = 1$: $\sin\theta = \frac{n\lambda}{d} = \frac{1 \times 6 \times 10^{-7}}{2 \times 10^{-6}} = 0.3$

Step 4: $\theta = \arcsin(0.3) \approx 17.5°$

Answer: The angle of the first-order maximum is approximately $17.5°$

1. Remember that $v = f\lambda$ applies to all waves
2. Sound intensity follows the inverse square law
3. For diffraction gratings, higher orders are only visible if $\sin\theta \leq 1$
4. EM spectrum: radio, microwave, infrared, visible, UV, X-ray, gamma (increasing energy)

1. A wave travels at $340\,\text{m/s}$ with wavelength $0.5\,\text{m}$. Find the frequency.
2. Two sound sources are $2\,\text{m}$ apart and vibrate in phase. Find the position of the first minimum between them for sound of wavelength $0.4\,\text{m}$.
3. What is the energy of a photon with wavelength $500\,\text{nm}$? ($h = 6.63 \times 10^{-34}\,\text{J s}$, $c = 3 \times 10^8\,\text{m/s}$)

Problem: A string of length $0.5\,\text{m}$ is fixed at both ends and vibrates in its third harmonic at $150\,\text{Hz}$. Find the wave speed.

Solution:

Step 1: For a string fixed at both ends, the $n$-th harmonic frequency is: $f_n = \frac{nv}{2L}$

Step 2: For the third harmonic ($n = 3$): $150 = \frac{3v}{2 \times 0.5} = \frac{3v}{1}$

Step 3: Solve for $v$: $v = \frac{150}{3} = 50\,\text{m/s}$

Answer: The wave speed is $50\,\text{m/s}$

Common mistake: Using $f_n = nv/L$ instead of $f_n = nv/(2L)$ for a string fixed at both ends. The factor of 2 arises because both ends are nodes.

Problem: A train sounding its horn at $400\,\text{Hz}$ approaches a stationary observer at $30\,\text{m/s}$. The speed of sound is $340\,\text{m/s}$. Find the frequency heard by the observer.

Solution:

Step 1: For a source approaching a stationary observer: $f' = f \cdot \frac{v}{v - v_s}$

Step 2: Substitute values: $f' = 400 \times \frac{340}{340 - 30} = 400 \times \frac{340}{310} = 400 \times 1.097 = 438.7\,\text{Hz}$

Answer: The observer hears a frequency of approximately $439\,\text{Hz}$

Common mistake: Using the wrong sign in the Doppler formula. When the source approaches, the denominator is $v - v_s$ (frequency increases). When the source recedes, it is $v + v_s$ (frequency decreases).

Problem: Light of wavelength $200\,\text{nm}$ strikes a metal with work function $3.5\,\text{eV}$. Find the maximum kinetic energy of the emitted electrons ($h = 6.63 \times 10^{-34}\,\text{J s}$, $c = 3 \times 10^8\,\text{m/s}$, $1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$).

Solution:

Step 1: Calculate photon energy: $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{200 \times 10^{-9}} = 9.945 \times 10^{-19}\,\text{J}$

Step 2: Convert to eV: $E = \frac{9.945 \times 10^{-19}}{1.6 \times 10^{-19}} = 6.22\,\text{eV}$

Step 3: Apply the photoelectric equation: $E_k = E - W_0 = 6.22 - 3.5 = 2.72\,\text{eV}$

Answer: The maximum kinetic energy is $2.72\,\text{eV}$

Common mistake: Forgetting to convert between joules and electron volts. Always check the units requested in the answer.

Problem: Find the fundamental frequency of a pipe of length $0.8\,\text{m}$ that is open at both ends ($v = 340\,\text{m/s}$).

Solution:

Step 1: For a pipe open at both ends, the fundamental frequency occurs when the length equals half a wavelength: $L = \frac{\lambda}{2} \implies \lambda = 2L = 2 \times 0.8 = 1.6\,\text{m}$

Step 2: Apply the wave equation: $f = \frac{v}{\lambda} = \frac{340}{1.6} = 212.5\,\text{Hz}$

Answer: The fundamental frequency is $212.5\,\text{Hz}$

Common mistake: Using $L = \lambda$ instead of $L = \lambda/2$ for the fundamental mode in a pipe open at both ends.

Problem: Two tuning forks produce frequencies of $256\,\text{Hz}$ and $260\,\text{Hz}$. Find the beat frequency and the time interval between successive maxima.

Solution:

Step 1: Beat frequency is the difference of the two frequencies: $f_{\text{beat}} = |f_1 - f_2| = |256 - 260| = 4\,\text{Hz}$

Step 2: Time interval between successive maxima: $T = \frac{1}{f_{\text{beat}}} = \frac{1}{4} = 0.25\,\text{s}$

Answer: The beat frequency is $4\,\text{Hz}$ and the time interval is $0.25\,\text{s}$

Common mistake: Confusing beat frequency with the average frequency. The beat frequency is the difference, not the sum or average.

Problem: Light traveling in glass ($n = 1.5$) strikes the glass-air boundary at an angle of incidence of $40°$. Find the angle of refraction and determine if total internal reflection occurs.

Solution:

Step 1: Check for total internal reflection. The critical angle is: $\sin C = \frac{n_2}{n_1} = \frac{1}{1.5} = 0.667$ $C = \arcsin(0.667) = 41.8°$

Step 2: Since the angle of incidence ($40°$) is less than the critical angle ($41.8°$), total internal reflection does not occur.

Step 3: Apply Snell’s law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$ $1.5 \times \sin 40° = 1 \times \sin\theta_2$ $\sin\theta_2 = 1.5 \times 0.643 = 0.964$ $\theta_2 = \arcsin(0.964) = 74.6°$

Answer: The angle of refraction is $74.6°$ and no total internal reflection occurs

Common mistake: Forgetting to check the critical angle before applying Snell’s law. If the angle of incidence exceeds the critical angle, all light is reflected back into the denser medium.

Waves are energy in motion without matter following it — think of a Mexican wave in a stadium where each person moves up and down while the pattern travels forward. Interference is what happens when two waves occupy the same space: they add together, creating regions of reinforcement and cancellation. Diffraction reveals that waves bend around obstacles, a behaviour that becomes more pronounced when the obstacle size approaches the wavelength. The Doppler effect is the reason a siren changes pitch as it passes you — the wavefronts compress ahead and stretch behind.

Confusing the wave speed equation variables. The equation v = f * lambda relates wave speed (m/s), frequency (Hz), and wavelength (m). Students often rearrange it incorrectly or mix up which variable to solve for. Remember: speed equals frequency times wavelength, always.

Using the wrong sign in the Doppler effect formula. When the source approaches the observer, the observed frequency increases: f’ = f v / (v - vs). When the source recedes, frequency decreases: f’ = f v / (v + vs). Students often use the wrong sign, giving the opposite effect to what actually occurs.

Forgetting that sound intensity follows the inverse square law. Sound intensity decreases as 1/r^2 with distance from the source. Doubling the distance reduces intensity to one-quarter, not one-half. Students often assume a linear decrease, leading to incorrect calculations of sound levels at different distances.

• Mechanics — Simple harmonic motion is the foundation for understanding oscillatory wave behaviour.
• Algebra — Logarithmic functions are used in decibel calculations and sound intensity levels.
• Calculus — Differentiation and integration are used in wave equations and standing wave analysis.
• Inorganic — Electromagnetic spectrum properties connect to atomic structure and electron transitions.