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 2.9 kg mass on a spring with k = 390 N/m is pulled 12 cm in the positive direction and released from rest at t = 0. Find its velocity v at t = 1.15 s (with sign).
m = 2.9 kgk = 390 N/mA = 12 cmt = 1.15 s
v = m/s
2.Springs & pendulumsHard
A 1.7 kg mass on a spring with k = 170 N/m is pulled 17 cm from equilibrium and released. Find its maximum speed vmax.
m = 1.7 kgk = 170 N/mA = 17 cm
vmax = m/s
3.Waves, sound & opticsMedium
A small source sends sound equally in all directions. The level is 95 dB at 10 m from it. Find the level β2 at 100 m, ignoring absorption and reflections.
β1 = 95 dBr1 = 10 mr2 = 100 m
β2 = dB
4.Waves, sound & opticsHard
Two slits 0.15 mm apart are lit with monochromatic light. On a screen 1.5 m away, bright fringe number 2 lies 10 mm from the central fringe. Find the wavelength λ.
d = 0.15 mmL = 1.5 my2 = 10 mmm = 2
λ = nm
3Solutions
1.Springs & pendulums
Answer: v = −0.9685 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 = √(390/2.9) = 11.6 rad/s
ω·t = 11.6·1.15 = 13.34 rad
Velocity
v = −A·ω·sin(ω·t) = −0.12·11.6·sin(13.34) = −0.9685 m/s
Answer
v = −0.9685 m/s
2.Springs & pendulums
Answer: vmax = 1.7 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 = √(170/1.7) = 10 rad/s
Maximum speed
vmax = A·ω = 0.17·10 = 1.7 m/s
Answer
vmax = 1.7 m/s
3.Waves, sound & optics
Answer: β2 = 75 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 = (10/100)² = 0.01
New level
β₂ = 95 + 10·log₁₀(0.01) = 75 dB
Answer
β2 = 75 dB
Every doubling of the distance lowers the level by about 6 dB.
4.Waves, sound & optics
Answer: λ = 500 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 = y·d/(mL) = 0.01·0.00015/(2·1.5) = 5 × 10⁻⁷ m = 500 nm
Answer
λ = 500 nm
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