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 3.7 kg mass on a spring with k = 50 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 = 0.6 s (with sign).
m = 3.7 kgk = 50 N/mA = 13 cmt = 0.6 s
x = cm
2.Springs & pendulumsHard
A body oscillates in simple harmonic motion with period 2 s and amplitude 12 cm. What is the shortest time t it needs to move from equilibrium to a displacement of 9.6 cm?
T = 2 sA = 12 cmx = 9.6 cm
t = s
3.Waves, sound & opticsMedium
An object 1 cm tall stands 30 cm in front of a thin converging lens with a focal length of 10 cm. Find the image height hi (negative for an inverted image).
ho = 1 cmdo = 30 cmf = 10 cm
hi = cm
4.Waves, sound & opticsHard
An optical fibre has a core with n₁ = 1.5 and a cladding with n₂ = 1.49. Find the largest angle θa to the fibre axis at which light from air can enter the end of the fibre and still be guided by total internal reflection.
n1 = 1.5n2 = 1.49n0 = 1.00
θa = °
3Solutions
1.Springs & pendulums
Answer: x = −7.71 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 = √(50/3.7) = 3.676 rad/s
ω·t = 3.676·0.6 = 2.206 rad
Displacement
x = A·cos(ω·t) = 13·cos(2.206) = −7.71 cm
Answer
x = −7.71 cm
2.Springs & pendulums
Answer: t = 0.2952 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 = 2π/T = 2π/2 = 3.142 rad/s
Solve for the time
0.8 = sin(ω·t) → ω·t = arcsin(0.8) = 0.9273 rad
t = 0.9273/3.142 = 0.2952 s
Answer
t = 0.2952 s
3.Waves, sound & optics
Answer: hi = −0.5 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 = 1/10 − 1/30
di = 15 cm
Magnification and height
m = −15/30 = −0.5
hi = −0.5·1 = −0.5 cm
Answer
hi = −0.5 cm
4.Waves, sound & optics
Answer: θa = 9.957 °
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.
Critical angle at the cladding
sin θc = 1.49/1.5 = 0.9933 → θc = 83.38°
Numerical aperture
NA = √(1.5² − 1.49²) = 0.1729
Acceptance angle
θa = arcsin(0.1729) = 9.957°
Answer
θa = 9.957°
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