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 ) = l sin θ x(\theta)= l\sin\theta x ( θ ) = l sin θ
y ( θ ) = − l cos θ y(\theta) = -l\cos\theta y ( θ ) = − l cos θ
now, differentiating with respect to t,
d θ d t = θ ˙ \frac{d\theta}{dt}=\dot{\theta} d t d θ = θ ˙
d ( x ( θ ) ) d t = l cos θ d θ d t \frac{d(x(\theta))}{dt}=l\cos\theta \frac{d\theta}{dt} d t d ( x ( θ )) = l cos θ d t d θ
⇒ d ( x ( θ ) ) d t = l θ ˙ cos θ \Rightarrow \frac{d(x(\theta))}{dt}=l\dot{\theta} \cos\theta ⇒ d t d ( x ( θ )) = l θ ˙ cos θ
d ( y ( θ ) ) d t = l sin θ d θ d t \frac{d(y(\theta))}{dt}=l\sin\theta \frac{d\theta}{dt} d t d ( y ( θ )) = l sin θ d t d θ
⇒ d ( y ( θ ) ) d t = l θ ˙ sin θ \Rightarrow \frac{d(y(\theta))}{dt}=l\dot{\theta}\sin\theta ⇒ d t d ( y ( θ )) = l θ ˙ sin θ
Now, applying the conservation of energy
L = T − V = K E − P E L=T-V=KE-PE L = T − V = K E − PE
= m l θ 2 ˙ 2 − m g l ( 1 − cos θ ) =\frac{ml\dot{\theta^2}}{2}-mgl(1-\cos\theta) = 2 m l θ 2 ˙ − m g l ( 1 − cos θ )
We know that the Lagrange equation for the generalized co-ordinate
d d t ( ( ∂ L ) d θ ˙ ) − ∂ L ∂ θ = 0 \frac{d}{dt}(\frac{(\partial L)}{d\dot{\theta}})-\frac{\partial L}{\partial{\theta}}=0 d t d ( d θ ˙ ( ∂ L ) ) − ∂ θ ∂ L = 0
⇒ θ ¨ + g l sin ( θ ) = 0 \Rightarrow \ddot{\theta}+\frac{g}{l}\sin(\theta)=0 ⇒ θ ¨ + l g sin ( θ ) = 0
Hence, natural frequency of the simple pendulum
ω = g l \omega = \sqrt{\frac{g}{l}} ω = l g
T = 2 π l g T=2\pi \sqrt{\frac{l}{g}} T = 2 π g l
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