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 0.7 kg mass on a spring with k = 170 N/m is pulled 14 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 1.5 s (with sign).
m = 0.7 kgk = 170 N/mA = 14 cmt = 1.5 s
x = cm
2.Springs & pendulumsHard
A 3.6 kg mass on a spring with k = 70 N/m is pulled 13 cm from equilibrium and released. Find its maximum speed vmax.
m = 3.6 kgk = 70 N/mA = 13 cm
vmax = m/s
3.Waves, sound & opticsMedium
An object 2 cm tall stands 50 cm in front of a thin converging lens with a focal length of 12 cm. Find the image height hi (negative for an inverted image).
ho = 2 cmdo = 50 cmf = 12 cm
hi = cm
4.Waves, sound & opticsHard
Light of wavelength 532 nm falls at right angles on a grating with 100 lines per millimetre. Find the angle θ3 of the order-3 maximum.
N = 100 mm⁻¹λ = 532 nmm = 3
θ3 = °
3Solutions
1.Springs & pendulums
Answer: x = −2.591 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.
Angular frequency
ω = √(170/0.7) = 15.58 rad/s
ω·t = 15.58·1.5 = 23.38 rad
Displacement
x = A·cos(ω·t) = 14·cos(23.38) = −2.591 cm
Answer
x = −2.591 cm
2.Springs & pendulums
Answer: vmax = 0.5732 m/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 = √(70/3.6) = 4.41 rad/s
Maximum speed
vmax = A·ω = 0.13·4.41 = 0.5732 m/s
Answer
vmax = 0.5732 m/s
3.Waves, sound & optics
Answer: hi = −0.6316 cm
Geometric optics: at a boundary, n₁·sin θ₁ = n₂·sin θ₂ (Snell's law), with angles measured from the normal and n = c/v. Going into a slower medium the ray bends towards the normal. From a denser medium (n₁ > n₂) beyond the critical angle θc = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/do + 1/di with the magnification m = −di/do. Sign convention: f > 0 for a converging lens and a concave mirror (f = R/2), f < 0 for a diverging lens and a convex mirror; di > 0 for a real image, di < 0 for a virtual one, and m < 0 for an inverted image.
Thin-lens equation
1/di = 1/f − 1/do = 1/12 − 1/50
di = 15.79 cm
Magnification and height
m = −15.79/50 = −0.3158
hi = −0.3158·2 = −0.6316 cm
Answer
hi = −0.6316 cm
4.Waves, sound & optics
Answer: θ3 = 9.184 °
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 = 1/100 mm = 1 × 10⁻⁵ m
Grating equation
sin θ = mλ/d = 3·532 × 10⁻⁹/(1 × 10⁻⁵) = 0.1596
θ = 9.184°
Answer
θ3 = 9.184°
ExamProblems · examproblems.com/courses/waves-and-optics