A Spring in Action?

📌 CONCEPT: When an external force is applied to a system undergoing simple harmonic motion, it can alter its amplitude, phase, or frequency. In this case, the upward impulse given to the 2 kg block attached to a horizontal spring will affect its subsequent oscillations.

📐 RULE / FORMULA: The impulse-momentum theorem states that impulse is equal to the change in momentum, which is the product of mass and velocity. This can be expressed as J = Δp = m * Δv. For a system undergoing simple harmonic motion, the velocity at the equilibrium position is zero, so the impulse will primarily affect the amplitude of the oscillations.

💡 WORKED EXAMPLE: Suppose the block is initially at rest at the equilibrium position. When the upward impulse is given, the block's velocity increases. Using the impulse-momentum theorem, we can calculate the new velocity: J = Δp = m * Δv => Δv = J / m = 5 Ns / 2 kg = 2.5 m/s. This increased velocity will result in a greater amplitude of oscillations.

⚠️ COMMON MISTAKE: Students often overlook the fact that the impulse will primarily affect the amplitude of the oscillations, rather than their frequency. They may mistakenly assume that the impulse will change the spring constant or the equilibrium position, which is not the case.