Simple harmonic motion equations derivation?

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SUMMARY

The discussion focuses on the derivation of energy equations in simple harmonic motion (SHM). It establishes that the total mechanical energy E in a spring-mass system is conserved and can be expressed as E = (1/2)kA^2 at maximum displacement (x = A) and E = (1/2)mω^2A^2 at equilibrium (x = 0). The key variables include the amplitude (A), spring constant (k), mass (m), and angular frequency (ω). The relationship between kinetic and potential energy in SHM is highlighted, confirming that both expressions for energy are equivalent under ideal conditions without friction or dissipation.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with energy conservation principles
  • Knowledge of spring constants (k) and mass (m)
  • Basic grasp of angular frequency (ω) and amplitude (A)
NEXT STEPS
  • Study the derivation of the equations for potential energy in springs, specifically E = (1/2)kx^2
  • Learn about kinetic energy in oscillatory systems, focusing on E = (1/2)mv^2
  • Explore the concept of angular frequency (ω) and its relationship to SHM
  • Investigate the effects of damping and friction on mechanical energy in oscillatory motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillations, as well as educators seeking to explain the principles of energy conservation in simple harmonic motion.

hey123a
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Well I was going through class lecture notes and my professor wrote this

When x = A(the maximum value), v=0: E=1/2kA^2
When v = wA, x=0: E=1/2mw^2A^2

where w = omega, A = amplitude, k = spring constant, m = mass, v = velocity

and apparently both equations are equal, i would like to know how and why
please and thank you
 
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If there is no friction or dissipation,kinetic and potential energy are alternately transformed into each other in SHM,but the total mechanical energy E=K+U is conserved.

Consider a spring with one end attached to the wall and other end having a block,going through its cycle of oscillation.

Total energy E = (1/2)mv2 + (1/2)kx2

At any position x, E = (1/2)mvx2 + (1/2)kx2

At the extreme position ,i.e at x=A the block has only potential energy ,so E = (1/2)kA2

At the equilibrium position ,at x=0 the block has only kinetic energy ,so E = (1/2)mvmax2

Now we know,vmax = ωA ,so at x=0 E=(1/2)mω2A2
 

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