Spring-mass System

Question

A block of mass 2 kg attached to a light spring of force constant 200 N per metre oscillates on a frictionless table, so what is its period?

RankGuru · Accepted Answer

For a horizontal spring-mass oscillator, NCERT Class 11, Chapter 14 (Oscillations) gives the period as $T = 2\pi\sqrt{\dfrac{m}{k}}$, where $m$ is the attached mass and $k$ is the spring's force constant. Here $m = 2$ kg and $k = 200$ N/m, so the ratio is $m/k = 2/200 = 0.01$ s$^2$ and its square root is $0.1$ s. Then $T = 2\pi 	imes 0.1 = 0.628$ s. The restoring force $F = -kx$ makes the motion simple harmonic, and notably the period is independent of amplitude. The option 0.314 s comes from wrongly dropping the factor of 2 (using $\pi$ instead of $2\pi$). The option 1.256 s doubles the correct answer, as if $m/k$ were mishandled. The option 3.14 s ignores the square-root step and just uses $\pi$. It is useful to note why the frictionless table matters: with no damping the amplitude stays constant and the motion is ideal simple harmonic motion, so a single period describes every cycle. The angular frequency here is $\omega = \sqrt{k/m} = \sqrt{200/2} = 10$ rad/s, and $T = 2\pi/\omega = 2\pi/10 = 0.628$ s, which independently reproduces the result and cross-checks the arithmetic. A plausibility check: a fairly stiff 200 N/m spring with a 2 kg block should oscillate quickly, and a period well under one second is physically sensible; the units of $\sqrt{m/k}$ work out as $\sqrt{	ext{kg}/(	ext{N/m})}$, which reduces to seconds, confirming the dimensions.

Spring-mass System

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