Simple Harmonic Motion Pendulum, can we use PE=1/2kAmplitude^2?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 1K views
hongiddong
Messages
64
Reaction score
1
I was wondering if we can somehow use the formula Potential Energy = 1/2K(x(x=Amplitude))^2 for a pendulum if we are only given the angle of displacement?

Would the problems normally just say the PE at the top of the pendulum is such and such, please find max Velocity,
Or also the max velocity is such and such please find max potential energy, or max amplitude.
 
Last edited:
Physics news on Phys.org
Your formula for potential energy is not the one usually given in descriptions of a harmonic oscillator, and it is generally unclear what you are asking about. Being more specific could help.
 
hongiddong said:
I was wondering if we can somehow use the formula Potential Energy = 1/2K(x(x=Amplitude))^2 for a pendulum if we are only given the angle of displacement?

Would the problems normally just say the PE at the top of the pendulum is such and such, please find max Velocity,
Or also the max velocity is such and such please find max potential energy, or max amplitude.
No, you can't. ##k## is the spring constant. There is no spring involved with a pendulum.
 
  • Like
Likes   Reactions: 1 person
The k in Hong's formula is an equivalent spring constant. The gravitational PE of the pendulum bob relative to its lowest position is, mgh, in which h is its height relative to its lowest point. It's easy to show by Pythagoras that for small displacements with horizontal component x, for a pendulum of length L, [itex]x^2 = 2hL[/itex]. So the PE of the bob is [itex]mgh = \frac{1}{2} \frac{mg}{L} x^2[/itex], so the effective spring constant, k, is [itex]\frac{mg}{L}[/itex].
 
  • Like
Likes   Reactions: 1 person