QJEEA-2024-04: JEE Advanced 2024 JEEA-2024-04 Solution: Q.4 A block of mass 5 kg moves along the x -direction subject…

Given:

• Mass (m): 5 kg
• Force (F): -20x + 10 N (where x is in meters)
• Initial Conditions (t = 0 s):Position (x) = 1 m
• Velocity (v) = 0 m/s (at rest)
• Target Time (t): π/4 s

Goal:

• Find the position (x) and momentum (p) of the block at t = π/4 s.

Step-by-Step Solution:

1. Find the Acceleration (a):

• We know F = ma (Newton's second law).
• Substitute the given force: -20x + 10 = 5a.
• Solve for acceleration: a = (-20x + 10) / 5 = -4x + 2 m/s².

1. Determine the Equilibrium Position (x_eq):

• At equilibrium, the net force is zero (F = 0), and therefore the acceleration is zero (a = 0).
• Set a = 0: -4x + 2 = 0.
• Solve for x: 4x = 2, so x = 0.5 m. This is the equilibrium position.

1. Rewrite Acceleration in Terms of Displacement from Equilibrium (X):

• Define the displacement from equilibrium as X = x - x_eq = x - 0.5.
• Therefore, x = X + 0.5.
• Substitute this into the acceleration equation: a = -4(X + 0.5) + 2.
• Simplify: a = -4X - 2 + 2 = -4X.

1. Recognize Simple Harmonic Motion (SHM):

• The equation a = -4X is in the form a = -ω²X, which is the characteristic equation for SHM.
• Therefore, ω² = 4, and the angular frequency ω = 2 rad/s.

1. Write the Equation for Position (x) as a Function of Time (t):

• The general equation for SHM is X = A cos(ωt + φ), where A is the amplitude and φ is the phase constant.
• since x = X + 0.5, then x = Acos(ωt+φ) + 0.5
• Use the initial conditions to find A and φ.
• At t = 0, x = 1: 1 = A cos(φ) + 0.5.
• Therefore, A cos(φ) = 0.5.
• At t = 0, v = 0. The velocity is the derivative of position: v = dx/dt = -Aω sin(ωt + φ).
• 0 = -A(2)sin(φ) therefore sin(φ) = 0.
• This means φ = 0.
• Therefore Acos(0) = 0.5. A = 0.5.
• The position equation is x = 0.5 cos(2t) + 0.5.

1. Calculate Position at t = π/4 s:

• Substitute t = π/4 into the position equation: x = 0.5 cos(2 * π/4) + 0.5.
• Simplify: x = 0.5 cos(π/2) + 0.5 = 0.5(0) + 0.5 = 0.5 m.

1. Calculate Velocity (v) as a Function of Time (t):

• Differentiate the position equation with respect to time: v = dx/dt = -0.5 * 2 sin(2t) = -sin(2t).

1. Calculate Velocity at t = π/4 s:

• Substitute t = π/4 into the velocity equation: v = -sin(2 * π/4) = -sin(π/2) = -1 m/s.

1. Calculate Momentum (p) at t = π/4 s:

• Momentum is p = mv.
• Substitute the values: p = 5 kg * (-1 m/s) = -5 kg m/s.

Conclusion:

• The position of the block at t = π/4 s is 0.5 m.
• The momentum of the block at t = π/4 s is -5 kg m/s