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 1.7 s? Use g = 9.81 m/s². Find L.
T = 1.7 sg = 9.81 m/s²
L = m
2.Springs & pendulumsHard
A 3.6 kg mass on a spring with k = 80 N/m is pulled 14 cm from equilibrium and released. Find its maximum acceleration amax.
m = 3.6 kgk = 80 N/mA = 14 cm
amax = m/s²
3.Waves, sound & opticsMedium
A pipe open at both ends must have its harmonic number 5 at 572 Hz. The speed of sound is 343 m/s. Find the length L of the pipe.
v = 343 m/sn = 5f5 = 572 Hz
L = m
4.Waves, sound & opticsHard
A sensor at rest sends ultrasound of 40 kHz towards a car that moves away at 20 m/s and mixes the echo with the emitted signal. Find the beat frequency fb (speed of sound 343 m/s).
f = 40 kHzu = 20 m/sv = 343 m/s
fb = Hz
3Solutions
1.Springs & pendulums
Answer: L = 0.7181 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·(1.7/2π)² = 0.7181 m
Answer
L = 0.7181 m
2.Springs & pendulums
Answer: amax = 3.111 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 = √(80/3.6) = 4.714 rad/s
Maximum acceleration
amax = A·ω² = 0.14·4.714² = 3.111 m/s²
Answer
amax = 3.111 m/s²
3.Waves, sound & optics
Answer: L = 1.499 m
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
λ = v/f = 343/572 = 0.5997 m
Length
L = nλ/2 = 5·0.5997/2 = 1.499 m
Answer
L = 1.499 m
4.Waves, sound & optics
Answer: fb = 4408 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 − 20)/343 = 37668 Hz
Frequency of the echo
f″ = f′·343/(343 + 20) = 35592 Hz
Beat frequency
fb = |f″ − f| = 4408 Hz
Answer
fb = 4408 Hz
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