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If a simple harmonic motion is represented by

\frac{d^{2}x}{dt^{2}}+\alpha x= 0, its time period is

  • Option 1)

    2\pi \alpha

  • Option 2)

    2\pi \sqrt{\alpha}

  • Option 3)

    2\pi /\alpha

  • Option 4)

    2\pi /\sqrt{\alpha}

 

Answers (1)

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 )

 

 

 

Standard differential equation of SHM is

\frac{d^{2}x}{dt^{2}}+\omega ^{2}x= 0

 

Given equ.is     \frac{d^{2}x}{dt^{2}}+\alpha x= 0

\therefore \omega ^{2}= \alpha

or \: \: \omega = \sqrt{\alpha }

\therefore T= \frac{2\pi }{\omega }= \frac{2\pi }{\sqrt{\alpha }}

Correct option is 4.


Option 1)

2\pi \alpha

This is an incorrect option.

Option 2)

2\pi \sqrt{\alpha}

This is an incorrect option.

Option 3)

2\pi /\alpha

This is an incorrect option.

Option 4)

2\pi /\sqrt{\alpha}

This is the correct option.

Posted by

Vakul

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