Spring-mass Oscillator

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A 0.8 kg block resting on a frictionless bench is attached to a light spring of stiffness 200 N/m and set into oscillation. What is the time period of the resulting motion?

RankGuru · Accepted Answer

A mass attached to an ideal spring executes SHM with time period $T = 2\pi\sqrt{m/k}$, where $m$ is the oscillating mass and $k$ is the spring constant. This follows because the spring supplies a linear restoring force $F = -kx$, giving angular frequency $\omega = \sqrt{k/m}$. Substituting $m = 0.8	ext{ kg}$ and $k = 200	ext{ N/m}$ yields $T = 2\pi\sqrt{0.8/200} = 2\pi\sqrt{0.004} = 2\pi 	imes 0.0632 \approx 0.40	ext{ s}$. The value $0.13	ext{ s}$ wrongly drops the factor $2\pi$ and reports only $\sqrt{m/k}$ scaled incorrectly. The value $0.63	ext{ s}$ comes from inverting the ratio to $\sqrt{k/m}$ inside the root in part. The value $1.26	ext{ s}$ doubles the correct period, a common slip from using $4\pi$ instead of $2\pi$. This is the textbook spring-block result from the NCERT chapter on oscillations. As a plausibility check, a stiff spring with a modest mass should oscillate quickly, and a sub-second period is consistent with the relatively large stiffness of 200 N/m.

Spring-mass Oscillator

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