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 1.3 s? Use g = 9.81 m/s². Find L.
T = 1.3 sg = 9.81 m/s²
L = m
2.Springs & pendulumsHard
A 12 g bullet moving at 450 m/s hits and stays in a 1 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 1500 N/m. Find the amplitude A of the resulting oscillation in cm.
m = 12 gv = 450 m/sM = 1 kgk = 1500 N/m
A = cm
3.Waves, sound & opticsMedium
A source of 300 Hz moves at 30 m/s away from a stationary observer. The speed of sound is 343 m/s. Find the frequency f′ that the observer hears.
f = 300 Hzv = 343 m/svs = 30 m/s
f′ = Hz
4.Waves, sound & opticsHard
An optical fibre has a core with n₁ = 1.62 and a cladding with n₂ = 1.54. Find the largest angle θa to the fibre axis at which light from air can enter the end of the fibre and still be guided by total internal reflection.
n1 = 1.62n2 = 1.54n0 = 1.00
θa = °
3Solutions
1.Springs & pendulums
Answer: L = 0.4199 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·(1.3/2π)² = 0.4199 m
Answer
L = 0.4199 m
2.Springs & pendulums
Answer: A = 13.86 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) = 0.012·450/1.012 = 5.336 m/s
Angular frequency
ω = √(k/(m + M)) = √(1500/1.012) = 38.5 rad/s
Amplitude
A = V/ω = 5.336/38.5 = 0.1386 m = 13.86 cm
Answer
A = 13.86 cm
3.Waves, sound & optics
Answer: f′ = 275.9 Hz
Standing waves and the Doppler effect: a string fixed at both ends, or a pipe open at both ends, fits whole half-wavelengths: L = nλ/2, fn = n·v/(2L), n = 1, 2, 3, … A pipe closed at one end has a node at the closed end and an antinode at the open end: L = nλ/4 with odd n only, fn = n·v/(4L). When source or observer move along the line between them, f′ = f·(v ± vo)/(v ∓ vs): use the upper signs for motion towards each other. Two waves of nearly equal frequency beat at |f₁ − f₂|.
Doppler formula
f′ = f·(v − vo)/(v + vs) = 300·(343 − 0)/(343 + 30)
f′ = 275.9 Hz
Answer
f′ = 275.9 Hz
4.Waves, sound & optics
Answer: θa = 30.18 °
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.
Critical angle at the cladding
sin θc = 1.54/1.62 = 0.9506 → θc = 71.92°
Numerical aperture
NA = √(1.62² − 1.54²) = 0.5028
Acceptance angle
θa = arcsin(0.5028) = 30.18°
Answer
θa = 30.18°
ExamProblems · examproblems.com/courses/waves-and-optics