Waves and Optics: cram pack

Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.

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1Formulas and methods

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, 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

  1. Identify the kind of problem: wave speed, sound level, standing wave, Doppler, refraction, lens or mirror, or interference.
  2. 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.
  3. Convert units: nm, mm and cm to m, and check whether angles are measured from the normal.
  4. 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
L = 1.7 m

A simple pendulum has a length of 1.7 m. Use g = 9.81 m/s². Find its period T for small swings.

L = 1.7 mg = 9.81 m/s²
T = s
2.Springs & pendulumsHard
2.7 kgk = 500 N/mA = 16 cm

A 2.7 kg mass on a spring with k = 500 N/m is pulled 16 cm from equilibrium and released. Find its maximum acceleration amax.

m = 2.7 kgk = 500 N/mA = 16 cm
amax = m/s²
3.Waves, sound & opticsMedium
P8 × 75 dB

One machine gives a sound level of 75 dB at a certain point. Find the level β at that point when 8 identical machines run at the same time (incoherent sources).

β1 = 75 dBn = 8
β = dB
4.Waves, sound & opticsHard
f1​ = 5 cmf2​ = 15 cm15 cm15 cm

Two thin lenses share an axis, 15 cm apart: lens 1 with f₁ = 5 cm and lens 2 with f₂ = 15 cm. An object stands 15 cm in front of lens 1. Find the position di2 of the final image measured from lens 2 (positive beyond lens 2, real-is-positive convention).

f1 = 5 cmf2 = 15 cms = 15 cmdo = 15 cm
di2 = cm

3Solutions

1.Springs & pendulums

Answer: T = 2.616 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.

Pendulum period

T = 2π·√(1.7/9.81) = 2.616 s

Answer

T = 2.616 s
2.Springs & pendulums

Answer: amax = 29.63 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.

Angular frequency

ω = √(500/2.7) = 13.61 rad/s

Maximum acceleration

amax = A·ω² = 0.16·13.61² = 29.63 m/s²

Answer

amax = 29.63 m/s²
3.Waves, sound & optics

Answer: β = 84.03 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 = 8·I₁

Level

β = 75 + 10·log₁₀(8) = 84.03 dB

Answer

β = 84.03 dB
4.Waves, sound & optics

Answer: di2 = −15 cm

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.

Image of lens 1

1/di1 = 1/5 − 1/15 → di1 = 7.5 cm

Object for lens 2

do2 = 15 − 7.5 = 7.5 cm

Final image

1/di2 = 1/15 − 1/7.5 → di2 = −15 cm

Answer

di2 = −15 cm

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