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  • Need explanation for: - Oscillations and Waves - JEE Main

The displacement of an object attached to a spring and executing simple harmonic motion is given by x= 2\times 10^{-2}\cos \pi \: t\: metre maximum speed first occurs is

  • Option 1)

    0.25 s

  • Option 2)

    0.5 s

  • Option 3)

    0.75 s

  • Option 4)

    0.125 s

 

Answers (1)

best_answer

As we learnt in

Phase constant -

The constant \delta appearing in eqation x=A\sin \left ( wt+\delta \right ) is called phase constant. It describes initial state.

 

- wherein

This constant depends on the choice of the instant t=0.

 

 

 

:    Given : displacement  x= 2\times 10^{-2}\cos \pi t

Velocity\: \: \nu = \frac{dx}{dt}= -2\times 10^{-2}\pi \sin \pi t

For\: the\: first \: time \: when\: \: \nu = \nu _{max},\sin \pi t= 1

or\: \: \: \sin \pi t= \sin \frac{\pi }{2}\: \: \: \: or\: \: \: \: \: \pi t= \frac{\pi }{2}

or\: \: \: t= \frac{1}{2}\: \: \: \: s= 0.5s

Correct option is 2.


Option 1)

0.25 s

This is an incorect option.

Option 2)

0.5 s

This is the correct option.

Option 3)

0.75 s

This is an incorect option.

Option 4)

0.125 s

This is an incorect option.

Posted by

prateek

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