Module 3: Waves & Thermodynamics

First find the speed: $v = \dfrac{60}{4.0} = 15\text{ m/s}$. Then $\lambda = \dfrac{v}{f} = \dfrac{15}{5.0} = 3.0\text{ m}$. (You must find the speed before you can use $v = f\lambda$.)

A distance graph shows wavelength directly: $\lambda = 0.8\text{ m}$. Then $v = f\lambda = 25 \times 0.8 = 20\text{ m/s}$, and in $0.10\text{ s}$ the wave travels $d = vt = 20 \times 0.10 = 2.0\text{ m}$. (Two steps: read the wavelength, then apply speed = distance/time.)

Careful: a displacement-time graph shows the period, not the wavelength. Here $T = 0.04\text{ s}$, so $f = \dfrac{1}{T} = 25\text{ Hz}$. Then $v = f\lambda = 25 \times 1.2 = 30\text{ m/s}$. (Reading 0.04 as a wavelength is the classic trap.)

By the principle of superposition, overlapping displacements add (with sign): $+4 + (-3) = +1\text{ cm}$. The pulses then pass through each other unchanged. (This partial cancellation is destructive interference; equal and opposite pulses would give zero.)

The wave speed is $v = \dfrac{\lambda}{T} = \dfrac{6.0}{2.5} = 2.4\text{ m/s}$. Time to cover $30\text{ m}$: $t = \dfrac{d}{v} = \dfrac{30}{2.4} = 12.5\text{ s}$. (The "every $2.5\text{ s}$" is the period, giving the speed together with the $6\text{ m}$ wavelength.)

In medium A, $v_A = f\lambda = 10 \times 2.0 = 20\text{ m/s}$, so $v_B = 10\text{ m/s}$. Frequency is set by the source and does not change across a boundary, so $\lambda_B = \dfrac{v_B}{f} = \dfrac{10}{10} = 1.0\text{ m}$. (Halving the speed halves the wavelength.)

For refraction, $\dfrac{\sin\theta_1}{\sin\theta_2} = \dfrac{v_1}{v_2}$. So $\sin\theta_2 = \sin 40^\circ \times \dfrac{4.0}{6.0} = 0.643 \times 0.667 = 0.429$, giving $\theta_2 \approx 25^\circ$. The wave slows and bends toward the normal.

At the critical angle the refracted ray grazes the surface ($\theta_2 = 90^\circ$), so $\sin\theta_c = \dfrac{n_{air}}{n_{glass}} = \dfrac{1}{1.5} = 0.667$, giving $\theta_c \approx 42^\circ$. Beyond this, all the light reflects back into the glass, the principle behind optical fibres.

Three loops mean $\lambda = \dfrac{2L}{3} = \dfrac{2 \times 1.2}{3} = 0.80\text{ m}$ (each loop is half a wavelength). Then $v = f\lambda = 250 \times 0.80 = 200\text{ m/s}$. (Read the wavelength from the pattern, then apply the wave equation.)

A closed pipe produces only odd harmonics: $f_1, 3f_1, 5f_1, \dots$, so consecutive resonances differ by $2f_1$. Here $500 - 300 = 200 = 2f_1$, giving $f_1 = 100\text{ Hz}$ (and $300\text{ Hz}$, $500\text{ Hz}$ are the 3rd and 5th harmonics). ($200\text{ Hz}$ cannot appear, because even harmonics are missing.)

A closed pipe fits a quarter wavelength (node at the closed end, antinode at the open end): $\lambda = 4L = 4 \times 0.50 = 2.0\text{ m}$. Then $f = \dfrac{v}{\lambda} = \dfrac{340}{2.0} = 170\text{ Hz}$. (Using $\lambda = 2L$, as for an open pipe, doubles the answer to $340\text{ Hz}$.)

Destructive interference needs a path difference of $(n + \tfrac{1}{2})\lambda$. The wavelength is largest for the smallest $n$ ($n = 0$): $\tfrac{1}{2}\lambda = 1.2\text{ m}$, so $\lambda = 2.4\text{ m}$. (Larger $n$ gives shorter wavelengths that also work.)

A $4\text{ Hz}$ beat means the string was at $252\text{ Hz}$ or $260\text{ Hz}$. Tightening a string raises its frequency; since the beat got larger (further from $256\text{ Hz}$), the string must have been above the fork, at $260\text{ Hz}$. (Had it been $252\text{ Hz}$, tightening it toward $256\text{ Hz}$ would have reduced the beat.)

Diffraction (bending around obstacles) is significant when the wavelength is comparable to the obstacle or gap. Sound wavelengths are around a metre, similar to the corner, so sound spreads around it. Light's wavelength is under a millionth of a metre, so it diffracts negligibly and travels essentially straight.

In $0.40\text{ s}$ the pulse travels down and back, a round trip of $1500 \times 0.40 = 600\text{ m}$. The depth is half of that: $d = \dfrac{600}{2} = 300\text{ m}$. (Forgetting the pulse travels there and back gives $600\text{ m}$.)

By the inverse-square law $I_1 r_1^2 = I_2 r_2^2$. So $r_2 = r_1\sqrt{\dfrac{I_1}{I_2}} = 2.0\sqrt{\dfrac{0.80}{0.050}} = 2.0\sqrt{16} = 8.0\text{ m}$. (The intensity fell to one-sixteenth, so the distance grew by a factor of 4.)

Rearrange $Q = mc\Delta T$: $c = \dfrac{Q}{m\Delta T} = \dfrac{1500}{0.10 \times 25} = \dfrac{1500}{2.5} = 600\text{ J}$/kg/°C. (This low value, compared with water's 4200, is typical of a metal.)

Heat lost by the hot water equals heat gained by the cool water; since both are water, $c$ cancels: $0.20(90 - T) = 0.30(T - 20)$. Then $18 - 0.20T = 0.30T - 6$, so $24 = 0.50T$ and $T = 48^\circ\text{C}$ (a mass-weighted average, nearer the larger portion).

Rearrange $Q = mL$: $m = \dfrac{Q}{L} = \dfrac{50000}{3.34 \times 10^5} \approx 0.15\text{ kg}$. (Remember $50\text{ kJ}$ = $50000\text{ J}$; the temperature stays at $0^\circ\text{C}$ the whole time.)

Each plateau's length is proportional to the energy absorbed at constant temperature (heat added at a steady rate). A longer boiling plateau means more energy is needed to vaporise the substance than to melt it, i.e. the latent heat of vaporisation exceeds the latent heat of fusion, which is true for water and most substances.

Melting the ice needs $mL = 0.050 \times 3.34\times10^5 = 16700\text{ J}$; the warm water can give up to $0.30 \times 4200 \times 30 = 37800\text{ J}$, so all the ice melts. Balancing heat: $0.30(4200)(30 - T) = 16700 + 0.050(4200)T$. This gives $37800 - 1260T = 16700 + 210T$, so $21100 = 1470T$ and $T \approx 14^\circ\text{C}$. (You must first check the ice fully melts, then include the melt-water warming up.)

The conduction rate is $\dfrac{Q}{t} = \dfrac{kA\Delta T}{d} = \dfrac{0.80 \times 2.0 \times 20}{0.0040} = \dfrac{32}{0.0040} = 8000\text{ W}$. (The thickness must be in metres: $4.0\text{ mm}$ = $0.0040\text{ m}$, which is where a factor of 1000 is easily lost.)

"Feeling cold" is about how fast heat leaves your hand. Metal is a good conductor, so it draws heat from your skin quickly and feels cold. Wood is a poor conductor (a good insulator), so heat leaves slowly and it feels warmer, even though both are at $10^\circ\text{C}$.

Heat lost by metal = heat gained by water: $0.50(400)(100 - T) = 1.0(4200)(T - 20)$. So $200(100 - T) = 4200(T - 20)$, i.e. $20000 - 200T = 4200T - 84000$, giving $104000 = 4400T$ and $T \approx 24^\circ\text{C}$. The water barely warms because it has far more heat capacity than the metal.

Heat needed: $Q = mc\Delta T = 2.0 \times 4200 \times 80 = 672000\text{ J}$. Power is energy per second, and 5.0 min = $300\text{ s}$, so $P = \dfrac{Q}{t} = \dfrac{672000}{300} = 2240\text{ W}$. (Use $\Delta T = 80$, and convert minutes to seconds.)