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  • Can someone explain - Oscillations and Waves - JEE Main-4

The P.E of a particle executing SHM at a distance x from its equillibrium position is 

  • Option 1)

    \frac{1}{2}m\omega^{2}x^{2}

  • Option 2)

    \frac{1}{2}m\omega^{2}a^{2}

  • Option 3)

    \frac{1}{2}m\omega^{2}(a^{2} - x^{2})

  • Option 4)

    zero

 

Answers (1)

^{1}/_{2}\cdot m\omega^{2}x^{2}

 

Potential energy in S.H.M. -

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

 

- wherein

Where K= m\omega ^{2}

 

 

 


Option 1)

\frac{1}{2}m\omega^{2}x^{2}

This is correct.

Option 2)

\frac{1}{2}m\omega^{2}a^{2}

This is incorrect.

Option 3)

\frac{1}{2}m\omega^{2}(a^{2} - x^{2})

This is incorrect.

Option 4)

zero

This is incorrect.

Posted by

Vakul

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