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 1.6 kg mass on a spring with k = 130 N/m is pulled 13 cm in the positive direction and released from rest at t = 0. Find its displacement x from equilibrium at t = 0.8 s (with sign).
m = 1.6 kgk = 130 N/mA = 13 cmt = 0.8 s
x = cm
2.Springs & pendulumsHard
A 2.7 kg mass on a spring with k = 320 N/m is pulled 13 cm from equilibrium and released. Find the total energy E of the oscillation.
m = 2.7 kgk = 320 N/mA = 13 cm
E = J
3.Waves, sound & opticsMedium
An object stands 40 cm in front of a thin diverging lens with a focal length of 20 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 = 40 cmf = −20 cm
di = cm
4.Waves, sound & opticsHard
A sensor at rest sends ultrasound of 40 kHz towards a car that approaches at 15 m/s and mixes the echo with the emitted signal. Find the beat frequency fb (speed of sound 343 m/s).
f = 40 kHzu = 15 m/sv = 343 m/s
fb = Hz
3Solutions
1.Springs & pendulums
Answer: x = 7.794 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.
Angular frequency
ω = √(130/1.6) = 9.014 rad/s
ω·t = 9.014·0.8 = 7.211 rad
Displacement
x = A·cos(ω·t) = 13·cos(7.211) = 7.794 cm
Answer
x = 7.794 cm
2.Springs & pendulums
Answer: E = 2.704 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 = √(320/2.7) = 10.89 rad/s
Energy
E = ½·k·A² = ½·320·0.13² = 2.704 J
Answer
E = 2.704 J
3.Waves, sound & optics
Answer: di = −13.33 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) − 1/40
di = −13.33 cm
Ray diagram
virtual, on the same side as the object
Answer
di = −13.33 cm
4.Waves, sound & optics
Answer: fb = 3659 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|.
Frequency at the car
f′ = 40000·(343 + 15)/343 = 41749 Hz
Frequency of the echo
f″ = f′·343/(343 − 15) = 43659 Hz
Beat frequency
fb = |f″ − f| = 3659 Hz
Answer
fb = 3659 Hz
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