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 = 1.2 m

A simple pendulum has a length of 1.2 m. Use g = 9.81 m/s². Find its period T for small swings.

L = 1.2 mg = 9.81 m/s²
T = s
2.Springs & pendulumsHard
1.7 kgk = 30 N/mA = 2 cm

A 1.7 kg mass on a spring with k = 30 N/m is pulled 2 cm from equilibrium and released. Find its maximum speed vmax.

m = 1.7 kgk = 30 N/mA = 2 cm
vmax = m/s
3.Waves, sound & opticsMedium
FFdo​ = 8 cmf = 25 cm

An object stands 8 cm in front of a thin converging lens with a focal length of 25 cm. Find the image distance di. Use the sign convention 1/f = 1/do + 1/di (f < 0 for a diverging lens, di < 0 for a virtual image).

do = 8 cmf = 25 cm
di = cm
4.Waves, sound & opticsHard
Fdo​ = 12 cmR = 20 cm

An object is 12 cm in front of a convex mirror with a radius of curvature of 20 cm. Find the magnification m (negative for an inverted image).

do = 12 cmR = 20 cm
m =

3Solutions

1.Springs & pendulums

Answer: T = 2.198 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·√(1.2/9.81) = 2.198 s

Answer

T = 2.198 s
2.Springs & pendulums

Answer: vmax = 0.08402 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 = √(30/1.7) = 4.201 rad/s

Maximum speed

vmax = A·ω = 0.02·4.201 = 0.08402 m/s

Answer

vmax = 0.08402 m/s
3.Waves, sound & optics

Answer: di = −11.76 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.

Thin-lens equation

1/di = 1/f − 1/do = 1/25 − 1/8
di = −11.76 cm
FF8 cmf = 25 cm

Ray diagram

virtual, on the same side as the object

Answer

di = −11.76 cm
4.Waves, sound & optics

Answer: m = 0.4545

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.

Focal length

f = −R/2 = −10 cm

Mirror equation

1/di = 1/f − 1/do = 1/(−10) − 1/12
di = −5.455 cm
F12 cmf = −10 cm

Magnification

m = −(−5.455)/12 = 0.4545

Answer

m = 0.4545

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