A mass of 2 kg is attached to a horizontal spring with spring constant 50 N/m. A sudden horizontal force of 100 N is applied and then removed. What is the maximum displacement of the mass from its equilibrium position during the resulting oscillation?

Step-by-step solution

1. Calculate the initial displacement caused by the applied force using Hooke's law $x_0=\frac{F}{k}$.
2. Insert $F=100\text{N}$ and $k=50\text{N/m}$ to obtain $x_0=\frac{100}{50}=2\text{m}$.
3. Recognize that after the force is removed the mass starts oscillating from the displaced position $x_0$, so the amplitude of the subsequent SHM is $A=x_0$.
4. Since amplitude equals maximum displacement from equilibrium, the maximum displacement is 2 m.

Concept tested: Amplitude of mass-spring SHM after impulsive force

Published 7 October 2026 · How this solution was prepared and checked