Energy changes in mass spring system

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SUMMARY

The discussion centers on the relationship between kinetic energy and velocity in a mass-spring system, specifically addressing the misconception that the kinetic energy vs. time curve is merely the modulus of the velocity time curve. The mathematical proof involves analyzing the harmonic oscillator's trajectory defined by the equation x(t) = A cos(ωt) + B sin(ωt). By differentiating this equation to obtain the velocity v(t) and substituting it into the kinetic energy formula K(t) = 1/2 m v(t)^2, the relationship is established definitively.

PREREQUISITES
  • Understanding of harmonic oscillators
  • Familiarity with calculus, specifically differentiation
  • Knowledge of kinetic energy formulas
  • Basic concepts of oscillatory motion
NEXT STEPS
  • Study the mathematical derivation of harmonic oscillator equations
  • Learn about energy conservation in oscillatory systems
  • Explore the implications of damping in mass-spring systems
  • Investigate the relationship between potential energy and kinetic energy in oscillators
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of mass-spring systems and energy transformations in oscillatory motion.

Asad Raza
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Why is the kinetic energy vs time curve of a mass spring system is just a modulus of respective velocity time curve.
How can we prove it mathematically ?
 
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Actually that's not true!
Step1) Consider the most general trajectory for a harmonic oscillator:## x(t)=A\cos \omega t+B\sin\omega t ##
Step2) Differentiate it w.r.t. time to get the velocity ##v(t)##.
Step3) Put ##v(t)## in ##K(t)=\frac 1 2 m v(t)^2 ##.
 
Thread 'A high school physics problem demonstrating relative motion'

Thread 'A high school physics problem demonstrating relative motion'

I remembered a pretty high school problem from kinematics. But it seems it can help even undergraduates to develop their understanding of what a relative motion is. Consider a railway circle of radius ##r##. Assume that a carriage running along this circle has a speed ##v##. See the picture. A fly ##M## flies in the opposite direction and has a speed ##u,\quad |OM|=b##. Find a speed of the fly relative to the carriage. The obvious incorrect answer is ##u+v## while the correct answer is...
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