Derive T for horizontal and vertical spring. Effect of cutting spring in half.

Time Period of a Spring-Mass System — T = 2π√(m/k)

Question

A block of mass $m$ is attached to a spring of spring constant $k$ on a frictionless horizontal surface. Find the time period of oscillation. What happens to the time period if we cut the spring in half and attach the same block?

Solution — Step by Step

When the block is displaced by $x$ from equilibrium, the spring exerts a restoring force $F = -kx$ . The negative sign is crucial — it tells us the force always opposes displacement, which is the definition of SHM.

So: $ma = -kx$ , which gives $a = -\frac{k}{m}x$ .

The standard SHM equation is $a = -\omega^2 x$ . Comparing directly:

\omega^2 = \frac{k}{m} \implies \omega = \sqrt{\frac{k}{m}}

Time period $T$ is the time for one complete oscillation. Since $\omega = \frac{2\pi}{T}$ :

T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}

When you cut the spring in half, the length halves — but the spring constant doubles. Here’s why: spring constant $k$ is inversely proportional to natural length $L$ (a shorter spring is stiffer). So $k_{\text{new}} = 2k$ .

T_{\text{new}} = 2\pi\sqrt{\frac{m}{2k}} = \frac{T}{\sqrt{2}} \approx 0.707\,T

The time period decreases by a factor of $\sqrt{2}$ .

\boxed{T = 2\pi\sqrt{\frac{m}{k}}}

$m$ = mass of block (kg)
$k$ = spring constant (N/m)
$T$ is independent of amplitude

Why This Works

The formula $T = 2\pi\sqrt{m/k}$ tells a clean physical story. A heavier mass ( $m \uparrow$ ) has more inertia, so it resists changes to its motion — it oscillates more slowly, and $T$ increases. A stiffer spring ( $k \uparrow$ ) pulls the mass back more aggressively — it oscillates faster, and $T$ decreases.

The most important insight for exams: $T$ is completely independent of amplitude. Whether the block oscillates 1 cm or 10 cm, the time period is the same. This is the defining property of SHM, and CBSE boards love asking it as a one-mark “true or false.”

For the vertical spring case, the math works out identically — gravity shifts the equilibrium point down by $\frac{mg}{k}$ , but the restoring force about the new equilibrium is still $-kx$ . So the same formula applies, no modification needed.

Alternative Method — Energy Approach

We can arrive at the same result using energy conservation without ever writing $F = ma$ .

At maximum displacement $A$ (amplitude), all energy is potential: $E = \frac{1}{2}kA^2$ .

At equilibrium, all energy is kinetic: $E = \frac{1}{2}mv_{\max}^2$ .

In SHM, $v_{\max} = A\omega$ . Equating the energies:

\frac{1}{2}kA^2 = \frac{1}{2}m(A\omega)^2 \implies \omega = \sqrt{\frac{k}{m}}

Same result. This energy method is faster in JEE MCQs when they give you $v_{\max}$ and $A$ directly.

Common Mistake

Confusing spring constant and spring length when cutting.

Most students write $k_{\text{new}} = \frac{k}{2}$ (thinking “half the spring = half the $k$ ”). It’s exactly backwards. A shorter spring is harder to stretch — $k$ is inversely proportional to length. Cutting to half the length gives $k_{\text{new}} = 2k$ .

A quick way to remember: imagine cutting a rubber band. The shorter piece is definitely stiffer.

One-liner for board exams: If a spring of constant $k$ is cut into $n$ equal pieces, each piece has spring constant $nk$ . If pieces are joined in series, the combined $k = \frac{k}{n}$ (original value). This combination fact appeared in CBSE 2023 as a 2-mark question.

Question

Solution — Step by Step

Why This Works

Alternative Method — Energy Approach

Common Mistake

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