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The displacement of a particle executing simple harmonic motion is given by 

y = A _ 0 + A \sin \omega t + B \ cos \omega t 

Then the amplitude of its oscillation is given by : 

 

  • Option 1)

    A _0 + \sqrt{ A^2 + B^2 }

  • Option 2)

    \sqrt{ A^2 + B^2 }

  • Option 3)

    \sqrt{A_0 ^2+ (A + B)^2 }

  • Option 4)

    A+ B

Answers (1)

best_answer

The particle execute SHM about its center at \alpha = A_0 \\\\ amplitude = \sqrt { A^2 + B^2 }


Option 1)

A _0 + \sqrt{ A^2 + B^2 }

Option 2)

\sqrt{ A^2 + B^2 }

Option 3)

\sqrt{A_0 ^2+ (A + B)^2 }

Option 4)

A+ B

Posted by

Aadil

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