Let $x(t)$ be the displacement of the block from its equilibrium position at time $t$. The free end of the right spring is being pulled with constant velocity $u$. Therefore, at time $t$, the position of the pulling end is $x_{pull} = ut$.

Forces acting on the block:
1. Force from the left spring (compressed/extended by $x$): $F_L = -kx$
2. Force from the right spring (extended by relative displacement): The extension is $(x_{pull} – x) = (ut – x)$.
$F_R = k(ut – x)$

Using Newton’s Second Law ($F_{net} = ma$): $$ m\ddot{x} = -kx + k(ut – x) $$ $$ m\ddot{x} = -2kx + kut $$ $$ \ddot{x} + \frac{2k}{m} x = \frac{k u}{m} t $$ Let $\omega^2 = \frac{2k}{m}$. The equation becomes: $$ \ddot{x} + \omega^2 x = \frac{\omega^2 u}{2} t $$

The solution consists of a Homogeneous part ($x_h$) and a Particular part ($x_p$).

1. Homogeneous Solution: $$ x_h(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t) $$ 2. Particular Solution: Since the driving term is linear in $t$, we assume $x_p(t) = At + B$. Substituting into the DE: $$ 0 + \omega^2 (At + B) = \frac{\omega^2 u}{2} t $$ $$ At + B = \frac{u}{2} t $$ Comparison gives $A = \frac{u}{2}$ and $B = 0$.
So, the general solution is: $$ x(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t) + \frac{u}{2}t $$

At $t=0$, the block is at rest ($x=0, \dot{x}=0$).
$x(0) = C_1 + 0 + 0 = 0 \Rightarrow C_1 = 0$.
Velocity $\dot{x}(t) = \frac{d}{dt} \left( C_2 \sin(\omega t) + \frac{u}{2}t \right) = C_2 \omega \cos(\omega t) + \frac{u}{2}$.
$\dot{x}(0) = C_2 \omega + \frac{u}{2} = 0 \Rightarrow C_2 = -\frac{u}{2\omega}$.

Final Velocity Equation: $$ v(t) = -\frac{u}{2} \cos(\omega t) + \frac{u}{2} = \frac{u}{2} \left[ 1 – \cos(\omega t) \right] $$ Final Displacement Equation: $$ x(t) = -\frac{u}{2\omega} \sin(\omega t) + \frac{u}{2}t $$

We need to find the distance moved when the block’s speed equals $u$. $$ v(t) = u $$ $$ \frac{u}{2} [1 – \cos(\omega t)] = u $$ $$ 1 – \cos(\omega t) = 2 \Rightarrow \cos(\omega t) = -1 $$ This implies $\omega t = \pi$, or $t = \frac{\pi}{\omega}$.

Substitute this time $t$ into the displacement equation $x(t)$: $$ x = -\frac{u}{2\omega} \sin(\pi) + \frac{u}{2} \left( \frac{\pi}{\omega} \right) $$ Since $\sin(\pi) = 0$: $$ x = \frac{u \pi}{2 \omega} $$ Substituting $\omega = \sqrt{\frac{2k}{m}}$: $$ x = \frac{u \pi}{2 \sqrt{\frac{2k}{m}}} = \frac{\pi u}{2} \sqrt{\frac{m}{2k}} $$