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 1.5 m long has a period of 2.58 s for small swings. Find the gravitational acceleration g on that planet.
L = 1.5 mT = 2.58 s
g = m/s²
2.Springs & pendulumsHard
A 20 g bullet moving at 400 m/s hits and stays in a 2 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 1800 N/m. Find the amplitude A of the resulting oscillation in cm.
m = 20 gv = 400 m/sM = 2 kgk = 1800 N/m
A = cm
3.Waves, sound & opticsMedium
A small loudspeaker radiates 0.01 W of sound power equally in all directions. Find the sound intensity level β at 8 m.
P = 0.01 Wr = 8 mI0 = 10⁻¹² W/m²
β = dB
4.Waves, sound & opticsHard
Light of wavelength 600 nm passes through a single slit 0.08 mm wide. Find the width w of the central bright maximum on a screen 1.5 m away (between the first dark fringes on either side).
λ = 600 nma = 0.08 mmL = 1.5 m
w = mm
3Solutions
1.Springs & pendulums
Answer: g = 8.896 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 = 4π²·1.5/2.58² = 8.896 m/s²
Answer
g = 8.896 m/s²
2.Springs & pendulums
Answer: A = 13.27 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.
Speed after impact
V = m·v/(m + M) = 0.02·400/2.02 = 3.96 m/s
Angular frequency
ω = √(k/(m + M)) = √(1800/2.02) = 29.85 rad/s
Amplitude
A = V/ω = 3.96/29.85 = 0.1327 m = 13.27 cm
Answer
A = 13.27 cm
3.Waves, sound & optics
Answer: β = 70.95 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
I = P/(4πr²) = 1.243 × 10⁻⁵ W/m²
Level
β = 10·log₁₀(1.243 × 10⁻⁵/10⁻¹²) = 70.95 dB
Answer
β = 70.95 dB
4.Waves, sound & optics
Answer: w = 22.5 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, …
First dark fringe
y₁ = λL/a = 600 × 10⁻⁹·1.5/(0.08 × 10⁻³) = 11.25 mm
Width of the central maximum
w = 2y₁ = 22.5 mm
Answer
w = 22.5 mm
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