Quartic Potential

Rules={m->1,ω->1};
U[x_,σ_]=((m (ω^2))/2) (x^2)+σ((m (ω^2))/2) (x^4);
P1=Plot[{U[x,-1]/.Rules,U[x,0]/.Rules,U[x,+1]/.Rules},{x,-(1/(Sqrt[2])),+(1/(Sqrt[2]))},AxesLabel->{"(Sqrt[\[LeftBracketingBar]σ\[RightBracketingBar]])x","\!\(\*FractionBox[\(\[LeftBracketingBar]σ\[RightBracketingBar]\), \(m\\ \*SuperscriptBox[\(ω\), \(2\)]\)]\)U[x]"},PlotStyle->Thick]

Solve[(∂_x)U[x,σ]==0,x]

$Assumptions={0<m,0<ω,0<y<x,σ∈Reals,-2σ (x^2)<1};
Solve[(1/2)m (v^2)+U[y,σ]==U[x,σ],v]//FullSimplify

v[y_,x_]=ω(Sqrt[(x-y) (x+y) (1+((x^2)+(y^2)) σ)]);
T[x_,σ_]=4(∫_0^x)(1/(v[y,x]))\[DifferentialD]y//FullSimplify

T[x_,σ_]=(4/ω) ((EllipticK[-1+(1/(1+(x^2) σ))])/(Sqrt[1+(x^2) σ]));
P2=Plot[{((T[x,-1])/(2π))/.Rules,((T[x,0])/(2π))/.Rules,((T[x,+1])/(2π))/.Rules},{x,-(1/(Sqrt[2])),+(1/(Sqrt[2]))},AxesLabel->{"(Sqrt[\[LeftBracketingBar]σ\[RightBracketingBar]])x","(ω/(2  π))T[x]"},PlotStyle->Thick,PlotRange->{0,4}];
GraphicsRow[{P1,P2}]

Pendulum Potential

Rules={m->1,L->1,g->1};
y[θ_]=L (1-Cos[θ]);
U[θ_]=m g y[θ];
P1=Plot[{U[θ]/.Rules,(1/2)m(Sqrt[(g/L)])(θ^2)/.Rules},{θ,-π,π},AxesLabel->{"θ","\!\(\*FractionBox[\(U[θ]\), \(m\\ g\\ L\)]\)"},PlotStyle->Thick,Ticks->{Table[n (π/2),{n,-2,2}],Automatic}]

$Assumptions={0<m,0<L,0<g,0<ϕ<θ<(π/2)};
Solve[(1/2)m (L^2) (ω^2)+U[ϕ]==U[θ],ω]//FullSimplify
ω[ϕ_,θ_]=(Sqrt[((2g)/L)]) (Sqrt[Cos[ϕ]-Cos[θ]]);
T[θ_]=4(∫_0^θ)(1/(ω[ϕ,θ]))\[DifferentialD]ϕ//FullSimplify

T[θ_]=4(Sqrt[(L/g)])EllipticK[((Sin[(θ/2)])^2)];
P2=Plot[{((T[θ])/(2π))/.Rules,(Sqrt[L/g])/.Rules},{θ,-π,π},AxesLabel->{"θ","((Sqrt[g/L])/(2  π))T[θ]"},PlotStyle->Thick,PlotRange->{0,4},Ticks->{Table[n (π/2),{n,-2,2}],Automatic}];
GraphicsRow[{P1,P2}]