Springs & pendulums
- Mass on a spring
- T = 2π·√(m/k)
- Pendulum
- T = 2π·√(L/g)
- Angular frequency
- ω = √(k/m) = 2π/T
- Maximum speed and acceleration
- vmax = A·ω, amax = A·ω²
- Energy
- E = ½·k·A²
Waves and Optics: cram pack
Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.
1Formulas and methods
Waves, sound & optics
- Wave speed
- v = fλ, T = 1/f
- String
- v = √(T/μ), μ = m/L
- Sound level
- β = 10·log10(I/I0), I0 = 10⁻¹² W/m²
- Point source
- I = P/(4πr²)
- Standing waves
- string or open pipe fn = n·v/(2L); closed pipe fn = n·v/(4L), n odd
- Doppler effect
- f′ = f·(v ± vo)/(v ∓ vs)
- Refractive index
- n = c/v, λ = λ0/n
- Snell’s law
- n1sin θ1 = n2sin θ2
- Critical angle
- sin θc = n2/n1
- Optical fibre
- NA = sin θa = √(n1² − n2²)
- Lenses and mirrors
- 1/f = 1/do + 1/di, m = −di/do, mirror f = R/2
- Double slit
- d·sin θ = mλ, ym = mλL/d
- Grating
- d = 1/N, d·sin θ = mλ, mmax ≤ d/λ
- Single slit
- a·sin θ = mλ (dark), w = 2λL/a
Method
- Identify the kind of problem: wave speed, sound level, standing wave, Doppler, refraction, lens or mirror, or interference.
- Write the relation first: v = fλ, β = 10·log(I/I₀), fₙ = n·v/(2L) or n·v/(4L), f′ = f(v ± vo)/(v ∓ vs), n₁sin θ₁ = n₂sin θ₂, 1/f = 1/d₀ + 1/dᵢ, d·sin θ = mλ.
- Convert units: nm, mm and cm to m, and check whether angles are measured from the normal.
- Check signs and plausibility: a virtual image has dᵢ < 0, an approaching source sounds higher, a closed pipe has only odd harmonics.
2Practice problems
1.Springs & pendulumsMedium
An astronaut on another planet finds that a simple pendulum 2 m long has a period of 2.98 s for small swings. Find the gravitational acceleration g on that planet.
L = 2 mT = 2.98 s
g = m/s²
2.Springs & pendulumsHard
A body oscillates in simple harmonic motion with period 1.9 s and amplitude 14 cm. What is the shortest time t it needs to move from its maximum displacement of 14 cm to a displacement of 10.5 cm?
T = 1.9 sA = 14 cmx = 10.5 cm
t = s
3.Waves, sound & opticsMedium
A string fixed at both ends is 0.5 m long; waves travel on it at 300 m/s. Find the frequency f1 of its fundamental (first harmonic).
v = 300 m/sn = 1L = 0.5 m
f1 = Hz
4.Waves, sound & opticsHard
Light of wavelength 650 nm falls on two slits 0.25 mm apart. The screen is 1.2 m away. Find the distance y1 from the central bright fringe to bright fringe number 1.
λ = 650 nmd = 0.25 mmL = 1.2 mm = 1
y1 = mm
3Solutions
1.Springs & pendulums
Answer: g = 8.891 m/s²
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Solve for g
g = 4π²·L/T² = 4π²·2/2.98² = 8.891 m/s²
Answer
g = 8.891 m/s²
2.Springs & pendulums
Answer: t = 0.2186 s
Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².
Angular frequency
ω = 2π/T = 2π/1.9 = 3.307 rad/s
Solve for the time
0.75 = cos(ω·t) → ω·t = arccos(0.75) = 0.7227 rad
t = 0.7227/3.307 = 0.2186 s
Answer
t = 0.2186 s
3.Waves, sound & optics
Answer: f1 = 300 Hz
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
Wavelength of this mode
λ = 2L/n = 2·0.5/1 = 1 m
Frequency
f = v/λ = 300/1 = 300 Hz
Answer
f1 = 300 Hz
4.Waves, sound & optics
Answer: y1 = 3.12 mm
Interference and diffraction: light from two slits a distance d apart reaches a point on the screen with a path difference d·sin θ. Bright fringes appear where d·sin θ = mλ (m = 0, ±1, ±2, …). For a screen at distance L ≫ d, the m-th bright fringe is at y = mλL/d and the fringe spacing is Δy = λL/d. A grating with N lines per millimetre has d = 1/N and sharp maxima at d·sin θ = mλ; since sin θ ≤ 1, the highest order is the largest whole number m ≤ d/λ. A single slit of width a has dark fringes at a·sin θ = mλ, m = ±1, ±2, …
Fringe spacing
Δy = λL/d = 650 × 10⁻⁹·1.2/(0.25 × 10⁻³) = 3.12 mm
Fringe number 1
y1 = 1·3.12 = 3.12 mm
Answer
y1 = 3.12 mm
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