Spring constant: static extension vs oscillation timing
A student determines the spring constant k of a light helical spring by two methods. Method A: the spring hangs vertically, a 200 g mass is hung from it, and the extension is read with a millimetre ruler as 4.9 cm. Method B: the same spring lies on a smooth horizontal bench, one end fixed and the other attached to a 200 g trolley; the trolley is pulled 5.0 cm from equilibrium and released, and 20 complete oscillations take 8.9 s. A partner objects that Method B is "invalid because the spring itself has mass". The teacher adds two facts: the ruler has a zero error of about 3 mm, and in Method A the 4.9 cm was measured from the top of the coil, a point that itself shifts as the spring stretches. Take g = 9.8 m/s², π² = 9.87, and the mass of the spring as 45 g.
1. Calculate k from each method and state whether the two values agree once their uncertainties are considered.
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Method A (static). At equilibrium the load balances the spring force:
k = mg/x = (0.200 × 9.8) / 0.0490 = 1.96 / 0.0490 ≈ 40 N/m
A 3 mm zero error means the true extension lies between 46 mm and 52 mm, giving k between 1.96/0.052 ≈ 37.7 and 1.96/0.046 ≈ 42.6 N/m, i.e. about ±6 %.
Method B (dynamic). T = 8.9/20 = 0.445 s, so
k = 4π²m/T² = (4 × 9.87 × 0.200) / (0.445)² = 7.896 / 0.198 ≈ 39.9 N/m
A reaction-time uncertainty of 0.2 s over 8.9 s is 2.2 % in T; since k ∝ T⁻², that is about 4.5 % in k (roughly 38–42 N/m).
Verdict: the two central values (40.0 and 39.9 N/m) agree almost exactly, and each lies inside the other's uncertainty band. Agreement here is therefore weaker evidence than it looks, because Method A's band is dominated by a systematic error rather than by scatter.
2. The partner objects that the oscillating spring's own mass invalidates Method B. Judge which of the two values is more trustworthy, and how the objection changes the comparison.
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The objection is physically correct but quantifiable, so it does not invalidate Method B - it shifts it. Only part of the spring moves at the trolley's speed, so the oscillating mass is m + m_spring/3:
m_eff = 0.200 + 0.045/3 = 0.200 + 0.015 = 0.215 kg
k = 4π²(0.215)/(0.445)² = (7.896 × 1.075)/0.198 ≈ 42.9 N/m
So the uncorrected Method B value is about 7 % too low, but the correction is a known, one-line calculation.
Method A's problem is different in kind. Reading extension from the top of the coil does not measure the spring's extension at all - the reference point moves during the measurement, and the size and even the sign of the error are unknown. A 3 mm zero error on a 49 mm extension is a further 6 %. Repeating Method A cannot remove either effect.
Judgement: Method B, corrected for spring mass (k ≈ 43 N/m), is more trustworthy. Its error is random and reduced by timing more oscillations; its systematic error is calculable. A measurement with a correctable systematic error beats one whose systematic error is unknown.
3. State the single best improvement to Method A and explain exactly what error it removes.
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The best change is not a better ruler but a better procedure: hang several different loads (say 100 g to 500 g), record the position of the same fixed point on the spring or its hook each time, and plot load (mg) against position. Take k from the gradient.
Why this works: any constant offset - the ruler's 3 mm zero error, the unloaded length, the position of the top of the coil - appears as an intercept, not a slope. Differencing consecutive points cancels it completely. The gradient also averages out random reading scatter over several points instead of relying on one 4.9 cm reading.
If only one load is allowed, clamp a pointer or fiducial mark to the bottom of the spring and sight it against a fixed scale attached to the stand, so that only the change in length is measured. That removes the moving-reference error, but the zero error only becomes harmless if you take the difference between loaded and unloaded readings rather than a single absolute reading. The graph method is preferred because it removes both problems at once.