Q.A spring is connected to a mass m suspended from it and its time period for vertical oscillation is T. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillation is a) $T'=T\sqrt2$ b) $T'=T/\sqrt2$ c) $T'=2T$ d) $T'=T/2$

You have a spring. You push it — it compresses easily. Now imagine cutting that spring into two equal halves and pushing just one of those halves. It feels much harder to compress, doesn't it? That's the core intuition: a shorter piece of the same spring is stiffer.

Why? Think of the spring as a collection of tiny coils, each contributing a little bit of the total stretch. When you pull the full spring by some amount, every coil stretches a little. If you cut the spring in half, the same force now acts on only half the number of coils. Each coil must stretch twice as much to achieve the same overall extension. Since force is proportional to stretch (Hooke's law), the half-spring requires twice the force for the same total extension — its force constant doubles.

Let's make this precise. For a spring of original length $L$ and force constant $k$, Hooke's law says $F = k x$, where $x$ is the total extension. Now imagine the spring is made of $n$ identical coils in series. Each coil has its own force constant $k_{\text{coil}}$. When the full spring stretches by $x$, each coil stretches by $x/n$. The force through every coil is the same $F$, so for one coil:

But for the whole spring, $F = k x$. Equating the two expressions for $F$:

Now cut the spring into $n$ equal pieces. Each piece is exactly one coil (or one set of coils that behaves identically). Its force constant is $k_{\text{coil}} = n k$.

A common exam trap: students think cutting a spring weakens it because there's less material. The opposite is true — fewer coils in series means less "give" per unit force. The force constant is inversely proportional to the number of coils (or length), so a shorter spring is stiffer. …