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
A simple pendulum has a length of 0.4 m. Use g = 9.81 m/s². Find its period T for small swings.
L = 0.4 mg = 9.81 m/s²
T = s
2.Springs & pendulumsHard
A 15 g bullet moving at 350 m/s hits and stays in a 1.5 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 1400 N/m. Find the amplitude A of the resulting oscillation in cm.
m = 15 gv = 350 m/sM = 1.5 kgk = 1400 N/m
A = cm
3.Waves, sound & opticsMedium
A source of 600 Hz moving at 34 m/s and an observer moving at 15 m/s approach each other. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
f = 600 Hzv = 343 m/svs = 34 m/svo = 15 m/s
f′ = Hz
4.Waves, sound & opticsHard
Light of wavelength 500 nm falls at right angles on a grating with 1200 lines per millimetre. Find the angle θ1 of the order-1 maximum.
N = 1200 mm⁻¹λ = 500 nmm = 1
θ1 = °
3Solutions
1.Springs & pendulums
Answer: T = 1.269 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.
Pendulum period
T = 2π·√(0.4/9.81) = 1.269 s
Answer
T = 1.269 s
2.Springs & pendulums
Answer: A = 11.4 cm
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.
Speed after impact
V = m·v/(m + M) = 0.015·350/1.515 = 3.465 m/s
Angular frequency
ω = √(k/(m + M)) = √(1400/1.515) = 30.4 rad/s
Amplitude
A = V/ω = 3.465/30.4 = 0.114 m = 11.4 cm
Answer
A = 11.4 cm
3.Waves, sound & optics
Answer: f′ = 695.1 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|.
Doppler formula
f′ = f·(v + vo)/(v − vs) = 600·(343 + 15)/(343 − 34)
f′ = 695.1 Hz
Answer
f′ = 695.1 Hz
4.Waves, sound & optics
Answer: θ1 = 36.87 °
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, …
Line spacing
d = 1/1200 mm = 8.333 × 10⁻⁷ m
Grating equation
sin θ = mλ/d = 1·500 × 10⁻⁹/(8.333 × 10⁻⁷) = 0.6
θ = 36.87°
Answer
θ1 = 36.87°
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