Angular frequency of the small oscillations of a pendulum

[Apashanka]

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=√(d²V_effect./mdθ²) about stable equilibrium.

The Attempt at a Solution

Solving for this gives ω=√(gl),am I missing out something
Since ω=√(g/l)

 

[ehild]

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=√(d²V_effect./mdθ²) about stable equilibrium.

The Attempt at a Solution

Solving for this gives ω=√(gl),am I missing out something
Since ω=√(g/l)

ω=√(d²V_effect./mdθ²) is not true. Check the dimension.

 

[Apashanka]

ω=√(d²V_effect./mdθ²) is not true. Check the dimension.

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

 

[ehild]

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

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.
The force constant is $k= \frac{d^2V}{dx^2}$, and in case of small displacements, x=Lθ, so $k= \frac{d^2V}{L^2dθ^2}$.