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A particle undergoing simple harmonic motion has a dependent displacement given by x(t) = x(t) = A\sin \frac{\pi t}{90}. The ratio of kinetic to potential energy of this particle at t = 210s will be:

 

  • Option 1)

     

  • Option 2)

    \frac{1}{3}

     

  • Option 3)

    \frac{1}{9}

  • Option 4)

     

    1

Answers (1)

best_answer

 

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}

 

 

Potential energy in S.H.M. -

P.E.= \frac{1}{2}Kx^{2}

 

- wherein

Where K= m\omega ^{2}

 

 

 

 


Option 1)

 

Option 2)

\frac{1}{3}

 

Option 3)

\frac{1}{9}

Option 4)

 

1

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