For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly ?

(graphs are schematic and not drawn to scale)

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

Answers (1)
V Vakul

As we learnt in

Kinetic energy in S.H.M. -

K.E.= \frac{1}{2}mu^{2}\\ \: \: \: = \frac{1}{2}m\left ( A^{2} -x^{2}\right )\omega ^{2}

 

- wherein

K.E.= \frac{1}{2}K\left ( A^{2}-x^{2} \right ) \\K= m\omega ^{2}

 

 For a simple pendulum variation K.E. and P.E. with displacement d is

K.E.=\frac{1}{2}m\omega^{2}(A^{2}-d^{2})

P.E.=\frac{1}{2}m\omega^{2}d^{2}

if d = 0  K.E.=\frac{1}{2}m\omega^{2}A^{2}        P.E.= 0

if  d\pm A     K.E. = 0         P.E.=\frac{1}{2}m\omega^{2}A^{2}

\therefore   Graph 2 represents the variation correctly. 

Correct option is 2.

 


Option 1)

This is an incorrect option.

Option 2)

This is the correct option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

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