Question #161653

What is the lagrange's equation of simple pendulum?


1
Expert's answer
2021-02-07T19:09:40-0500

Let the length of the pendulum be l, then we can represent the co-ordinates of the motion of the pendulum as per the given question


x(θ)=lsinθx(\theta)= l\sin\theta

y(θ)=lcosθy(\theta) = -l\cos\theta

now, differentiating with respect to t,

dθdt=θ˙\frac{d\theta}{dt}=\dot{\theta}


d(x(θ))dt=lcosθdθdt\frac{d(x(\theta))}{dt}=l\cos\theta \frac{d\theta}{dt}


d(x(θ))dt=lθ˙cosθ\Rightarrow \frac{d(x(\theta))}{dt}=l\dot{\theta} \cos\theta


d(y(θ))dt=lsinθdθdt\frac{d(y(\theta))}{dt}=l\sin\theta \frac{d\theta}{dt}

d(y(θ))dt=lθ˙sinθ\Rightarrow \frac{d(y(\theta))}{dt}=l\dot{\theta}\sin\theta

Now, applying the conservation of energy

L=TV=KEPEL=T-V=KE-PE


=mlθ2˙2mgl(1cosθ)=\frac{ml\dot{\theta^2}}{2}-mgl(1-\cos\theta)


We know that the Lagrange equation for the generalized co-ordinate

ddt((L)dθ˙)Lθ=0\frac{d}{dt}(\frac{(\partial L)}{d\dot{\theta}})-\frac{\partial L}{\partial{\theta}}=0


θ¨+glsin(θ)=0\Rightarrow \ddot{\theta}+\frac{g}{l}\sin(\theta)=0

Hence, natural frequency of the simple pendulum

ω=gl\omega = \sqrt{\frac{g}{l}}

T=2πlgT=2\pi \sqrt{\frac{l}{g}}


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