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
A simple pendulum has a length of 1.5 m. Use g = 9.81 m/s². Find its period T for small swings.
L = 1.5 mg = 9.81 m/s²
T = s
2.Springs & pendulumsHard
A 1 kg mass on a spring with k = 340 N/m is pulled 5 cm from equilibrium and released. Find its maximum acceleration amax.
m = 1 kgk = 340 N/mA = 5 cm
amax = m/s²
3.Waves, sound & opticsMedium
An object 1 cm tall stands 12 cm in front of a thin converging lens with a focal length of 20 cm. Find the image height hi (negative for an inverted image).
ho = 1 cmdo = 12 cmf = 20 cm
hi = cm
4.Waves, sound & opticsHard
An object is 25 cm in front of a convex mirror with a radius of curvature of 50 cm. Find the image distance di, positive for an image in front of the mirror (real) and negative for one behind it (virtual).
do = 25 cmR = 50 cm
di = cm
3Solutions
1.Springs & pendulums
Answer: T = 2.457 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·√(1.5/9.81) = 2.457 s
Answer
T = 2.457 s
2.Springs & pendulums
Answer: amax = 17 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 = √(340/1) = 18.44 rad/s
Maximum acceleration
amax = A·ω² = 0.05·18.44² = 17 m/s²
Answer
amax = 17 m/s²
3.Waves, sound & optics
Answer: hi = 2.5 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 = 1/20 − 1/12
di = −30 cm
Magnification and height
m = −(−30)/12 = 2.5
hi = 2.5·1 = 2.5 cm
Answer
hi = 2.5 cm
4.Waves, sound & optics
Answer: di = −12.5 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.
Focal length
f = −R/2 = −25 cm
Mirror equation
1/di = 1/f − 1/do = 1/(−25) − 1/25
di = −12.5 cm
Ray diagram
virtual image behind the mirror
Answer
di = −12.5 cm
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