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A point mass oscillates along the x -axis according to law x= x_{0}\cos \left ( \omega t-\pi /4 \right ). If the acceleration of the particle is written as a= A\cos \left ( \omega t+\delta \right ) then

  • Option 1)

    A=x_{0}\omega ^{2},\delta = 3\pi/4

  • Option 2)

    A= x_{0},\delta = -\pi /4

  • Option 3)

    A=x_{0}\omega ^{2},\delta = \pi/4

  • Option 4)

    A= x_{0}\omega ^{2},\delta = -\pi /4

 

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 :x= x_{0}\cos \left ( \omega t-\frac{\pi }{4} \right )\cdots \cdots \cdots (i)

Acceleration\: \: a= A\: \cos \left ( \omega t +\delta \right )\cdots \cdots (ii)

Velocity\: \: \nu = \frac{dx}{dt}

\nu = -x_{0}\omega \sin \left ( \omega t-\frac{\pi }{4} \right )\cdots \cdots \cdots (iii)

Acceleration\: \: a= \frac{d\nu }{dt}

= x_{0}\omega ^{2}\cos \left [ \omega t+\frac{3\pi }{4} \right ]

Compare \: \: (iv)\: \: with\: \: (ii),\: \: we\: \: get

A= x_{0}\omega ^{2},\delta = \frac{3\pi }{4}

Correct option is 1.

 


Option 1)

A=x_{0}\omega ^{2},\delta = 3\pi/4

This is the correct option.

Option 2)

A= x_{0},\delta = -\pi /4

This is an incorrect option.

Option 3)

A=x_{0}\omega ^{2},\delta = \pi/4

This is an incorrect option.

Option 4)

A= x_{0}\omega ^{2},\delta = -\pi /4

This is an incorrect option.

Posted by

prateek

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