A particle of mass  m  is attached to a spring (of spring constant k)  and has a natural angular frequency  \omega _{0}  . An external force   F(t)   proportional to \cos \omega t\left ( \omega \neq \omega _{0} \right )  is applied to the oscillator . The time displacement of the oscillator will be proportional to

  • Option 1)

    \frac{m}{\omega {_{0}}^{2}-\omega ^{2}}

  • Option 2)

    \frac{1}{m\left ( {\omega _{0}}^{2}-\omega ^{2} \right )}

  • Option 3)

    \frac{1}{m\left ( {\omega _{0}}^{2}+\omega ^{2} \right )}

  • Option 4)

    \frac{m}{\omega {_{0}}^{2}+\omega ^{2}}

 

Answers (1)
V Vakul

As we discussed in @7249

x=a \sin \left ( \omega t+ \phi \right )

where     a= \frac{F_{o}/m}{{\omega_{0}}^{2}- \omega ^{2}}

\therefore   x is proportional to    \frac{1}{m\left ({\omega_{0}}^{2} - \omega ^{2} \right )}


Option 1)

\frac{m}{\omega {_{0}}^{2}-\omega ^{2}}

This answer is incorrect.

Option 2)

\frac{1}{m\left ( {\omega _{0}}^{2}-\omega ^{2} \right )}

This answer is correct.

Option 3)

\frac{1}{m\left ( {\omega _{0}}^{2}+\omega ^{2} \right )}

This answer is incorrect.

Option 4)

\frac{m}{\omega {_{0}}^{2}+\omega ^{2}}

This answer is incorrect.

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