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The displacement of a particle varies according to the relation x=4\left ( \cos \pi t+\sin \pi t \right ). The amplitude of the particle is

  • Option 1)

    -4

  • Option 2)

    4

  • Option 3)

    4\sqrt{2}

  • Option 4)

    8

 

Answers (1)

best_answer

As we learnt in

Equation of S.H.M. -

a=-\frac{d^{2}x}{dt^{2}}= -w^{2}x

w= \sqrt{\frac{k}{m}}

 

- wherein

x= A\sin \left ( wt+\delta \right )

 

 

 

:x= 4\left ( \cos \pi t+\sin \pi t \right )

= 4\times \sqrt{2}\left [ \frac{1}{\sqrt{2}}\cos \pi t+\frac{1}{\sqrt{2}} \sin \pi t\right ]

or        x= 4 \sqrt{2}\left [ \sin \frac{\pi }{4}\cos \pi t+\cos \frac{\pi }{4} \sin \pi t\right ]

= 4 \sqrt{2}\sin \left ( \pi t+\frac{\pi }{4} \right )

  Hence\: \: amplitude \: \: = 4\sqrt{2}

Correct option is 3.


Option 1)

-4

The is an incorrect option.

Option 2)

4

The is an incorrect option.

Option 3)

4\sqrt{2}

This is the correct option.

Option 4)

8

The is an incorrect option.

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