A 10 kg collar attached to a spring with force constant 1000 N/m slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 20 cm and released. What is the period of oscillation? A) 0.854 s B) 0.243 s C) 0.521 s D) 0.628 s

What matters for the period

For a frictionless horizontal mass-spring system, the period depends only on the mass $m$ and spring constant $k$, not on the release displacement (amplitude) as long as Hooke's law applies.

Use the simple harmonic motion period formula

For a mass on a spring, $$ T=2\pi\sqrt{\frac{m}{k}}. $$ Here, $m=10\,\text{kg}$ and $k=1000\,\text{N/m}$.

Compute $T$ carefully

$$ T=2\pi\sqrt{\frac{10}{1000}} =2\pi\sqrt{0.01} =2\pi(0.1) =0.2\pi \approx 0.628\,\text{s}. $$

Match to the multiple-choice options

$0.628\,\text{s}$ corresponds to option D.