If a simple harmonic oscillator has got a displacement of 0.02 m and acceleration equal to 2.0 ms^{-2} at any time, the angular frequency of the oscillator is equal to

  • Option 1)

    10 rad s^{-1}

  • Option 2)

    0.1 rad s^{-1}

  • Option 3)

    100 rad s^{-1}

  • Option 4)

    1 rad s^{-1}

 

Answers (1)

As we discussed in concept

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 )

 

 a=-w^{2}x                                                    => 2=w^{2}\times0.02

w=\sqrt{\frac{2}{0.02}}\:\:\:=\sqrt{100}=10\:rad\:s^{-1}


Option 1)

10 rad s^{-1}

This option is correct.

Option 2)

0.1 rad s^{-1}

This option is incorrect.

Option 3)

100 rad s^{-1}

This option is incorrect.

Option 4)

1 rad s^{-1}

This option is incorrect.

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