Spring-mass System

For a horizontal spring-mass oscillator, NCERT Class 11, Chapter 14 (Oscillations) gives the period as $T=2\pi \sqrt{\frac{m}{k}}$, 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 $T=2\pi \times 0.1=0.628$ 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 $\sqrt{\text{kg}/(\text{N/m})}$, which reduces to seconds, confirming the dimensions.