| Symbol | Quantity | Unit | Typical Formula |
|---|---|---|---|
| x | Displacement from equilibrium | m | ‑ |
| x_0 | Amplitude | m | ‑ |
| t | Time | s | ‑ |
| v | Velocity | m s⁻¹ | v = dx/dt |
| a | Acceleration | m s⁻² | a = dv/dt = d²x/dt² |
| ω | Angular frequency | rad s⁻¹ | ω = √(k/m) = √(g/L) … |
| f | Frequency | Hz | f = ω/2π |
| T | Period | s | T = 1/f = 2π/ω |
| φ | Phase constant | rad | ‑ |
| k | Spring constant | N m⁻¹ | ‑ |
| m | Mass | kg | ‑ |
| g | Acceleration due to gravity | m s⁻² | ≈9.81 |
| I | Moment of inertia | kg m² | ‑ |
| Q | Electric charge | C | ‑ |
| V | Potential difference | V | ‑ |
| R | Resistance | Ω | ‑ |
| C | Capacitance | F | ‑ |
| μ₀, ε₀ | Permeability & permittivity of free space | H m⁻¹, F m⁻¹ | ‑ |
| Equation | When to use |
|---|---|
| v = u + at | constant acceleration |
| s = ut + ½at² | displacement with known u, a |
| v² = u² + 2as | no time needed |
Example: A car accelerates from rest at 2 m s⁻² for 5 s. Find its final speed and distance travelled.
Example: A 3 kg block slides down a 30° incline with μk = 0.15. Find acceleration.
Component of weight down the plane: mg sin 30° = 3·9.81·0.5 = 14.7 N.
Friction: μk N = 0.15·mg cos 30° = 0.15·3·9.81·0.866 ≈ 3.8 N.
Net force = 14.7 – 3.8 = 10.9 N ⇒ a = F/m = 10.9/3 ≈ 3.6 m s⁻².
| Quantity | Formula | Units |
|---|---|---|
| Work | W = F·s·cosθ | J |
| Kinetic energy | K = ½mv² | J |
| Gravitational potential energy | U = mgh | J |
| Power | P = W/t = Fv | W |
Example: A 0.5 kg ball is dropped from 2 m. Find speed just before impact (ignore air resistance).
mgh = ½mv² ⇒ v = √(2gh) = √(2·9.81·2) ≈ 6.26 m s⁻¹.
Example (elastic): 0.2 kg bullet (v = 400 m s⁻¹) embeds in 0.8 kg block initially at rest. Find final speed.
p_initial = 0.2·400 = 80 kg m s⁻¹.
p_final = (0.2+0.8)v ⇒ v = 80/1.0 = 80 m s⁻¹.
| Quantity | Formula | Units |
|---|---|---|
| Stress | σ = F/A | Pa |
| Strain | ε = ΔL/L | ‑ (dimensionless) |
| Young’s modulus | E = σ/ε | Pa |
| Hooke’s law | F = –k x | N |
Example: A steel wire (A = 2 mm², E = 2×10¹¹ Pa) is stretched by 1 mm. Find the force.
ΔL/L = 0.001/0.5 m = 2×10⁻³ (assuming original length 0.5 m).
σ = E·ε = 2×10¹¹·2×10⁻³ = 4×10⁸ Pa.
F = σA = 4×10⁸·2×10⁻⁶ = 800 N.
Example: A string fixed at both ends vibrates at its fundamental frequency of 120 Hz with wavelength 2 m. Find wave speed.
v = fλ = 120·2 = 240 m s⁻¹.
| Quantity | Formula |
|---|---|
| Charge | Q = It |
| Current density | J = σE |
| Coulomb’s law | F = k_e Q₁Q₂/r² |
Example: A current of 3 A flows for 5 s. Find charge transferred.
Q = It = 3·5 = 15 C.
Example (parallel): Two resistors, 4 Ω and 6 Ω, are connected across a 12 V battery. Find total current.
1/R_eq = 1/4 + 1/6 = (3+2)/12 = 5/12 ⇒ R_eq = 12/5 = 2.4 Ω.
I = V/R_eq = 12/2.4 = 5 A.
Example: A sample contains 1.0 g of ^{60}Co (t_{½}=5.27 yr). How many atoms remain after 10 yr?
Number initially N₀ = (1.0 g / 60 g mol⁻¹)·6.02×10²³ ≈ 1.00×10²².
λ = ln2 / 5.27 yr = 0.131 yr⁻¹.
N = N₀ e^{-λ·10} ≈ 1.00×10²²·e^{-1.31} ≈ 2.7×10²¹ atoms.
$$\frac{d^{2}x}{dt^{2}} = -\omega^{2}x \qquad\text{(acceleration opposite to displacement)}
$$x(t)=x_{0}\sin(\omega t+\phi)$$
$$v(t)=\frac{dx}{dt}= \omega x_{0}\cos(\omega t+\phi)$$
$$a(t)=\frac{d^{2}x}{dt^{2}}= -\omega^{2}x_{0}\sin(\omega t+\phi)= -\omega^{2}x(t)$$
| Quantity | Expression |
|---|---|
| Maximum kinetic energy | K_{max}=½mω²x₀² |
| Maximum potential (elastic) energy | U_{max}=½k x₀² = ½mω²x₀² |
| Total mechanical energy | E = K + U = ½k x₀² (constant) |
Problem: A 0.5 kg mass attached to a spring (k = 200 N m⁻¹) is pulled 0.10 m from equilibrium and released from rest. Find its displacement after 0.05 s.
Example: A car rounds a curve of radius 50 m at 20 m s⁻¹. Find centripetal acceleration.
a_c = v²/r = 20²/50 = 8 m s⁻².
| Quantity | Formula |
|---|---|
| Gravitational force | F = G M m / r² |
| Field strength | g = GM / r² |
| Orbital speed (circular) | v = √(GM/r) |
| Period of satellite | T = 2π√(r³/GM) |
Example: 1 mol of an ideal gas is heated from 300 K to 400 K at constant volume. Find ΔU.
ΔU = (3/2)nRΔT = 1.5·8.31·(400–300) ≈ 1245 J.
Equation of motion (damped):
$$m\ddot{x}+b\dot{x}+kx=0$$Natural frequency ω₀ = √(k/m); damping ratio ζ = b/(2√{mk}).
Resonance occurs when driving frequency ω ≈ ω₀ (small ζ).
| Quantity | Formula |
|---|---|
| Electric field | E = F/q = V/d (for uniform field) |
| Magnetic field (Biot‑Savart) | B = μ₀I/2πr (long straight wire) |
| Force on moving charge | F = q(v×B) |
| Step | Action |
|---|---|
| 1 | Set up a light string of length L = 0.50 m with a small bob; attach a stopwatch. |
| 2 | Displace the bob by ≤ 5° and release; measure the time for 20 complete oscillations. Repeat three times. |
| 3 | Average the three total times, then divide by 20 to obtain T. |
| 4 | Calculate g using g = 4π²L/T² and propagate uncertainties from L (±0.001 m) and T (stopwatch ±0.2 s). |
Typical result: T ≈ 1.42 s → g ≈ 9.8 m s⁻² (within experimental uncertainty).
¨x = –ω²x and explain its physical meaning?ω for a mass‑spring system and for a simple pendulum?x(t)=x₀ sin(ωt+φ) and determine φ from any pair of initial conditions?f, ω, T and use the correct units?a = –ω²x directly to find displacement, velocity or acceleration at a given time?Mass‑spring system: a block of mass m attached to a horizontal spring (constant k) displaced a distance x from equilibrium. Arrows indicate the restoring force F = –kx and the resulting acceleration a = –ω²x toward the equilibrium position.
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