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 0.8 kg mass on a spring with k = 450 N/m is pulled 10 cm from equilibrium and released. Find the total energy E of the oscillation.
m = 0.8 kgk = 450 N/mA = 10 cm
E = J
3.Waves, sound & opticsMedium
A pipe open at both ends is 1.2 m long. The speed of sound is 343 m/s. Find the frequency f4 of its harmonic number 4.
v = 343 m/sn = 4L = 1.2 m
f4 = Hz
4.Waves, sound & opticsHard
An optical fibre has a core with n₁ = 1.48 and a cladding with n₂ = 1.4. 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 = 1.48n2 = 1.4n0 = 1.00
θa = °
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: E = 2.25 J
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 = √(450/0.8) = 23.72 rad/s
Energy
E = ½·k·A² = ½·450·0.1² = 2.25 J
Answer
E = 2.25 J
3.Waves, sound & optics
Answer: f4 = 571.7 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|.
Wavelength of this mode
λ = 2L/n = 2·1.2/4 = 0.6 m
Frequency
f = v/λ = 343/0.6 = 571.7 Hz
Answer
f4 = 571.7 Hz
4.Waves, sound & optics
Answer: θa = 28.69 °
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 = 1.4/1.48 = 0.9459 → θc = 71.08°
Numerical aperture
NA = √(1.48² − 1.4²) = 0.48
Acceptance angle
θa = arcsin(0.48) = 28.69°
Answer
θa = 28.69°
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