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A particle starts SHM from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is 

  • Option 1)

    \frac{A}{2}

  • Option 2)

    \frac{A}{\sqrt2}

  • Option 3)

    \frac{A\sqrt3}{2}

  • Option 4)

    \frac{2A}{\sqrt3}

 

Answers (2)

best_answer

\\*v_{max} = \omega A \Rightarrow v = \frac{\omega A}{2} = \omega\sqrt{A^{2}- y^{2}} \\*\Rightarrow A^{2} - y^{2} = \frac{A^{2}}{4} \Rightarrow y^{2} = \frac{3}{4}A^{2} \Rightarrow y = \frac{\sqrt{3}}{2}A

 

Mean Position -

A position  during oscillation where the particle is at equilibrium position, i.e. net force on particle at this position is zero.

- wherein

Force acting on particle always point towards mean position.

 

 

 


Option 1)

\frac{A}{2}

This is incorrect.

Option 2)

\frac{A}{\sqrt2}

This is incorrect.

Option 3)

\frac{A\sqrt3}{2}

This is correct.

Option 4)

\frac{2A}{\sqrt3}

This is incorrect.

Posted by

Aadil

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Posted by

Koli hitesh

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