Q.A block of mass $0.25\ \text{kg}$ is attached to a spring of force constant $k = 100\ \text{N/m}$ and set into oscillation on a frictionless horizontal surface. Find the time period of the resulting S.H.M.

The Spring-Mass System: Why a Weight on a Spring Oscillates

Imagine hanging a weight from a spring and giving it a gentle tug downward. It bounces back up, overshoots, comes down again, and keeps going. That rhythmic up-and-down motion is simple harmonic motion, and the time it takes to complete one full bounce — down, up, and back to the start — is called the period $T$.

The key question: what determines how fast or slow this bouncing happens? Intuition says two things matter: how heavy the weight is, and how stiff the spring is.

The Intuition

A heavier mass is harder to accelerate. It lumbers along, so the oscillation is slower — the period gets longer. A stiffer spring (one with a larger spring constant $k$) pulls back harder for the same stretch. That stronger restoring force whips the mass back faster, so the period gets shorter.

So the period $T$ should increase with mass $m$ and decrease with stiffness $k$. The exact relationship turns out to be:

The $2\pi$ factor comes from the geometry of circular motion (which underlies all simple harmonic motion), and the square root tells us that doubling the mass only multiplies the period by $\sqrt{2} \approx 1.4$, not by 2.

Where Does This Formula Come From?

For a mass on an ideal spring, the restoring force is Hooke's law: $F = -kx$, where $x$ is the displacement from equilibrium. Newton's second law $F = ma$ gives:

This is a differential equation whose solution is a sine or cosine wave. The angular frequency $\omega$ (how fast the oscillation goes in radians per second) comes out as:

Since period $T$ is the time for one complete cycle, and one cycle corresponds to $2\pi$ radians, we have $T = 2\pi / \omega$. Substituting $\omega$ gives the formula above.

What the Formula Tells You

• Mass and period: Double the mass → period increases by $\sqrt{2}$ (about 1.4 times). Quadruple the mass → period doubles.
• Stiffness and period: Double the spring constant → period decreases by $\sqrt{2}$ (about 0.7 times). Quadruple the stiffness → period halves.
• Independence from amplitude: The period does not depend on how far you pull the mass initially. A small bounce and a big bounce take exactly the same time. This is the hallmark of simple harmonic motion for an ideal spring. …