A block of mass $5$ kg moves along the $x$-direction subject to the force $F = (-20x + 10)$ N, with the value of $x$ in metre. At time $t = 0$ s, it is at rest at position $x = 1$ m. The position and momentum of the block at $t = (\pi/4)$ s are...

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JEE Advanced 2024 Paper 1 Physics Question 4 Solution

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.