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A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. 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. Then
1. K2 = 2K1
2. \({K_2} = \frac{{{K_1}}}{2}\)
3. \({K_2} = \frac{{{K_1}}}{4}\)
4. K2 = K1

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Best answer
Correct Answer - Option 1 : K2 = 2K1

Concept:

The kinetic energy would be KE = ½mv2,

where m is the mass of the pendulum,

and v is the speed of the pendulum.

At its highest point (Point A) the pendulum is momentarily motionless.

All of the energy in the pendulum is gravitational potential energy and there is no kinetic energy.

At the lowest point (Point D) the pendulum has its greatest speed. All of the energy in the pendulum is kinetic energy and there is no gravitational potential energy. However, the total energy is constant as a function of time.

Calculation:

Kinetic energy of a pendulum is maximum at its mean position. Also, maximum kinetic energy of pendulum

\(= {{\rm{K}}_{{\rm{max}}}} = \frac{1}{2}{\rm{m}}{{\rm{\omega }}^2}{{\rm{\alpha }}^2}\)

where, angular frequency

\({\rm{\omega }} = \frac{{2{\rm{\pi }}}}{{\rm{T}}} = \frac{{2{\rm{\pi }}}}{{2{\rm{\pi }}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}} }}\)

or \({\rm{\omega }} = \sqrt {\frac{{\rm{g}}}{{\rm{l}}}} {\rm{\;or\;}}{{\rm{\omega }}^2} = \frac{{\rm{g}}}{{\rm{l}}}\)

and a = amplitude

As amplitude is same in both cases so;

Kmax ∝ ω2

\(\Rightarrow {K_{{\rm{max}}}} \propto \frac{1}{l}\) 

[∵ g is constant]

According to given data. \({K_1} \propto \frac{1}{l}{\rm{\;and\;}}{{\rm{K}}_2} \propto \frac{1}{{2{\rm{l}}}}\)

\(\therefore \frac{{{{\rm{K}}_1}}}{{{{\rm{K}}_2}}} = \left( {\frac{{1/{\rm{l}}}}{{1/2{\rm{l}}}}} \right) = 2\) 

⇒ K1 = 2K2

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