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 1.4 kg mass on a spring with k = 20 N/m is pulled 5 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 0.8 s (with sign).
m = 1.4 kgk = 20 N/mA = 5 cmt = 0.8 s
x = cm
2.Springs & pendulumsHard
A 3.5 kg mass on a spring with k = 420 N/m is pulled 13 cm from equilibrium and released. Find the total energy E of the oscillation.
m = 3.5 kgk = 420 N/mA = 13 cm
E = J
3.Waves, sound & opticsMedium
An object stands 20 cm in front of a thin converging lens with a focal length of 25 cm. Find the magnification m (negative for an inverted image).
do = 20 cmf = 25 cm
m =
4.Waves, sound & opticsHard
Light of wavelength 480 nm falls on two slits 0.1 mm apart. The screen is 3 m away. Find the distance y3 from the central bright fringe to bright fringe number 3.
λ = 480 nmd = 0.1 mmL = 3 mm = 3
y3 = mm
3Solutions
1.Springs & pendulums
Answer: x = −4.965 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 = √(20/1.4) = 3.78 rad/s
ω·t = 3.78·0.8 = 3.024 rad
Displacement
x = A·cos(ω·t) = 5·cos(3.024) = −4.965 cm
Answer
x = −4.965 cm
2.Springs & pendulums
Answer: E = 3.549 J
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 = √(420/3.5) = 10.95 rad/s
Energy
E = ½·k·A² = ½·420·0.13² = 3.549 J
Answer
E = 3.549 J
3.Waves, sound & optics
Answer: m = 5
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 = 1/25 − 1/20
di = −100 cm
Magnification
m = −di/do = −(−100)/20 = 5
Answer
m = 5
4.Waves, sound & optics
Answer: y3 = 43.2 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 = 480 × 10⁻⁹·3/(0.1 × 10⁻³) = 14.4 mm
Fringe number 3
y3 = 3·14.4 = 43.2 mm
Answer
y3 = 43.2 mm
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