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 0.7 kg mass on a spring with k = 400 N/m oscillates on a frictionless surface with an amplitude of 15 cm. Find its speed v when it is 7 cm from equilibrium.
m = 0.7 kgk = 400 N/mA = 15 cmx = 7 cm
v = m/s
2.Springs & pendulumsHard
A 2.4 kg mass on a spring with k = 500 N/m is pulled 10 cm from equilibrium and released. Find the total energy E of the oscillation.
m = 2.4 kgk = 500 N/mA = 10 cm
E = J
3.Waves, sound & opticsMedium
A small source sends sound equally in all directions. The level is 60 dB at 5 m from it. Find the level β2 at 10 m, ignoring absorption and reflections.
β1 = 60 dBr1 = 5 mr2 = 10 m
β2 = dB
4.Waves, sound & opticsHard
A parking sensor at rest sends ultrasound of 40 kHz towards a car that moves away at 5 m/s. Find the frequency f″ of the echo received back at the sensor (speed of sound 343 m/s).
f = 40 kHzu = 5 m/sv = 343 m/s
f″ = Hz
3Solutions
1.Springs & pendulums
Answer: v = 3.171 m/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.
Angular frequency
ω = √(400/0.7) = 23.9 rad/s
Speed at x
v = ω·√(A² − x²) = 23.9·√(0.15² − 0.07²) = 3.171 m/s
Answer
v = 3.171 m/s
2.Springs & pendulums
Answer: E = 2.5 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 = √(500/2.4) = 14.43 rad/s
Energy
E = ½·k·A² = ½·500·0.1² = 2.5 J
Answer
E = 2.5 J
3.Waves, sound & optics
Answer: β2 = 53.98 dB
Waves: a periodic wave moves one wavelength in one period, so v = fλ with T = 1/f. On a stretched string v = √(T/μ), with T the tension and μ = m/L the mass per metre. Sound in air travels at about 343 m/s at 20 °C, light in vacuum at c = 3.00 × 10⁸ m/s. Loudness is measured on a logarithmic scale: β = 10·log₁₀(I/I₀) dB with I₀ = 10⁻¹² W/m², so 10 times the intensity adds 10 dB. A point source spreads its power P over a sphere: I = P/(4πr²).
Intensity ratio
I₂/I₁ = (5/10)² = 0.25
New level
β₂ = 60 + 10·log₁₀(0.25) = 53.98 dB
Answer
β2 = 53.98 dB
Every doubling of the distance lowers the level by about 6 dB.
4.Waves, sound & optics
Answer: f″ = 38851 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 − 5)/343 = 39417 Hz
Frequency of the echo
f″ = f′·343/(343 + 5) = 38851 Hz
Answer
f″ = 38851 Hz
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