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Step-by-Step Solution
Step 1: Understand the initial conditions
A block of mass $m$ is attached to a spring and performs simple harmonic motion with amplitude $A$ on a frictionless surface. At the equilibrium position, the entire energy of the system is in the form of kinetic energy. If the angular frequency of the motion is $\omega_1$, then at the equilibrium position, the speed of the block $v_0$ is:
$v_0 = \omega_1 A = \sqrt{\frac{k}{m}} \, A$
Step 2: Mass breaks off at equilibrium
As soon as the block passes through the equilibrium position, half of its mass (i.e., $\frac{m}{2}$) breaks off. The remaining mass is $\frac{m}{2}$. Since there is no external horizontal force (the plane is frictionless), the velocity at that instant remains the same for the remaining mass. Let the new amplitude of oscillation be $A^1$. The angular frequency for the remaining mass is
$\omega = \sqrt{\frac{k}{\frac{m}{2}}} = \sqrt{\frac{2k}{m}}.$
Step 3: Equate the kinetic energies (speeds) at that instant
The speed of the remaining mass at the instant the mass breaks off is the same as the initial speed $v_0$. After the mass breaks off, the block (of mass $\frac{m}{2}$) will oscillate with amplitude $A^1$, for which its speed at equilibrium can be written as:
$v = \omega A^1 = \sqrt{\frac{2k}{m}} \, A^1.
Since the velocity does not suddenly change at that instant, we have $v = v_0$, so
$\sqrt{\frac{k}{m}} \, A = \sqrt{\frac{2k}{m}} \, A^1.
Step 4: Solve for the new amplitude
Divide both sides by $\sqrt{\frac{k}{m}}$ to obtain:
$A = \sqrt{2}\,A^1 \quad \Longrightarrow \quad A^1 = \frac{A}{\sqrt{2}}.
Therefore, the factor $f$ by which the amplitude changes is:
$\displaystyle f = \frac{A^1}{A} = \frac{1}{\sqrt{2}}.
Answer
The value of $f$ is $\displaystyle \frac{1}{\sqrt{2}}$.
