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 3.5 kg mass on a spring with k = 280 N/m is pulled 11 cm in the positive direction and released from rest at t = 0. Find its velocity v at t = 1.05 s (with sign).
m = 3.5 kgk = 280 N/mA = 11 cmt = 1.05 s
v = m/s
2.Springs & pendulumsHard
A body oscillates in simple harmonic motion with period 2.9 s and amplitude 4 cm. What is the shortest time t it needs to move from its maximum displacement of 4 cm to a displacement of 3.2 cm?
T = 2.9 sA = 4 cmx = 3.2 cm
t = s
3.Waves, sound & opticsMedium
A small source sends sound equally in all directions. The level is 85 dB at 1 m from it. Find the level β2 at 8 m, ignoring absorption and reflections.
β1 = 85 dBr1 = 1 mr2 = 8 m
β2 = dB
4.Waves, sound & opticsHard
An object is 15 cm in front of a convex mirror with a radius of curvature of 16 cm. Find the image distance di, positive for an image in front of the mirror (real) and negative for one behind it (virtual).
do = 15 cmR = 16 cm
di = cm
3Solutions
1.Springs & pendulums
Answer: v = −0.03275 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 = √(280/3.5) = 8.944 rad/s
ω·t = 8.944·1.05 = 9.391 rad
Velocity
v = −A·ω·sin(ω·t) = −0.11·8.944·sin(9.391) = −0.03275 m/s
Answer
v = −0.03275 m/s
2.Springs & pendulums
Answer: t = 0.297 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/2.9 = 2.167 rad/s
Solve for the time
0.8 = cos(ω·t) → ω·t = arccos(0.8) = 0.6435 rad
t = 0.6435/2.167 = 0.297 s
Answer
t = 0.297 s
3.Waves, sound & optics
Answer: β2 = 66.94 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 = (1/8)² = 0.01563
New level
β₂ = 85 + 10·log₁₀(0.01563) = 66.94 dB
Answer
β2 = 66.94 dB
Every doubling of the distance lowers the level by about 6 dB.
4.Waves, sound & optics
Answer: di = −5.217 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.
Focal length
f = −R/2 = −8 cm
Mirror equation
1/di = 1/f − 1/do = 1/(−8) − 1/15
di = −5.217 cm
Ray diagram
virtual image behind the mirror
Answer
di = −5.217 cm
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