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
How long must a simple pendulum be to have a period of 1.5 s? Use g = 9.81 m/s². Find L.
T = 1.5 sg = 9.81 m/s²
L = m
2.Springs & pendulumsHard
A mass on a spring with k = 200 N/m oscillates on a frictionless surface with an amplitude of 4 cm. Find the energy U stored in the spring when the mass is 2 cm from equilibrium.
k = 200 N/mA = 4 cmx = 2 cm
U = J
3.Waves, sound & opticsMedium
Light travels in ethanol (n = 1.36) towards a flat boundary with air (n = 1.00). Find the critical angle θc beyond which it is totally reflected.
n1 = 1.36n2 = 1.00
θc = °
4.Waves, sound & opticsHard
Light of wavelength 532 nm falls on two slits 0.4 mm apart. The screen is 1.2 m away. Find the distance y4 from the central bright fringe to bright fringe number 4.
λ = 532 nmd = 0.4 mmL = 1.2 mm = 4
y4 = mm
3Solutions
1.Springs & pendulums
Answer: L = 0.5591 m
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 the length
L = g·(T/2π)² = 9.81·(1.5/2π)² = 0.5591 m
Answer
L = 0.5591 m
2.Springs & pendulums
Answer: U = 0.04 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.
Total energy
E = ½·k·A² = ½·200·0.04² = 0.16 J
Spring energy at x
U = ½·k·x² = ½·200·0.02² = 0.04 J
Answer
U = 0.04 J
3.Waves, sound & optics
Answer: θc = 47.33 °
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.
Refraction at 90°
sin θc = n₂/n₁ = 1.00/1.36 = 0.7353
Critical angle
θc = arcsin(0.7353) = 47.33°
Answer
θc = 47.33°
4.Waves, sound & optics
Answer: y4 = 6.384 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 = 532 × 10⁻⁹·1.2/(0.4 × 10⁻³) = 1.596 mm
Fringe number 4
y4 = 4·1.596 = 6.384 mm
Answer
y4 = 6.384 mm
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