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 1 m long has a period of 1.26 s for small swings. Find the gravitational acceleration g on that planet.
L = 1 mT = 1.26 s
g = m/s²
2.Springs & pendulumsHard
A 0.6 kg mass is attached to the lower end of an unstretched vertical spring with k = 420 N/m and released from rest. Use g = 9.81 m/s². Find the greatest extension xmax of the spring in cm.
m = 0.6 kgk = 420 N/mg = 9.81 m/s²
xmax = cm
3.Waves, sound & opticsMedium
An object 5 cm tall stands 4 cm in front of a thin diverging lens with a focal length of 12 cm. Find the image height hi (negative for an inverted image).
ho = 5 cmdo = 4 cmf = −12 cm
hi = cm
4.Waves, sound & opticsHard
A diffraction grating has 800 lines per millimetre. Light of wavelength 405 nm falls on it at right angles. What is the highest order mmax of bright maximum that can be seen?
N = 800 mm⁻¹λ = 405 nm
mmax =
3Solutions
1.Springs & pendulums
Answer: g = 24.87 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 = 4π²·1/1.26² = 24.87 m/s²
Answer
g = 24.87 m/s²
2.Springs & pendulums
Answer: xmax = 2.803 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.
Equilibrium and amplitude
x₀ = m·g/k = 0.6·9.81/420 = 0.01401 m = A
Lowest point
xmax = 2A = 0.02803 m = 2.803 cm
Answer
xmax = 2.803 cm
3.Waves, sound & optics
Answer: hi = 3.75 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) − 1/4
di = −3 cm
Magnification and height
m = −(−3)/4 = 0.75
hi = 0.75·5 = 3.75 cm
Answer
hi = 3.75 cm
4.Waves, sound & optics
Answer: mmax = 3
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 = 1/800 mm = 1.25 µm
Highest order
m ≤ d/λ = 1.25 µm/0.405 µm = 3.086 → mmax = 3
Answer
mmax = 3
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