AP Unit 14 Solutions

The mark moves transversely about equilibrium as the pulse passes, while the pulse transfers energy right. The tone consists of repeating longitudinal pressure compressions and rarefactions in air.

The maximum departure from equilibrium is 101328−101325=3 Pa (also 101325−101322=3 Pa), so the initial pressure amplitude is 3 Pa. Doubling makes it 6 Pa. The initial peak-to-trough range 6 Pa is not the initial amplitude.

Larger pressure departures mean greater energy transfer and generally greater loudness. Linear-wave intensity scales as pressure amplitude squared, so it becomes four times as large. Warming the same air increases sound speed.

μP = 0.0040 kg/m and μQ = 0.016 kg/m. Using √(Fₜ/μ), vP = √4000 = 63.2 m/s and vQ = √1000 = 31.6 m/s.

tP = 3.0/63.2 = 0.0474 s; tQ = 3.0/31.6 = 0.0949 s.

Use the same string and marked travel distance, vary known tension, time repeated small pulses, and calculate v. Plot v² against Fₜ; a line through the origin has slope 1/μ.

Estimate timing scatter and tension uncertainty; avoid changing stretch so much that μ changes appreciably.

String speed depends on tension and mass per length; sound in the same medium travels faster when temperature increases. Radio and visible electromagnetic waves both travel at c = 3.00 × 10⁸ m/s in vacuum.

A = 0.015 m, λ = 0.80 m, T = 0.20 s, f = 5.0 Hz. Then v = fλ = 4.0 m/s.

$y(t)=0.015\cos(10\pi t)\ \mathrm{m}$ starts at maximum displacement.

Place extremes at ±0.015 m. Spatial crests repeat every 0.80 m; temporal crests repeat every 0.20 s. A vertical rescaling changes only amplitude.

TR = 1/6 s and TS = 1/12 s; λR = 18/6 = 3.0 m and λS = 18/12 = 1.5 m. Equal amplitude and medium imply greater energy transfer for S; sinusoidal-string average power is four times R.

S at 960 Hz has the higher pitch. Increase R’s sound pressure amplitude at fixed 480 Hz to make it louder without changing pitch.

v₁ = √(8/0.020) = 20 m/s and v₂ = √(8/0.0050) = 40 m/s. Frequency remains 20 Hz, giving λ₁ = 1.0 m and λ₂ = 2.0 m.

Entering the faster string produces a noninverted reflected pulse, traveling back in string 1.

A partial boundary reflection can coexist with transmission. In a lossless junction the incident energy is divided between reflected and transmitted waves; reflection does not imply all energy returns.

The incident speed is now 40 m/s in the light string. Transmission enters the heavy string at 20 m/s; its lower speed at equal tension means the reflected pulse is inverted, so its displacement is downward.

The reflected pulse returns through the light string at 40 m/s. Reflection changes its direction and displacement phase, not the wave speed of the medium it remains in.

The first polarizer gives 40 W/m². The analyzer gives 40 cos²60° = 10 W/m². Pavg = IS = (10)(0.015) = 0.15 W.

Intensity averages power per area over oscillations; it is not an instantaneous peak power density.

Horizontal is perpendicular to the transmitted vertical E, so ideal transmission is zero. Longitudinal air sound has no transverse oscillation plane to select, so crossed slots do not provide the same polarization mechanism.

Reflection can preferentially select electric-field orientations, making the beam polarized or partly polarized. A rotating analyzer then selects different components, changing intensity. Refraction through a medium or passage through suitable openings can also select transverse orientations.

Air sound is longitudinal and has no such transverse orientation choices, so this light observation does not extend the polarization mechanism to sound.

f = (3.00 × 10⁸)/0.030 = 1.0 × 10¹⁰ Hz. B may point along −y for propagation +x. A constant-phase wavefront is a y–z plane.

No. The wave consists of fields and does not require vibrating material particles. Its vacuum speed remains c.

The requested order is radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. Within visible it is red, orange, yellow, green, blue, violet.

They have the same speed c and travel time D/c. Gamma has greater f because its λ is smaller. The ordering supports that comparison without a memorized numerical boundary.

Light often means visible electromagnetic radiation, but the term can also refer to electromagnetic radiation across all wavelengths.

Before passage, advancing emission points crowd fronts on the listener’s side, giving higher observed f. After passage, fronts arriving behind the departing source are farther apart, giving lower f.

Faster source motion changes each emission position by more, so front compression/expansion and frequency shifts are greater. More amplitude affects loudness, not front repetition.

Same velocity: the effects cancel and observed f equals rest f. Siren faster: their separation shrinks and the observer encounters fronts at a higher rate than source timing after both motion effects. Listener faster: separation grows and observed f is lower.

No. Source velocity changes wavelength in air; observer velocity changes how fast fronts are crossed. Equal closing speed alone does not fix both quantities. A quantitative formula is not required to identify this distinction.

At every listed point P + Q = 0, so the net displacement graph lies on equilibrium at this instant. In the linear model both pulses subsequently reappear with their original shapes and directions.

Q becomes [0,−1,−2,−1,0] mm, giving net [0,1,2,1,0] mm. The signs oppose wherever nonzero, so the overlap is destructive even though cancellation is incomplete.

Q is [0,1,2,1,0] mm, giving net [0,3,6,3,0] mm. Both displacements have the same sign wherever nonzero, so this is constructive interference; the net peak is 6 mm.

The fixed string requires L = nλ/2: fundamental λ = 1.20 m, f = 100 Hz; third λ = 0.40 m, f = 300 Hz. The third has nodes at 0,0.20,0.40,0.60 m and antinodes halfway between.

A closed–open pipe requires odd quarter wavelengths: f₁ = 120/(4 × 0.60) = 50 Hz, then 150 Hz, 250 Hz, … .

fbeat = |504 − 500| = 4 Hz. The fork phases repeatedly align and oppose, giving loudness maxima four times per second.

The fixed–loose string has a displacement node at x=0 and an antinode at x=L, so its fundamental has λ=4L=2.40 m and f=120/2.40=50 Hz.

The open–open pipe has displacement antinodes at both ends and one interior node in its fundamental; λ=2L=1.20 m and f=100 Hz.

A loose–loose string has displacement antinodes at both ends, like the open–open pipe; its fundamental also has λ=2L and f=100 Hz at the same wave speed.

y₁ = λL/a = 0.0050 m and y₂ = 0.010 m. The symmetric values are ±5.0 mm and ±10 mm. Central width is 10 mm.

The largest |y|/L is 0.010, so θ ≈ 0.57°, safely below 10°.

At u=0 the slit contributions reinforce, so the center is a maximum. Its bounding minima are u=±1, making central width 2 in u; a side lobe lies between consecutive nonzero minima and has width 1 in u. This maps to twice the neighboring dark-order interval on the small-angle screen.

Edge-to-edge path difference is a sin θ = λ. Points separated by a/2 differ in path by λ/2, so corresponding contributions from the two halves oppose.

Sound wavelengths can be comparable to a doorway and spread strongly around its edges. Visible wavelengths are much smaller, so most doorway transmission travels nearly straight. A narrow slit spreads most across its narrow width and makes bands; a circular pinhole makes a circular pattern.

Use one laser and fixed L; measure symmetric first-minimum positions for several known a values. Plot half the central width against 1/a; slope predicts λL. Repeat measurements and use the band width/center uncertainty as horizontal-position uncertainty.

Keep incidence normal, brightness suitable, and angles small; terminate the beam below eye level.

Δy = λL/d = 3.0 mm. The envelope first zero is at |y| = λL/a = 12 mm.

The fourth interference maximum would be at 4 × 3.0 = 12 mm, coinciding with an envelope zero, so m = ±4 is missing.

Draw fine approximately equally spaced peaks under a broad central envelope, fading toward zeros at ±12 mm. Interference requires phase-sensitive superposition, a wave property demonstrated by Young’s experiment.

The extra path from one slit is the projection of separation onto the outgoing direction: $\Delta=d\sin\theta$. In-phase slit contributions reinforce when $\Delta=m\lambda$.

For small θ, sin θ ≈ tan θ = ymax/L, giving $d(y_{\max}/L)\approx m\lambda$. Opposing phase occurs at $\Delta=(m+\tfrac12)\lambda$.

d = 1/(4.00 × 10⁵) = 2.50 × 10⁻⁶ m. θred = asin(0.260) = 15.1°; θviolet = asin(0.160) = 9.21°.

m = 0 sends all wavelengths straight ahead, so the center is white.

mmax is the largest integer ≤ d/λ: red 3, violet 6. Adding slits at unchanged spacing sharpens the maxima but preserves d sin θ = mλ.

The air→film reflection reverses by π; the film→air reflection does not. With one relative reversal, 2nt = λ₀/2 gives t = 532/(4 × 1.33) = 100 nm.

At each interface some incident light reflects and some transmits. Absorption is neglected here, so energy carried by reflected and transmitted light accounts for the incident energy at a lossless boundary.

One-reversal destructive phase uses 2nt = mλ₀. The smallest positive case m = 1 gives t = 532/(2 × 1.33) = 200 nm.

For white light different thicknesses satisfy reinforcement for different λ₀; the observed reflected color therefore varies across the film.

Air→oil and oil→water both increase index, so both reflections reverse and their relative reflection phase is zero. The soap film has only one reversal. The constructive/destructive travel conditions therefore swap between these two stacks, but varying oil thickness still changes 2nt/λ₀ and can reinforce different reflected colors.

For a steady interface, incident power is reflected plus transmitted plus absorbed. Thus Pabs=100−12−68=20 mW.

Both reflections are lower-to-higher index and reverse by π; relative reflection phase is zero. Set 2nt = λ₀/2, so t = 560/(4 × 1.40) = 100 nm.

1.40 = √(1.00 × 1.96), giving matching ideal reflected amplitudes.

Refraction/transmission adds no abrupt reflection reversal in this model. Travel within the coating accumulates the propagation phase used in the optical-path calculation.

Changing λ₀ changes 2nt/λ₀; changing angle changes the optical travel path. The two reflected fields generally cease to oppose exactly, so the reflection can increase. Numerical AP thin-film analysis here is limited to normal incidence.

v = 2Lf₁ and v² = Fₜ/μ give f₁² = Fₜ/(4L²μ). Plot f₁² against tension. μ = 0.0100 kg/m, so slope = 1/(4 × 1.00² × 0.0100) = 25.0 Hz²/N.

The data give (4,100), (9,225), (16,400), all on a line of slope 25.0 through the origin.

Sweep frequency on both sides of the amplitude maximum, record the interval of near-resonance behavior, and repeat. Measure length and tension independently. A nominal end that moves appreciably violates a displacement node; the mode spacing or node placement can differ from the fixed–fixed prediction.

λ = Δyd/L = (0.0030)(0.00020)/1.20 = 5.0 × 10⁻⁷ m = 500 nm. a = λL/0.012 = 5.0 × 10⁻⁵ m = 0.050 mm.

m = 12/3 = 4 is missing. The 500 nm light has f = 6.0 × 10¹⁴ Hz, greater than the 650 nm light’s 4.62 × 10¹⁴ Hz.

Use the separation of symmetric ±m features divided by 2, and use differences between neighboring maxima. A common center offset cancels in these differences. A one-sided y measured from the wrong center biases λ or a; L and slide tolerances remain separate uncertainties.

f = c/λ₀ = 5.0 × 10¹⁴ Hz. λfilm = λ₀/n = 400 nm. Both coating reflections reverse, and t = λfilm/4 = 100 nm; √(1 × 2.25) = 1.50 matches amplitudes.

The polarizer changes transmitted intensity/orientation; a stationary interface changes speed and wavelength but preserves frequency.

sin θ = 600 nm/2000 nm = 0.300, so θ = 17.5°. Rotating the filter changes brightness but not wavelength, so it does not move the principal grating direction if light remains transmitted.