Combination Of Springs

Question

An engineer mounts a 2 kg instrument between two identical springs, each of stiffness 300 N/m, with both springs connected on the same side in parallel. What is the oscillation frequency of the instrument?

RankGuru · Accepted Answer

When two springs act in parallel on the same mass, each contributes its own restoring force for a common displacement, so the effective stiffness adds: $k_{eff} = k_1 + k_2$. Here $k_{eff} = 300 + 300 = 600	ext{ N/m}$. The oscillation frequency is $f = \frac{1}{2\pi}\sqrt{k_{eff}/m}$, which gives $f = \frac{1}{2\pi}\sqrt{600/2} = \frac{1}{2\pi}\sqrt{300} = \frac{17.32}{6.283} \approx 2.76	ext{ Hz}$. The value $1.95	ext{ Hz}$ results from treating the springs as a single 300 N/m spring, ignoring the parallel combination. The value $3.90	ext{ Hz}$ doubles the correct frequency by mistakenly using $4k$. The value $0.38	ext{ Hz}$ inverts the result, reporting a period-like number in place of frequency. This applies the NCERT rule that parallel springs add their constants while series springs add reciprocals. As a sanity check, adding a second spring stiffens the system, so the frequency must exceed the single-spring value of about $1.95	ext{ Hz}$, and $2.76	ext{ Hz}$ correctly reflects the factor $\sqrt{2}$ increase.

Combination Of Springs

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