1. Write the equation of motion

Given $F=-20x+10$ and mass $m=5\text{ kg}.$

Using Newton's second law: $m\ddot x = -20x+10$ $5\ddot x = -20x+10$ $\ddot x +4x=2$

So the motion is simple harmonic about the equilibrium position.

2. Find the equilibrium position

At equilibrium, acceleration is zero: $-20x+10=0$ $x=\frac{1}{2}$

Let $y=x-\frac{1}{2}$ Then $\ddot y = \ddot x$

So the equation becomes $\ddot y +4y=0$

This is SHM with angular frequency $\omega=2\ \text{rad/s}.$

3. Apply initial conditions

At $t=0$, $x=1 \Rightarrow y=1-\frac{1}{2}=\frac{1}{2}$

Also the block is at rest, so $\dot x(0)=0 \Rightarrow \dot y(0)=0$

General solution: $y=A\cos 2t + B\sin 2t$

Using $y(0)=\frac{1}{2}$: $A=\frac{1}{2}$

Using $\dot y(0)=0$: $\dot y=-2A\sin 2t +2B\cos 2t$ At $t=0$, $2B=0 \Rightarrow B=0$

Hence $y=\frac{1}{2}\cos 2t$

Therefore $x=\frac{1}{2}+\frac{1}{2}\cos 2t$

4. Find position at $t=\frac{\pi}{4}$

$2t=\frac{\pi}{2}$ so $\cos\frac{\pi}{2}=0$

Thus $x=\frac{1}{2}+\frac{1}{2}(0)=\frac{1}{2}\text{ m}$

5. Find momentum at $t=\frac{\pi}{4}$

Velocity: $\dot x=\dot y=-\sin 2t$

At $t=\frac{\pi}{4}$, $\dot x=-\sin\frac{\pi}{2}=-1\text{ m/s}$

Momentum: $p=mv=5\times(-1)=-5\text{ kg m/s}$

6. Match with options

At $t=\frac{\pi}{4}\text{ s}$,

• Position = $0.5\text{ m}$
• Momentum = $-5\text{ kg m/s}$

This matches Option C.

7. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.