- #31
pb23me
- 204
- 0
thanx...that was a long one
What Is the Maximum Extension of the Spring in Simple Harmonic Motion?
- Thread starter pb23me
- Start date
Click For Summary
SUMMARY
The discussion focuses on calculating the maximum extension of a spring in simple harmonic motion (SHM) and the corresponding acceleration of a mass attached to it. The spring constant is given as k=25 N/m, and a 1 kg mass is attached. The maximum extension is determined to be 0.392 m, derived from the equation mgx = 1/2 kx². The angular frequency (ω) is calculated as 5 rad/s, leading to an acceleration of -4.9 m/s² at maximum extension.
PREREQUISITES- Understanding of Hooke's Law (F = -kx)
- Knowledge of simple harmonic motion (SHM) equations
- Familiarity with energy conservation principles in mechanics
- Basic calculus concepts, particularly derivatives
- Study the derivation of the equations of motion for simple harmonic oscillators.
- Explore the relationship between angular frequency (ω) and spring constant (k) using T = 2π√(m/k).
- Learn about energy conservation in oscillatory systems, specifically how potential and kinetic energy interchange.
- Investigate the effects of damping on simple harmonic motion and how it alters maximum extension.
Students and educators in physics, particularly those focusing on mechanics and oscillatory motion, as well as anyone preparing for exams involving simple harmonic motion calculations.