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Use the result of Problem 8 to prove that if the pendulum in Figure 4.18 on page 208 is released from rest at the angle 0<α<π, then |θ(t)|≤αfor all t.

Short Answer

Expert verified

Therefore, the given statement is true. The pendulum in Figure 4.18 is released from rest at the angle 0<α<π, then |θ(t)|≤αfor all t is true.

Step by step solution

01

The Energy Integral Lemma

Let ytbe a solution to the differential equation y"=f(y), where f(y) is a continuous function that does not depend on y’ or the independent variable t.

Let F(y) is an indefinite integral of f(y)i.e. f(y)=ddyF(y). Then the quantityE(t)=12y'(t)2-F(y(t)) is constant; i.e.role="math" localid="1655365549886" ddtE(t)=0 .

02

Prove the given equation

Referring from Problem 8: 12θ'(t)2-gâ„“³¦´Ç²õθ=constant â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€¦(1)

To prove: the pendulum is released from rest at the angle 0<α<π, then|θ(t)|≤αfor all t.

The initial conditions are θ(0)=α,θ'(0)=0.

Let us take role="math" localid="1655365792527" constant=k. Then,

role="math" localid="1655365833327" 12θ'(t)2-gâ„“³¦´Ç²õθ=k â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â€‰â€‰â€‰â€‰â€¦(2)

Now, implement the initial conditions here.

role="math" localid="1655365997812" 12θ'(t)2-gâ„“³¦´Ç²õθ=k12θ'(0)2-gâ„“³¦´Ç²õθ(0)=k0-gâ„“³¦´Ç²õα=kk=-gâ„“³¦´Ç²õα

Now, substitute the value of k in equation (2).

12θ'(t)2-gℓcosθ=k12θ'(t)2-gℓcosθ=-gℓcosα

Here (θ')2can’t be a negative number. So, we can write this condition as:

gâ„“³¦´Ç²õθ≥gâ„“³¦´Ç²õ᳦´Ç²õθ≥³¦´Ç²õα

Multiplycos−1 on both sides.

cos−1(³¦´Ç²õθ)≤cos−1(³¦´Ç²õα)|θ(t)|≤α

Hence it is proved that |θ(t)|≤α.

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