Question 4

The force is $F = -20x + 10 = -20(x - 0.5)$. This is a restoring force for SHM about the equilibrium position $x_{eq} = 0.5$ m. The equation of motion is $m \frac{d^2x}{dt^2} = -k(x - x_{eq})$, where $m = 5$ kg and $k = 20$ N/m. The angular frequency is $\omega = \sqrt{k/m} = \sqrt{20/5} = 2$ rad/s. The position as a function of time is $x(t) = 0.5 + A \cos(\omega t + \phi)$. At $t = 0$, $v = 0$, which implies $\phi = 0$. At $t = 0$, $x = 1$ m, so $1 = 0.5 + A \cos(0) \Rightarrow A = 0.5$ m. Thus, $x(t) = 0.5 + 0.5 \cos(2t)$. At $t = \pi/4$ s: $x(\pi/4) = 0.5 + 0.5 \cos(2 \cdot \pi/4) = 0.5 + 0.5 \cos(\pi/2) = 0.5$ m. The velocity is $v(t) = \frac{dx}{dt} = -0.5(2) \sin(2t) = -\sin(2t)$. At $t = \pi/4$ s, $v(\pi/4) = -\sin(\pi/2) = -1$ m/s. The momentum is $p = mv = 5 \times (-1) = -5$ kg m/s.