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 1.3 kg mass on a spring with k = 400 N/m oscillates on a frictionless surface with an amplitude of 12 cm. Find its speed v when it is 11 cm from equilibrium.
m = 1.3 kgk = 400 N/mA = 12 cmx = 11 cm
v = m/s
2.Springs & pendulumsHard
A 10 g bullet moving at 350 m/s hits and stays in a 0.5 kg block resting on a frictionless surface. The block is attached to a horizontal spring with k = 1800 N/m. Find the amplitude A of the resulting oscillation in cm.
m = 10 gv = 350 m/sM = 0.5 kgk = 1800 N/m
A = cm
3.Waves, sound & opticsMedium
A string fixed at both ends must have its harmonic number 4 at 800 Hz; waves travel on it at 240 m/s. Find its length L.
v = 240 m/sn = 4f4 = 800 Hz
L = m
4.Waves, sound & opticsHard
A parking sensor at rest sends ultrasound of 25 kHz towards a car that approaches at 20 m/s. Find the frequency f″ of the echo received back at the sensor (speed of sound 343 m/s).
f = 25 kHzu = 20 m/sv = 343 m/s
f″ = Hz
3Solutions
1.Springs & pendulums
Answer: v = 0.8412 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 = √(400/1.3) = 17.54 rad/s
Speed at x
v = ω·√(A² − x²) = 17.54·√(0.12² − 0.11²) = 0.8412 m/s
Answer
v = 0.8412 m/s
2.Springs & pendulums
Answer: A = 11.55 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.
Speed after impact
V = m·v/(m + M) = 0.01·350/0.51 = 6.863 m/s
Angular frequency
ω = √(k/(m + M)) = √(1800/0.51) = 59.41 rad/s
Amplitude
A = V/ω = 6.863/59.41 = 0.1155 m = 11.55 cm
Answer
A = 11.55 cm
3.Waves, sound & optics
Answer: L = 0.6 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 = 240/800 = 0.3 m
Length
L = nλ/2 = 4·0.3/2 = 0.6 m
Answer
L = 0.6 m
4.Waves, sound & optics
Answer: f″ = 28096 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′ = 25000·(343 + 20)/343 = 26458 Hz
Frequency of the echo
f″ = f′·343/(343 − 20) = 28096 Hz
Answer
f″ = 28096 Hz
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