Is the Velocity Zero in Equilibrium Points of a Simple Harmonic Oscillator?

AI Thread Summary
In a simple harmonic oscillator, the velocity is indeed zero at the equilibrium points, where the potential energy is at a minimum and kinetic energy is at a maximum. Conversely, at maximum displacement, which represents unstable equilibrium points, the potential energy is at its maximum and the velocity is zero. The total energy of the system remains constant, with kinetic energy and potential energy exchanging values as the oscillator moves. The discussion also touches on the concept of unstable equilibrium, such as a pendulum at its highest point, where it can theoretically remain but is prone to disturbances. Overall, the relationship between kinetic and potential energy in oscillatory motion is crucial for understanding these dynamics.
omri3012
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Hallo,

Does the velocity i simple harmonic oscillator is zero in equilibrium points? if it's true

how does it make sense with the fact that i suppose to get a maximum kinetic

Energy in those points (stable ones)

i would really appreciate if someone could clear this issue for me.

Thanks,

Omri
 
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The velocity of a simple harmonic oscillator is a maximum when it passes through its equilibrium point. Is that what you were asking?
 
diazona said:
The velocity of a simple harmonic oscillator is a maximum when it passes through its equilibrium point. Is that what you were asking?

yes,

but what about the non stable equilibrium point (maximum points)?

thanks
omri
 
Simple Harmonic Scillators have stable equilibria (or they wouldn't oscillate).

The extremes in position are not equilibrium points.
 
JazzFusion said:
Simple Harmonic Scillators have stable equilibria (or they wouldn't oscillate).

The extremes in position are not equilibrium points.

sorry for that... i ment that if i have a potential with maximum points does the velocity
is zero there?
 
In general, the total energy of a pendulum is T + V (kinetic plus potential energy). T + V = constant. One is a maximum when the other is minimum, and vice versa. There is one other quasistable equilibrium point for a pendulum, when the pendulum is exactly upside down. In principle, it should stay there forever, barring vibration and air currents. In actually, it is unstable, because of the Heisenberg Uncertainty Principle. If the tip of the pendulum has a momentum uncertainty Δ p, and the uncertainty in position is Δ x, then the product is Δp Δx <=h/2 pi.

Based on this uncertainty (and barring friction), the pendulum will begin swinging (as I recall) in a few seconds.

Γ Δ Θ Λ Ξ Π Σ Φ Ψ Ω
 
omri3012 said:
sorry for that... i ment that if i have a potential with maximum points does the velocity
is zero there?
If I'm understanding your question right, YES. PE is a maximum when KE is zero, and vice versa.

The farthest extent of position is the maximum of potential energy, and V is zero.
 

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