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Physical phenomena and models

Learning objectives

  • Outcome

    Distinguish a phenomenon from a model: given a list of five statements, identify which are observed phenomena and which are model idealizations, justifying each choice in one sentence.

  • Outcome

    State at least three idealization assumptions for a given phenomenon and write down the model equation built on those assumptions.

  • Outcome

    Assess the range of applicability: compute the model prediction, find the relative deviation δ=∣xmodel−xexp∣/xexp\delta=|x_{model}-x_{exp}|/x_{exp} from the measurement, and name the condition under which the model fails.

Understand the idea

Physical Phenomena and Models: How Scientific Descriptions Are Built

Text

A physical phenomenon is a regularity of nature observed in experiment: free fall, heating, electrification. To describe a phenomenon quantitatively, physics builds a model — a simplified image of the object in which some properties are discarded and the rest are replaced by idealizations (point mass, ideal gas, harmonic oscillator). A model always introduces assumptions: for a falling stone one neglects air resistance and treats gg as constant. The model links quantities by equations, and its predictions are compared with measurements: for example, the distance travelled in uniformly accelerated motion from rest is

s=12at2.s=\frac{1}{2}at^2 .

Agreement within experimental uncertainty does not prove the model 'true'; it only confirms that the model is applicable under these conditions. Hence every model has a range of applicability: a point mass works when the body's size is small compared with the distance travelled, an ideal gas works at low density and far from condensation. Scientific description is a cycle: observation → idealization → equation → prediction → test → refinement.

Worked examples

Direct

Problem

Classify each item as a physical phenomenon (P) or a model (M): (a) free fall of a stone; (b) point mass; (c) ideal gas; (d) electrification of a comb by friction; (e) harmonic oscillator. For each model, name the property that was discarded.

Solution

Phenomena: (a) and (d) — they are observed and reproduced in experiment.
Models: (b), (c), (e).
Discarded properties: (b) point mass — size, shape and rotation of the body; (c) ideal gas — the molecules' own volume and their interaction; (e) harmonic oscillator — nonlinearity of the restoring force and energy losses.
Conclusion: a phenomenon is given in observation, while a model is constructed by a human and always contains assumptions.

Application

Problem

A student measures the fall time of a steel ball from a height h=2.0h=2.0 m and obtains texp=0.66t_{exp}=0.66 s (in seconds). Build the model 'point mass without air resistance', obtain the prediction tmodt_{mod}, compute the relative deviation δ\delta, and list the model assumptions.

Solution

Assumptions: the body is a point mass; only gravity acts; there is no air resistance; the initial speed is v0=0v_0=0; g=9.8g=9.8 (in m/s²); the height is small compared with the Earth's radius.
From h=12gt2h=\frac{1}{2}gt^2 it follows that

tmod=2hg=2⋅2.09.8≈0.64.t_{mod}=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\cdot 2.0}{9.8}}\approx 0.64 .

Deviation from experiment:

δ=∣tmod−texp∣texp=∣0.64−0.66∣0.66≈0.03=3%.\delta=\frac{|t_{mod}-t_{exp}|}{t_{exp}}=\frac{|0.64-0.66|}{0.66}\approx 0.03=3\%.

A deviation of 3% lies within the scatter of a hand measurement, so under these conditions the model is applicable. With a light ball or a larger height, air resistance becomes noticeable and the model leaves its range of applicability — then the drag force must be added to the equation.

Transfer

Problem

For a pendulum of length L=1.00L=1.00 m the small-oscillation model gives T0=2πL/gT_0=2\pi\sqrt{L/g}. Measurements give T=2.01T=2.01 s at amplitude θ0=5∘\theta_0=5^\circ and T=2.09T=2.09 s at θ0=45∘\theta_0=45^\circ. Compute T0T_0, find δ\delta for both cases, check the correction T≈T0(1+θ0216)T\approx T_0\left(1+\frac{\theta_0^2}{16}\right) (angle in radians), and state the range of applicability of the small-oscillation model.

Solution

Model prediction:

T0=2πLg=2π1.009.8≈2.006.T_0=2\pi\sqrt{\frac{L}{g}}=2\pi\sqrt{\frac{1.00}{9.8}}\approx 2.006 .

Case θ0=5∘\theta_0=5^\circ: δ=∣2.006−2.01∣2.01≈0.002=0.2%\delta=\frac{|2.006-2.01|}{2.01}\approx 0.002=0.2\% — the model performs excellently.
Case θ0=45∘\theta_0=45^\circ: θ0=0.785\theta_0=0.785 rad, the correction gives T≈2.006(1+0.785216)≈2.08T\approx 2.006\left(1+\frac{0.785^2}{16}\right)\approx 2.08, close to the measured 2.092.09; but the plain T0T_0 formula gives δ=∣2.006−2.09∣2.09≈0.04=4%\delta=\frac{|2.006-2.09|}{2.09}\approx 0.04=4\%.
Conclusion: from the correction T≈T0(1+θ0216)T\approx T_0\left(1+\frac{\theta_0^2}{16}\right) the relative period error is approximately θ02/16\theta_0^2/16 (angle in radians). The 1%1\% level is reached at θ0≈0.16=0.40\theta_0\approx\sqrt{0.16}=0.40 rad ≈23∘\approx 23^\circ: at θ0=20∘\theta_0=20^\circ (0.3490.349 rad) the error is ≈0.349216≈0.8%\approx\frac{0.349^2}{16}\approx 0.8\%, still below 1%1\%. Hence the 1%1\% boundary of the small-oscillation model is about 23∘23^\circ; the rule 'θ0≲10∘\theta_0\lesssim 10^\circ' is only a conservative rule of thumb with a safety margin, not the 1%1\% boundary. At larger amplitudes the model leaves its range of applicability, and the refined formula with the amplitude correction is required.

Practice

Text

A safe self-study plan in a simulation (no real weights and no sharp objects).

  1. Open any pendulum simulation (for example, Pendulum Lab) and fix L=1.00L=1.00 m, the mass and gg at their default values.
  2. For amplitudes θ0=2∘,5∘,10∘,20∘,45∘\theta_0=2^\circ,5^\circ,10^\circ,20^\circ,45^\circ, measure the time of 5 full oscillations and find the period TT by dividing by 5.
  3. Compare each measurement with the model T0=2πL/gT_0=2\pi\sqrt{L/g} and compute δ\delta.
  4. Plot δ(θ0)\delta(\theta_0) and mark the 1% level.
  5. Record in your notebook: which assumptions the model contains, at which amplitude it leaves its range of applicability, and which correction must be introduced.
    All measurements are made in the simulation; carry out real pendulum measurements only with a fixed support and no risk of the load falling.

Visual reference

Purpose

To show the cycle of building a scientific description and the range of applicability: how an observed phenomenon maps onto an idealized model, which assumptions are discarded, and how the deviation grows beyond the range of applicability.

Representation

Diagram (flowchart): 'Phenomenon → idealization (list of assumptions) → model equations → prediction → comparison with experiment → refinement of the model', with a feedback loop from the comparison back to the idealization. Next to it, a graph: the X axis is the amplitude θ0\theta_0 from 00 to 50∘50^\circ, the Y axis is the relative deviation δ\delta in percent; the measured points lie on a curve growing roughly as θ02\theta_0^2, and the horizontal dashed line at 1%1\% marks the range of applicability of the small-oscillation model (it crosses the curve near 23∘23^\circ).

Real-world examples

A 100-metre sprinter and the point model

Scenario

A runner covers 100 m100\ \text{m} in 10.5 s10.5\ \text{s}. To estimate the average speed, the coach divides distance by time: v=100 m10.5 s≈9.5 m/s.v=\frac{100\ \text{m}}{10.5\ \text{s}}\approx 9.5\ \text{m/s}. The runner's height (1.8 m1.8\ \text{m}), arm swing and body sway are not included in the calculation.

Connection

This is a direct test of replacing a real object with a point mass: the model keeps only the properties needed for the quantity v=s/tv=s/t, and discards everything else as irrelevant to the stated task.

The phone's accelerometer during a fall

Scenario

A phone of mass 0.20 kg0.20\ \text{kg} rests on a table and then falls from a height of 1.2 m1.2\ \text{m}. The sensor reads an acceleration near 9.8 m/s29.8\ \text{m/s}^2 in free fall and a support force F≈2.0 NF\approx 2.0\ \text{N} at rest. In the calculation the phone is treated as a rigid body with the sensor at a single point.

Connection

This checks that a model can be matched against measurements: the prediction F=mg=0.20 kg⋅9.8 m/s2≈2.0 NF=mg=0.20\ \text{kg}\cdot 9.8\ \text{m/s}^2\approx 2.0\ \text{N} is compared with the sensor reading, and any discrepancy marks the limits of the model's validity.

Load on a 1.2 km bridge

Scenario

A bridge span of length L=1200 mL=1200\ \text{m} carries a convoy of trucks with a total mass m=3.0⋅105 kgm=3.0\cdot 10^{5}\ \text{kg}. The engineer computes the weight of the load P=mg=3.0⋅105 kg⋅9.8 m/s2≈2.9 MNP=mg=3.0\cdot 10^{5}\ \text{kg}\cdot 9.8\ \text{m/s}^2\approx 2.9\ \text{MN} and distributes it along the length. The shape of the span, wind gusts of 15 m/s15\ \text{m/s} and a temperature of +30 ∘C+30\ ^\circ\text{C} are left out of this estimate and checked with separate models.

Connection

The example shows a complex phenomenon split into separate models: one describes only the weight of the load, another thermal expansion at +30 ∘C+30\ ^\circ\text{C}, a third the wind; together they cover what a single model cannot.