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A particle is oscillating in SHM. What fraction of total energy is kinetic when the particle is at \frac{A}{2} from the mean position (A is the amplitude)

  • Option 1)

    \frac{3}{}4 

  • Option 2)

    \frac{2}{}4

  • Option 3)

    \frac{4}{}7

  • Option 4)

    \frac{5}{}7

 

Answers (1)

best_answer

\\*k_{f} = \frac{k_{e}\; at\; \frac{A}{2}}{k_{total}} = \frac{\frac{1}{2}m\omega^{2}(A^{2}-\frac{A^{2}}{4})}{\frac{1}{2}m\omega^{2}A^{2}} \\* \Rightarrow k_{f} = \frac{3}{4}

 

Total energy in S.H.M. -

Total Energy = Kinetic + Potential Energy

- wherein

Total Energy =\frac{1}{2}K\left ( A^{2}-x^{2} \right )+\frac{1}{2}kx^{2}= \frac{1}{2}kA^{2}

Hence total energy in S.H.M. is constant

 

 

 


Option 1)

\frac{3}{}4 

This is correct.

Option 2)

\frac{2}{}4

This is incorrect.

Option 3)

\frac{4}{}7

This is incorrect.

Option 4)

\frac{5}{}7

This is incorrect.

Posted by

Aadil

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