The discussion revolves around the angular frequency of small oscillations of a pendulum, specifically focusing on the potential energy expression and its implications for the frequency calculation. Participants are examining the relationship between potential energy, angular displacement, and the derived formulas for angular frequency.
The discussion is ongoing, with participants providing insights into the dimensional analysis of the equations involved. Some participants have pointed out potential discrepancies in the expressions used and are exploring the implications of these on the calculations for angular frequency.
There are indications of confusion regarding the correct formulation of potential energy and its derivatives, as well as the dimensions associated with the derived quantities. Participants are encouraged to reconsider their assumptions and the definitions being used in their calculations.
ω=√(d2Veffect./mdθ2) is not true. Check the dimension.Apashanka said:Homework Statement
One silly thing may be I am missing for small oscillations of a pendulum the potential energy is -mglcosθ ,for θ=0 is the point of stable equilibrium (e.g minimum potential energy) .Homework Equations
Small oscillations angular frequency
ω=√(d2Veffect./mdθ2) about stable equilibrium.
The Attempt at a Solution
Solving for this gives ω=√(gl),am I missing out something
Since ω=√(g/l)
If the potential V(θ) is expanded about a local Maxima or minima point θ0 then V(θ)=V(θ0)+(1/2)(d2V(θ)/dθ2)(θ-θ0)2ehild said:ω=√(d2Veffect./mdθ2) is not true. Check the dimension.
The force constant is defined as force divided by displacement, so its dimension is Force/Length. The second derivative of the potential with respect to displacement is equal to the force constant. The second derivative of the potential with respect the angle has dimension of work.Apashanka said:If the potential V(θ) is expanded about a local Maxima or minima point θ0 then V(θ)=V(θ0)+(1/2)(d2V(θ)/dθ2)(θ-θ0)2
V(θ)~(1/2)kdθ2
or ω=√(k/m)
From that I am telling this