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 1.5 kg mass on a spring with k = 170 N/m is pulled 4 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 0.9 s (with sign).
m = 1.5 kgk = 170 N/mA = 4 cmt = 0.9 s
x = cm
2.Springs & pendulumsHard
A 0.9 kg mass on a spring with k = 310 N/m is pulled 5 cm from equilibrium and released. Find its maximum speed vmax.
m = 0.9 kgk = 310 N/mA = 5 cm
vmax = m/s
3.Waves, sound & opticsMedium
A small source sends sound equally in all directions. The level is 75 dB at 2 m from it. Find the level β2 at 10 m, ignoring absorption and reflections.
β1 = 75 dBr1 = 2 mr2 = 10 m
β2 = dB
4.Waves, sound & opticsHard
A grating with 200 lines per millimetre shows the order-3 maximum of a spectral line at 23.2° from the straight-through direction. Find the wavelength λ of the line.
N = 200 mm⁻¹θ3 = 23.2°m = 3
λ = nm
3Solutions
1.Springs & pendulums
Answer: x = −3.951 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 = √(170/1.5) = 10.65 rad/s
ω·t = 10.65·0.9 = 9.581 rad
Displacement
x = A·cos(ω·t) = 4·cos(9.581) = −3.951 cm
Answer
x = −3.951 cm
2.Springs & pendulums
Answer: vmax = 0.928 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 = √(310/0.9) = 18.56 rad/s
Maximum speed
vmax = A·ω = 0.05·18.56 = 0.928 m/s
Answer
vmax = 0.928 m/s
3.Waves, sound & optics
Answer: β2 = 61.02 dB
Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).
Intensity ratio
I₂/I₁ = (2/10)² = 0.04
New level
β₂ = 75 + 10·log₁₀(0.04) = 61.02 dB
Answer
β2 = 61.02 dB
Every doubling of the distance lowers the level by about 6 dB.
4.Waves, sound & optics
Answer: λ = 656.6 nm
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, …
Wavelength
λ = d·sin θ/m = 5 × 10⁻⁶·sin 23.2°/3 = 656.6 nm
Answer
λ = 656.6 nm
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