1. Initial SHM parameters

For a mass $m$ attached to a spring of constant $k$, the angular frequency is

$\omega = \sqrt{\frac{k}{m}}.$

Given amplitude $A$, when the block passes through the equilibrium position, its speed is maximum:

$v_{\max} = \omega A = A\sqrt{\frac{k}{m}}.$

2. Mass breaks at equilibrium position

At the equilibrium point, the spring is neither stretched nor compressed, so the spring force is zero.

When half the mass breaks off, the remaining mass becomes

$m' = \frac{m}{2}.$

At that instant:

• displacement $x=0$
• velocity of the remaining mass is still the same as just before breaking, because the separation happens instantaneously:

$v' = A\sqrt{\frac{k}{m}}.$

3. New angular frequency

For the remaining mass $m/2$, the new angular frequency is

$\omega' = \sqrt{\frac{k}{m/2}} = \sqrt{\frac{2k}{m}}.$

4. Find new amplitude

Since the mass is at the new equilibrium position ($x=0$) with speed $v'$, the amplitude for the new SHM is

$A' = \frac{v'}{\omega'}.$

Substitute values:

$A' = \frac{A\sqrt{k/m}}{\sqrt{2k/m}} = \frac{A}{\sqrt{2}}.$

Hence,

$f = \frac{A'}{A} = \frac{1}{\sqrt{2}}.$

5. Option check

• A: $1$ ❌
• B: $\frac{1}{2}$ ❌
• C: $\sqrt{2}$ ❌
• D: $\frac{1}{\sqrt{2}}$ ✅

Therefore, the correct answer is D.