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
How long must a simple pendulum be to have a period of 1.8 s? Use g = 9.81 m/s². Find L.
T = 1.8 sg = 9.81 m/s²
L = m
2.Springs & pendulumsHard
A 2.2 kg mass is attached to the lower end of an unstretched vertical spring with k = 170 N/m and released from rest. Use g = 9.81 m/s². Find the greatest speed vmax of the mass.
m = 2.2 kgk = 170 N/mg = 9.81 m/s²
vmax = m/s
3.Waves, sound & opticsMedium
Light travels in glycerol (n = 1.47) towards a flat boundary with water (n = 1.33). Find the critical angle θc beyond which it is totally reflected.
n1 = 1.47n2 = 1.33
θc = °
4.Waves, sound & opticsHard
Light of wavelength 500 nm passes through a single slit 0.15 mm wide. Find the width w of the central bright maximum on a screen 3 m away (between the first dark fringes on either side).
λ = 500 nma = 0.15 mmL = 3 m
w = mm
3Solutions
1.Springs & pendulums
Answer: L = 0.8051 m
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 the length
L = g·(T/2π)² = 9.81·(1.8/2π)² = 0.8051 m
Answer
L = 0.8051 m
2.Springs & pendulums
Answer: vmax = 1.116 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.
Equilibrium and amplitude
x₀ = m·g/k = 2.2·9.81/170 = 0.127 m = A
Greatest speed
The speed is largest at equilibrium.
ω = √(170/2.2) = 8.79 rad/s
vmax = A·ω = 0.127·8.79 = 1.116 m/s
Answer
vmax = 1.116 m/s
3.Waves, sound & optics
Answer: θc = 64.79 °
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.
Refraction at 90°
sin θc = n₂/n₁ = 1.33/1.47 = 0.9048
Critical angle
θc = arcsin(0.9048) = 64.79°
Answer
θc = 64.79°
4.Waves, sound & optics
Answer: w = 20 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 = 500 × 10⁻⁹·3/(0.15 × 10⁻³) = 10 mm
Width of the central maximum
w = 2y₁ = 20 mm
Answer
w = 20 mm
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