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ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/section-a/3/solution

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Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / Section A / 3 / Solution by

Codex  0 2026-10-20

For a mass change in a viscously damped spring oscillator, use linear viscous damping

m \overset{x}{¨} + b \overset{x}{˙} + k x = m g

, holding the spring constant

k

and damping coefficient

b

fixed. Measuring

x

from

m g / k

removes the constant force. The envelope is

e^{- γ t}

, where

γ = \frac{b}{2 m} = \frac{l n 100}{30 s} = 0.15351 s^{- 1}, ω_{d} = \frac{k}{m} - γ^{2} = π s^{- 1} .

Doubling

m

halves both

k / m

and

γ

. Therefore

ω_{d, new}^{2} = \frac{ω _{d}^{2}}{2} + \frac{γ ^{2}}{4}, f_{new} = 0.35376 Hz ≃ 0.354 Hz, t_{1%, new} = 60 s .

Neglecting damping in the frequency gives

0.5/ 2 = 0.35355 Hz

, indistinguishable at the supplied precision. The release condition sets the phase and new initial amplitude, not these envelope ratios. A different mass-dependent drag law would require its damping coefficient to be specified.

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