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Question: A pendulum is executing simple harmonic motion, and its maximum kinetic energy is \[{K_1}\]. If the ...

A pendulum is executing simple harmonic motion, and its maximum kinetic energy is K1{K_1}. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2{K_2}, Then?

Explanation

Solution

In this question, we first find the maximum kinetic energy of the simple pendulum, and then we double the length of the pendulum, after which we find the maximum kinetic energy is K2{K_2}.

Complete step by step answer:
The initial maximum kinetic energy of a simple pendulum =K1 = {K_1}
The maximum kinetic energy of a simple pendulum when the length is doubled =K2 = {K_2}
We know the maximum kinetic energy of a simple pendulum at a point is given by the formula
KE=12mω2A2(i)KE = \dfrac{1}{2}m{\omega ^2}{A^2} - - (i)

Here A is the linear amplitude of a simple pendulum as shown in the diagram, which can be written as
A=lθA = l\theta
Now substitute the value of the linear amplitude of simple pendulum in equation (i)

KE=12mω2(lθ)2 =12mω2l2θ2  KE = \dfrac{1}{2}m{\omega ^2}{\left( {l\theta } \right)^2} \\\ = \dfrac{1}{2}m{\omega ^2}{l^2}{\theta ^2} \\\

Now since we know the angular frequency is given asω=gl\omega = \sqrt {\dfrac{g}{l}} , hence we can further write

KE=12mgll2θ2ω =12mglθ2ω  KE = \dfrac{1}{2}m\dfrac{g}{l}{l^2}{\theta ^2}\omega \\\ = \dfrac{1}{2}mgl{\theta ^2}\omega \\\

Hence we can write K1=12mglθ2ω(ii){K_1} = \dfrac{1}{2}mgl{\theta ^2}\omega - - (ii)
Now it is said that the length of the pendulum is doubled so l=2ll' = 2l
So the maximum kinetic energy of the pendulum becomes

K2=12mg2lθ2 =mglθ2  {K_2} = \dfrac{1}{2}mg2l{\theta ^2} \\\ = mgl{\theta ^2} \\\

Hence we can write
K2=mglθ2(iii){K_2} = mgl{\theta ^2} - - (iii)
Now to find the change in Kinetic energy, divide equation (ii) by equation (iii)
K1K2=12mglθ2ωmglθ2\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{{\dfrac{1}{2}mgl{\theta ^2}\omega }}{{mgl{\theta ^2}}}
By solving
K1K2=12\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{1}{2}
Hence the kinetic energy K2=2K1{K_2} = 2{K_1}

Note:
Students must note that if the length of the simple pendulum is doubled, then its maximum kinetic energy also gets doubled. Maximum kinetic energy at a point is given by the formula KE=12mω2A2KE = \dfrac{1}{2}m{\omega ^2}{A^2}, where AA is the linear amplitude, ω\omega is the angular frequency.