To find the position and momentum of the block at t = π 4 s t = \frac{\pi}{4}\text{ s} t = 4 π s
The force acting on the block of mass m = 5 kg m = 5\text{ kg} m = 5 kg F ( x ) = − 20 x + 10 N F(x) = -20x + 10\text{ N} F ( x ) = − 20 x + 10 N
Using F = m d 2 x d t 2 F = m \frac{d^2 x}{dt^2} F = m d t 2 d 2 x 5 d 2 x d t 2 = − 20 x + 10 5 \frac{d^2 x}{dt^2} = -20x + 10 5 d t 2 d 2 x = − 20 x + 10
Dividing both sides by 5 5 5 d 2 x d t 2 = − 4 x + 2 = − 4 ( x − 0.5 ) \frac{d^2 x}{dt^2} = -4x + 2 = -4\left(x - 0.5\right) d t 2 d 2 x = − 4 x + 2 = − 4 ( x − 0.5 )
Let X = x − 0.5 X = x - 0.5 X = x − 0.5 t t t d 2 X d t 2 = d 2 x d t 2 \frac{d^2 X}{dt^2} = \frac{d^2 x}{dt^2} d t 2 d 2 X = d t 2 d 2 x X X X d 2 X d t 2 = − 4 X \frac{d^2 X}{dt^2} = -4X d t 2 d 2 X = − 4 X
This is the differential equation of Simple Harmonic Motion (SHM) about the mean position x mean = 0.5 m x_{\text{mean}} = 0.5\text{ m} x mean = 0.5 m ω 2 = 4 ⟹ ω = 2 rad/s \omega^2 = 4 \implies \omega = 2\text{ rad/s} ω 2 = 4 ⟹ ω = 2 rad/s
The general solution for X ( t ) X(t) X ( t ) X ( t ) = A cos ( ω t + ϕ ) X(t) = A \cos(\omega t + \phi) X ( t ) = A cos ( ω t + ϕ )
Given initial conditions at t = 0 s t = 0\text{ s} t = 0 s
The block is at x ( 0 ) = 1 m ⟹ X ( 0 ) = 1 − 0.5 = 0.5 m x(0) = 1\text{ m} \implies X(0) = 1 - 0.5 = 0.5\text{ m} x ( 0 ) = 1 m ⟹ X ( 0 ) = 1 − 0.5 = 0.5 m
The block is at rest, so v ( 0 ) = 0 m/s ⟹ d X d t ∣ t = 0 = 0 v(0) = 0\text{ m/s} \implies \left.\frac{dX}{dt}\right|_{t=0} = 0 v ( 0 ) = 0 m/s ⟹ d t d X t = 0 = 0
From these initial conditions, the amplitude is A = 0.5 m A = 0.5\text{ m} A = 0.5 m ϕ = 0 \phi = 0 ϕ = 0
Thus, the position as a function of time t t t x ( t ) = 0.5 + X ( t ) = 0.5 + 0.5 cos ( 2 t ) m x(t) = 0.5 + X(t) = 0.5 + 0.5 \cos(2t)\text{ m} x ( t ) = 0.5 + X ( t ) = 0.5 + 0.5 cos ( 2 t ) m
The velocity of the block as a function of time is:
v ( t ) = d x d t = − 0.5 × 2 sin ( 2 t ) = − sin ( 2 t ) m/s v(t) = \frac{dx}{dt} = -0.5 \times 2 \sin(2t) = -\sin(2t)\text{ m/s} v ( t ) = d t d x = − 0.5 × 2 sin ( 2 t ) = − sin ( 2 t ) m/s
Now, evaluate position and momentum at t = π 4 s t = \frac{\pi}{4}\text{ s} t = 4 π s
Position:
x ( π 4 ) = 0.5 + 0.5 cos ( 2 × π 4 ) = 0.5 + 0.5 cos ( π 2 ) = 0.5 m x\left(\frac{\pi}{4}\right) = 0.5 + 0.5 \cos\left(2 \times \frac{\pi}{4}\right) = 0.5 + 0.5 \cos\left(\frac{\pi}{2}\right) = 0.5\text{ m} x ( 4 π ) = 0.5 + 0.5 cos ( 2 × 4 π ) = 0.5 + 0.5 cos ( 2 π ) = 0.5 m
Momentum:
p ( π 4 ) = m ⋅ v ( π 4 ) = 5 × [ − sin ( 2 × π 4 ) ] = 5 × [ − sin ( π 2 ) ] = − 5 kg m/s p\left(\frac{\pi}{4}\right) = m \cdot v\left(\frac{\pi}{4}\right) = 5 \times \left[-\sin\left(2 \times \frac{\pi}{4}\right)\right] = 5 \times \left[-\sin\left(\frac{\pi}{2}\right)\right] = -5\text{ kg m/s} p ( 4 π ) = m ⋅ v ( 4 π ) = 5 × [ − sin ( 2 × 4 π ) ] = 5 × [ − sin ( 2 π ) ] = − 5 kg m/s
Thus, at t = π 4 s t = \frac{\pi}{4}\text{ s} t = 4 π s 0.5 m 0.5\text{ m} 0.5 m − 5 kg m/s -5\text{ kg m/s} − 5 kg m/s
This corresponds to Option (C).