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 3.9 kg mass on a spring with k = 210 N/m is pulled 9 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 0.85 s (with sign).
m = 3.9 kgk = 210 N/mA = 9 cmt = 0.85 s
x = cm
2.Springs & pendulumsHard
A 2.7 kg mass on a spring with k = 120 N/m is pulled 12 cm from equilibrium and released. Find the total energy E of the oscillation.
m = 2.7 kgk = 120 N/mA = 12 cm
E = J
3.Waves, sound & opticsMedium
One machine gives a sound level of 80 dB at a certain point. Find the level β at that point when 10 identical machines run at the same time (incoherent sources).
β1 = 80 dBn = 10
β = dB
4.Waves, sound & opticsHard
Light of wavelength 550 nm falls on two slits 0.25 mm apart. The screen is 3 m away. Find the distance y3 from the central bright fringe to bright fringe number 3.
λ = 550 nmd = 0.25 mmL = 3 mm = 3
y3 = mm
3Solutions
1.Springs & pendulums
Answer: x = 8.991 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 = √(210/3.9) = 7.338 rad/s
ω·t = 7.338·0.85 = 6.237 rad
Displacement
x = A·cos(ω·t) = 9·cos(6.237) = 8.991 cm
Answer
x = 8.991 cm
2.Springs & pendulums
Answer: E = 0.864 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 = √(120/2.7) = 6.667 rad/s
Energy
E = ½·k·A² = ½·120·0.12² = 0.864 J
Answer
E = 0.864 J
3.Waves, sound & optics
Answer: β = 90 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).
Total intensity
I = 10·I₁
Level
β = 80 + 10·log₁₀(10) = 90 dB
Answer
β = 90 dB
4.Waves, sound & optics
Answer: y3 = 19.8 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 = 550 × 10⁻⁹·3/(0.25 × 10⁻³) = 6.6 mm
Fringe number 3
y3 = 3·6.6 = 19.8 mm
Answer
y3 = 19.8 mm
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